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Lecture-note text with no author shown, dated July 15, 2014. It reviews linear algebra (bases, Einstein convention, change of basis, eigenbases), then covers multilinear forms, inner products and reciprocal bases, and general (p,q) tensors. The applications chapter treats the inertia, stress, strain, elasticity and conductivity tensors, with exercises and a solutions chapter.
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Multilinear Algebra
and Applications
July 15, 2014.
Contents
Chapter 1. Introduction 1
Chapter 2. Review of Linear Algebra 5
2.1. Vector Spaces and Subspaces 5
2.2. Bases 7
2.3. The Einstein convention 10
2.3.1. Change of bases, revisited 12
2.3.2. The Kronecker delta symbol 13
2.4. Linear Transformations 14
2.4.1. Similar matrices 18
2.5. Eigenbases 19
Chapter 3. Multilinear Forms 23
3.1. Linear Forms 23
3.1.1. Definition, Examples, Dual and Dual Basis 23
3.1.2. Transformation of Linear Forms under a Change of Basis 26
3.2. Bilinear Forms 30
3.2.1. Definition, Examples and Basis 30
3.2.2. Tensor product of two linear forms on V 32
3.2.3. Transformation of Bilinear Forms under a Change of Basis 33
3.3. Multilinear forms 34
3.4. Examples 35
3.4.1. A Bilinear Form 35
3.4.2. A Trilinear Form 36
3.5. Basic Operation on Multilinear Forms 37
Chapter 4. Inner Products 39
4.1. Definitions and First Properties 39
4.1.1. Correspondence Between Inner Products and Symmetric Po sitive
Definite Matrices 40
4.1.1.1. From Inner Products to Symmetric Positive Definite Matrices 42
4.1.1.2. From Symmetric Positive Definite Matrices to Inner Products 42
4.1.2. Orthonormal Basis 42
4.2. Reciprocal Basis 46
4.2.1. Properties of Reciprocal Bases 48
4.2.2. Change of basis from a basis Bto its reciprocal basis Bg50
III
IV CONTENTS
4.2.3. Isomorphisms Between a Vector Space and its Dual 52
4.2.4. Geometric Interpretation 54
Chapter 5. Tensors 57
5.1. Generalities 57
5.1.1. Canonical isomorphism between Vand (V∗)∗. 57
5.1.2. Towards general tensors 58
5.1.3. Tensor product of (1 ,0)-tensors on V∗59
5.1.4. Components of a (2 ,0)-tensor and their contravariance 60
5.2. Tensors of type ( p,q) 61
5.3. Tensor product 61
Chapter 6. Applications 65
6.1. Inertia tensor 65
6.1.1. Moment of inertia with respect to the axis determined by the an gular
velocity 65
6.1.2. Moment of inertia about any axis through the fixed point. 70
6.1.3. Moment of inertia with respect to an eigenbasis of the inertia te nsor 72
6.1.4. Angular momentum 73
6.2. Stress tensor (Spannung) 75
6.2.1. Special forms of the stress tensor (written with respect to an
orthonormal eigenbasis or another special basis) 80
6.2.2. Contravariance of the stress tensor 82
6.3. Strain tensor (Verzerrung) 83
The antisymmetric case 84
The symmetric case 85
6.3.1. Special forms of the strain tensor 87
6.4. Elasticity tensor 87
6.5. Conductivity tensor 88
6.5.1. Electrical conductivity 88
6.5.2. Heat conductivity 90
Chapter 7. Solutions 93
CHAPTER 1
Introduction
The main protagonists of this course are tensors andmultilinear maps , just
like the main protagonists of a Linear Algebra course are vectors an d linear maps.
Tensors are geometric objects that describe linear relations amon g objects in
space, and are represented by multidimensional arrays of numbers :
The indices can be upper or lower or, in tensor of order at least 2, so me of them
can be upper and some lower. The numbers in the arrays are called components
of the tensor and give the representation of the tensor with respect to a given basis .
There are two natural questions that arise:
(1) Why do we need tensors?
(2) What are the important features of tensors?
(1) Scalars are no enough to describe directions, for which we need to resort to
vectors. At the same time, vectors might not be enough, in that th ey lack the ability
to “modify” vectors.
Example 1.1.We denote by Bthe magnetic fluid density measured in Volt ·sec/m2
and byHthe megnetizing intensity measured in Amp /m. They are related by the
formula
B=µH,
whereµis the permeability of the medium in H /m. In free space, µ=µ0=
4π×10−7H/m is a scalar, so that the flux density and the magnetization are vect ors
differ only by their magnitude.
Other material however have properties that make these terms d iffer both in
magnitude and direction. In such materials the scalar permeability is r eplaced by
the tensor permeability µand
B=µ·H.
Being vectors, BandHare tensors of order 1, and µis a tensor of order 2. We will
see that they are of different type, and in fact the order of H“cancels out” with the
order ofµto give a tensor of order 1 . /square
(2) Physical laws do not change with different coordinate systems, hence tensors
describing them must satisfy some invariance properties. So tensors must have
invariance properties with respect to changes of bases, but their coordinates will of
course not stay invariant.
1
2 1. INTRODUCTION
Here is an example of a familiar tensor:
Example 1.2.We recall here the familiar transformation property that vectors
enjoy according to which they are an example of a contravariant tensor of first
order. We use here freely notions and properties that will be recalled in the next
chapter.
LetB:={b1,b2,b3}and/tildewideB:={˜b1,˜b2,˜b3}be two basis of a vector space V. A
vectorv∈Vcan be written as
v=v1b1+v2b2+v3b3,
or
v= ˜v1˜b1+ ˜v2˜b2+ ˜v3˜b3,
wherev1,v2,v3(resp. ˜v1,˜v2,˜v3) are the coordinate of vwith respect to the basis B
(resp./tildewideB).
Warning: Please keep the lower indices as lower indices and the upper o nes as
upper ones. You will see later that there is a reason for it!
We use the following notation:
[v]B=
v1
v2
v3
and [v]/tildewideB=
˜v1
˜v2
˜v3
, (1.1)
and we are interested in finding the relation between the coordinate s ofvin the two
bases.
The vectors ˜bj,j= 1,2,3, in the basis /tildewideBcan be written as a linear combination
of vectors inBas follows:
˜bj=L1
jb1+L2
jb2+L3
jb3,
for someLi
j∈R. We consider the matrix of the change of basis from Bto/tildewideB,
L:=L/tildewideBB=
L1
1L1
2L1
3
L2
1L2
2L2
3
L3
1L3
2L3
3
whosejth-column consists of the coordinates of the vectors ˜bjwith respect to the
basisB. The equalities
˜b1=L1
1b1+L2
1b2+L3
1b3
˜b2=L1
2b1+L2
2b2+L3
2b3
˜b3=L1
3b1+L2
3b2+L3
3b3
can simply be written as
/parenleftbig˜b1˜b2˜b3/parenrightbig
=/parenleftbigb1b2b3/parenrightbig
L. (1.2)
1. INTRODUCTION 3
(Check this symbolic equation using the rules of matrix multiplication.) A nalo-
gously, writing basis vectors in a row and vector coordinates in a colu mn, we can
write
v=v1b1+v2b2+v3b3=/parenleftbigb1b2b3/parenrightbig
v1
v2
v3
(1.3)
and
v= ˜v1˜b1+ ˜v2˜b2+ ˜v3˜b3=/parenleftbig˜b1˜b2˜b3/parenrightbig
˜v1
˜v2
˜v3
=/parenleftbigb1b2b3/parenrightbig
L
˜v1
˜v2
˜v3
, (1.4)
where we used (1.2) in the last equality. Comparing the expression of vin (1.3) and
in (1.4), we conclude that
L
˜v1
˜v2
˜v3
=
v1
v2
v3
or equivalently
˜v1
˜v2
˜v3
=L−1
v1
v2
v3
We say that the components of a vector varecontravariant1because they change
byL−1when the basis changes by L. A vectorvis hence a contravariant 1-tensor
ortensor of order (1,0). /square
Example 1.3 (A numerical example) .Let
B={e1,e2,e3}=
1
0
0
,
0
1
0
,
0
0
1
(1.5)
be the standard basis or R3and let
/tildewideB={˜b1,˜b2,˜b3}=
1
2
3
,
4
5
6
,
7
8
0
1In Latin contrameans “contrary’, against”.
4 1. INTRODUCTION
be another basis of R3. The vector2v=
1
1
1
has coordinates
[v]B=
1
1
1
and [v]/tildewideB=
−1
31
3
0
.
Since it is easy to check that
˜b1= 1·e1+4·e2+7·e3
˜b2= 2·e1+5·e2+8·e3
˜b3= 3·e1+6·e2,
the matrix of the change of coordinates from Bto/tildewideBis
L=
1 4 7
2 5 8
3 6 0
.
It is easy to check that
−1
31
3
0
=L−1
1
1
1
or equivalently
L
−1
31
3
0
=
1
1
1
.
/square
2The vector vhere is meant here as an element in R3. As such, it is identified by three real
numbers that we write in column surrounded by square brackets. T his should not be confused
with the coordinates of vwith respect to a basis B, that are indicated by round parentheses as in
(1.1), while [·]Bindicates the “operation” of taking the vector vand looking at its coordinates in
the basisB. Of course with this convention there is the – slightly confusing – fac t that ifBis the
basis in (1.5), then v=
1
1
1
and [v]B=
1
1
1
.
CHAPTER 2
Review of Linear Algebra
2.1. Vector Spaces and Subspaces
Definition 2.1.Avector space VoverRis asetVequipped withtwo operations:
(1)Vector addition: V×V→V, (v,w)/ma√sto→v+w, and
(2)Multiplication by a scalar: R×V→V, (α,v)/ma√sto→αv,
satisfying the following properties:
(1) (associativity) ( u+v)+w=u+(v+w) for every u,v,w∈V;
(2) (commutativity) u+v=v+ufor everyu,v∈V;
(3) (existence of the zero vector) there exists 0 ∈Vsuch thatv+ 0 =vfor
everyv∈V;
(4) (existence of additive inverse) For every v∈V, there exists wv∈Vsuch
thatv+wv= 0. The vector wvis denoted by−v.
(5)α(βv) = (αβ)vfor everyα,β∈Rand everyv∈R;
(6) 1v=vfor everyv∈V;
(7)α(u+w) =αu+αvfor allα∈Randu,v∈V;
(8) (α+β)v=αu+βvfor allα,β∈Randv∈V.
An element of the vector space is called a vector.
Example 2.2 (Prototypical example) .The Euclidean space Rn,n= 1,2,3,..., is
a vector space with componentwise addition and multiplication by scala rs. Vectors
inRnare denoted by v=
x1
...
xn
, withx1,...,x n∈R. /square
Examples 2.3 (Other examples) .(1) The set of real polynomials of degree ≤
nis a vector space, denoted by
V=R[x]n:={a0xn+a1xn−1+···+an−1x+an:aj∈R}.
(2) The set of real matrices of size m×n,
V=Mm×n(R) :=
a11... a 1m
......
an1... a nm
:aij∈R
.
(3) The space of solutions of a homogeneous linear (ordinary or par tial) differ-
ential equation.
5
6 2. REVIEW OF LINEAR ALGEBRA
(4) The space{f:W→R}, whereWis a vector space.
/square
Exercise 2.4.Are the following vector spaces?
(1) The set Vof all vectors in R3perpendicular to the vector
1
2
3
.
(2) The set of invertible 2 ×2 matrices, that is
V:=/braceleftbigg/bracketleftbigg
a b
c d/bracketrightbigg
:ad−bc=/ne}ationslash= 0/bracerightbigg
.
(3) The set of polynomials of degree exactly n, that is
V:={a0xn+a1xn−1+···+an−1x+an:aj∈R,an/ne}ationslash= 0}.
(4) The set Vof 2×4 matrices with last column zero, that is
V:=/braceleftbigg/bracketleftbigg
a b c 0
d e f 0/bracketrightbigg
:a,b,c,d,e,f∈R/bracerightbigg
(5) The set of solutions f:R→Rof the equation f′= 5, that is
V:={f:R→R:f(x) = 5x+C, C∈R}.
(6) The set of all linear transformations T:R2→R3.
Before we pass to the notion of subspace, recall that a linear combination of
vectorsv1,...,v n∈Vis a vector of the form α1v1+···+αnvnforα1,...,α n∈R.
Definition 2.5.A subsetWof a vector space Vthat is itself a vector space is a
subspace .
In other words, a subset W⊆Vis a subspace if the following conditions are
verified:
(1) The 0 element is in V;
(2)Wisclosed under addition , that isv+w∈Wfor everyv,w∈W;
(3)Wisclosed under multiplication by scalars , that isαv∈Wfor everyα∈R
and everyv∈W.
Condition (1) in fact follows from (2) and (3), but it is often emphasiz ed because
it is an easy way to check that a subset is not a subspace. In any cas e the above
three conditions are equivalent to the following ones:
(1)’Wis nonempty;
(2)’Wisclosed under linear combinations , that isαv+βw∈Wfor allα,β∈R
and allv,w∈W.
2.2. BASES 7
2.2. Bases
The key yo study vector spaces is the concept of basis.
Definition 2.6.The vectors{b1,...,b n}∈Vform abasisofVif:
(1) they are linearly independent and
(2) thespanV.
Warning: We consider only vector spaces that have bases consisting of a finite
number of elements.
We recall here the notions of liner dependence/independence and t he notion of
span.
Definition 2.7.The vectors{b1,...,b n}∈Varelinearly independent ifα1b1+
···+αnbn= 0 implies that α1=···=αn= 0. In other words if the only linear
combination that represents zero is the trivial one.
Example 2.8.The vectors
b1=
1
2
3
, b 2=
4
5
6
, b 3=
7
8
0
are linearly independent in R3. In fact,
µ1b1+µ2b2+µ3b3= 0⇐⇒
µ1+4µ2+7µ3= 0
2µ1+5µ2+8µ30
3µ1+6µ2= 0⇐⇒...⇐⇒µ1=µ2=µ3= 0.
(If you are unsure how to fill in the dots look at Example 2.13.) /square
Example 2.9.The vectors
b1=
1
2
3
, b 2=
4
5
6
, b 3=
7
8
9
are linearly dependent in R3. In fact,
µ1b1+µ2b2+µ3b3= 0⇐⇒
µ1+4µ2+7µ3= 0
2µ1+5µ2+8µ30
3µ1+6µ2+9µ3= 0⇐⇒...⇐⇒/braceleftbiggµ1=µ2
µ2=−2µ3,
so
b1−2b2+b3= 0
andb1,b2,b3are not linearly independent. For example b1= 2b2−b3is a non-trivial
linear relation between the vectors b1,b2andb3. /square
Definition 2.10.The vectors{b1,...,b n}∈VspanVif every vector v∈Vcan
be written as a linear combination v=α1b1+···+αnbn, for someα1,...,α n∈R.
8 2. REVIEW OF LINEAR ALGEBRA
Example 2.11.The vectors in Example 2.8 span R3, while the vectors in Exam-
ple 2.9 do not span R3. To see this, we recall the following facts about bases. /square
Facts about bases: LetVbe a vector space:
(1) All bases of Vhave the same number of elements. This number is called
thedimension ofVand indicated with dim V.
(2) IfB:={b1,...,b n}form a basis of V, there is a unique way of writing vas
a linear combination
v=v1b1+...vnbn
of elements inB. We denote by
[v]B=
v1
...
vn
the coordinate vector of vwith respect toB.
(3) If we know that dim V=n, then:
(a) More than nvectors inVmust be linearly dependent;
(b) Fewer than nvectors inVcannot span V;
(c) Anynlinearly independent vectors span V;
(d) Anynvectors that span Vmust be linearly independent;
(e) Ifkvectors span V, thenk≥nand some subset of those kvectors
must be a basis of V;
(f) If a set of mvectors is linearly independent, then m≤nand we can
always complete the set to form a basis of V.
Example 2.12.The vectorsinExample 2.8formabasis of R3since theyarelinearly
independent and they are exactly as many as the dimension of R3./square
Example 2.13 (Gauss-Jordan elimination) .We are going to compute here the co-
ordinates of v=
1
1
1
with respect to the basis B={b1,b2,b3}in Example 2.8. The
seeked coordinates [ v]B=
v1
...
vn
must satisfy the equation
v1
1
2
3
+v2
4
5
6
+v3
7
8
0
=
1
1
1
,
2.2. BASES 9
so to find them we have to solve the following system of linear equation s:
v1+4v2+7v3= 1
2v1+5v2+8v3= 1
3v1+6v2= 1
or, equivalently, reduced the following augmented matrix
1 4 7 1
2 5 8 1
3 6 0 1
in echelon form using the Gauss–Jordan elimination method. We are go ing to per-
form both calculations in parallel, which will also point out that they are indeed
seemingly different incarnation of the same method.
By multiplying the first equation/row by 2 (reps. 3) and subtracting it from the
second (reps. third) equation/row we obtain
v1+4v2+7v3= 1
−3v2−6v3=−1
−6v2−21v3=−2/squiggleleftright
1 4 7 1
0−3−6−1
0−6−21−2
.
By multiplying the second equation/row by −1/3 and by adding to the first (resp.
third)equation/rowthesecondequation/rowmultiplied by −4/3(resp. 2)weobtain
v1−v3= 1
v2+2v3=1
3
−9v3= 0/squiggleleftright
1 0−11
0 1 21
3
0 0−90
.
The last equation/row shows that v3= 0, hence the above becomes
v1= 1
v2=1
3
v3= 0/squiggleleftright
1 0 0 1
0 1 01
3
0 0 1 0
.
/square
Exercise 2.14.LetVbe the vector space consisting of all 2 ×2 matrices with trace
zero, namely
V:=/braceleftbigg/bracketleftbigg
a b
c d/bracketrightbigg
:a,b,c,d∈Randa+d= 0/bracerightbigg
.
10 2. REVIEW OF LINEAR ALGEBRA
(1) Show that
B:=/braceleftbigg/bracketleftbigg
1 0
0−1/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
b1,/bracketleftbigg
0 1
0 0/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
b2,/bracketleftbigg
0 0
1 0/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
b3/bracerightbigg
is a basis of V.
(2) Show that
/tildewideB:=/braceleftbigg/bracketleftbigg
1 0
0−1/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
˜b1,/bracketleftbigg
0−1
1 0/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
˜b2,/bracketleftbigg
0 1
1 0/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
˜b3/bracerightbigg
is another basis of V.
(3) Compute the coordinates of
v=/bracketleftbigg
2 1
7−2/bracketrightbigg
with respect toBand with respect to /tildewideB.
2.3. The Einstein convention
We start by setting a notation that will turn out to be useful later o n. Recall
that ifB={b1,b2,b3}is a basis of a vector space V, any vector v∈Vcan be written
as
v=v1b1+v2b2+v3b3 (2.1)
for appropriate v1,v2,v3∈R.
Notation. From now on, expressions like the one in (2.1) will be written as
v=✭✭✭✭✭✭✭✭✭✭ ❤❤❤❤❤❤❤❤❤❤ v1b1+v2b2+v3b3=vjbj. (2.2)
That is, from now on when an index appear twice(that is, once as a subscript and
once as a superscript ) in a term, we know that it implies that there is a summation
over all possible values of that index. The summation symbol will not b e displayed.
Analogously, indices that are not repeated in expressions like aijxkyjare free
indices not subject to summation.
Examples 2.15.For indices ranging over {1,2,3}, i.e.n= 3:
(1) The expression aijxiykmeans
aijxiyk=a1jx1yk+a2jx2yk+a3jx3yk,
and could be called Rk
j(meaning that Rk
jandaijxiykboth depend on the
indicesjandk).
(2) Likewise,
aijxkyj=ai1xky1+ai2xky2+ai3xky3=:Qk
i.
2.3. THE EINSTEIN CONVENTION 11
(3) Further
aijxiyj=a11x1y1+a12x1y2+a13x1y3
+a21x2y1+a22x2y2+a23x2y3
+a31x3y1+a32x3y2+a33x3y3=:P
(4) An expression like
AiBj
kℓCℓ=:Dij
k
makessense. Heretheindices i,j,karefree(i.e. freetorangein {1,2,...,n})
andℓis a summation index.
(5) On the other hand an expression like
EijFℓjkGℓ=Hjk
i
does not make sense because the expression on the left has only tw o free
indices,iandk, whilejandℓare summation indices and neither of them
can appear on the right hand side.
Notation. Sincevjbjdenotes a sum, the generic term of a sum will be denoted
withcapital letters . For example we write vIbIand the above expressions could
have been written as
(1)
aijxiyk=3/summationdisplay
i=1aIJxIyK=a1jx1yk+a2jx2yk+a3jx3yk,
(2)
aijxkyj=3/summationdisplay
j=1aIJxKyJ=ai1xky1+ai2xky2+ai3xky3.
(3)
aijxiyj=3/summationdisplay
j=13/summationdisplay
i=1aIJxIyJ=
=a11x1y1+a12x1y2+a13x1y3
+a21x2y1+a22x2y2+a23x2y3
+a31x3y1+a32x3y2+a33x3y3.
/square
12 2. REVIEW OF LINEAR ALGEBRA
2.3.1. Change of bases, revisited. LetBand/tildewideBbe two bases of a vector
spaceVand let
L:=L/tildewideBB=
L1
1... L1
n......
Ln
1... Ln
n
(2.3)
be the matrix of the change of basis from Bto/tildewideB. [Recall that the entries of the
j-th column of Lare the coordinates of the ˜bjs with respect to the basis B.] With
the Einstein convention we can write
˜bj=Li
jbi,. (2.4)
or, equivalently
/parenleftbig˜b1...˜bn/parenrightbig
=/parenleftbigb1... b n/parenrightbig
L.
If Λ =L−1denotes the matrix of the change of basis from /tildewideBtoB, then
/parenleftbigb1... b n/parenrightbig
=/parenleftbig˜b1...˜bn/parenrightbig
Λ.
Equivalently, this can be written in compact form using the Einstein no tation as
bj= Λi
j˜bi.
Analogously, the corresponding relations for the vector coordina tes are
v1
...
vi
...
vn
=
L1
1... L1
n......
Li
1... Li
n......
Ln
1... Ln
n
˜v1
...
˜vn
and
˜v1
...
˜vi
...
˜vn
=
Λ1
1...Λ1
n......
Λi
1...Λi
n......
Λn
1...Λn
n
v1
...
vn
and these can be written with the Einstein convention respectively a s
vi=✭✭✭✭✭✭✭✭✭✭ ❤❤❤❤❤❤❤❤❤❤Li
1˜v1+···+Li
n˜vn=Li
j˜vjand ˜vi=✭✭✭✭✭✭✭✭✭✭ ❤❤❤❤❤❤❤❤❤❤Λi
1v1+···+Λi
nvn= Λi
jvj, (2.5)
or, in matrix notation,
[v]B=L/tildewideBB[v]/tildewideBand [v]/tildewideB= (L/tildewideBB)−1[v]B=LB/tildewideB[v]B.
Example 2.16.We consider the following two bases of R2
B=/braceleftbigg/bracketleftbigg
1
0/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
b1,/bracketleftbigg
2
1/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
b2/bracerightbigg
/tildewideB=/braceleftbigg/bracketleftbigg
3
1/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
˜b1,/bracketleftbigg
−1
−1/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
˜b2/bracerightbigg (2.6)
2.3. THE EINSTEIN CONVENTION 13
and we look for the matrix of the change of basis. Namely we look for a matrixL
such that/bracketleftbigg
3−1
1−1/bracketrightbigg
=/parenleftbig˜b1˜b2/parenrightbig
=/parenleftbigb1b2/parenrightbig
L=/bracketleftbigg
1 2
0 1/bracketrightbigg
L.
There are two alternative ways of finding L:
(1)With matrix inversion: Recall that
/bracketleftbigg
a b
c d/bracketrightbigg−1
=1
D/bracketleftbigg
d−b
−c a/bracketrightbigg
, (2.7)
whereD= det/parenleftbigg/bracketleftbigg
a b
c d/bracketrightbigg/parenrightbigg
. Thus
L=/bracketleftbigg
1 2
0 1/bracketrightbigg−1/bracketleftbigg
3−1
1−1/bracketrightbigg
=/bracketleftbigg
1−2
0 1/bracketrightbigg/bracketleftbigg
3−1
1−1/bracketrightbigg
=/bracketleftbigg
1 1
1−1/bracketrightbigg
.
(2)With the Gauss-Jordan elimination:
/bracketleftbigg
1 23−1
0 11−1/bracketrightbigg
/squiggleleftright/bracketleftbigg
1 01 1
0 11−1/bracketrightbigg
/square
2.3.2. The Kronecker delta symbol.
Notation. TheKronecker delta symbol δi
jis defined as
δi
j:=/braceleftigg
1 ifi=j
0 ifi/ne}ationslash=j.(2.8)
Examples 2.17.IfLis a matrix, the ( i,j)-entry ofLis the coefficient in the i-th
row andj-th column, and is denoted by Li
j.
(1) Then×nidentity matrix
I=
1...0
.........
0...1
has (i,j)-entry equal to δi
j.
(2) LetLand Λ be two square matrices. The ( i,j)-th entry of the product Λ L
ΛL=
Λ1
1...Λ1
n......
Λi
1...Λi
n......
Λn
1...Λn
n
L1
1...L1
j... L1
n
.........
Ln
1...Ln
j... Ln
n
14 2. REVIEW OF LINEAR ALGEBRA
equals the dot product of the i-th row and j-th column,
/parenleftbigΛi
1...Λi
n/parenrightbig
·
L1
j
...
Ln
j
= Λi
1L1
j+···+Λi
nLn
j,
or, using the Einstein convention,
Λi
kLk
j
Notice that since in general Λ L/ne}ationslash=LΛ, it follows that
Λi
kLk
j/ne}ationslash=Li
kΛk
j= Λk
jLi
k.
On the other hand, if Λ = L−1, that is ΛL=LΛ =I, then we can write
Λi
kLk
j=δi
j=Li
kΛk
j.
/square
Remark 2.18.Using the Kronecker delta symbol we can check that the notations
in (2.5) are all consistent. In fact, from (2.2) we should have
vibi=v= ˜vi˜bi, (2.9)
and, in fact, using (2.5),
˜vi˜bi= Λi
jvjLk
ibk=δk
jvjbk=vjbj,
where we used that Λi
jLk
i=δk
jsince Λ =L−1.
Two words of warning:
•The two expressions vjbjandvkbkare identical, as the indices jandkare
dummy indices.
•Whenmultiplying twodifferent expressions inEinstein notation, yoush ould
becarefultodistinguishbydifferentlettersdifferentsummationind ices. For
example, if ˜ vi= Λi
jvjand˜bi=Lj
ibj, in order to perform the multiplication
˜vi˜biwe have to make sure to replace one of the dummy indices in the
two expressions. So, for example, we can write ˜bi=Lk
ibk, so that ˜vi˜bi=
Λi
jvjLk
ibk.
2.4. Linear Transformations
LetT:V→Vbe a linear transformation, that is a transformation that satisfies
the property
T(αv+βw) =αT(v)+βT(w),
for allα,β∈Rand allv,w∈V. Once we choose a basis of V, the transformation
Tis represented by a matrix Awith respect to that basis, and that matrix gives
2.4. LINEAR TRANSFORMATIONS 15
the effect of Ton the coordinate vectors. In other words, if T(v) is the effect of the
transformation Ton the vector v, with respect a basis Bwe have that
[v]B/ma√sto−→[T(v)]B=A[v]B. (2.10)
If/tildewideBis another basis, we have also
[v]/tildewideB/ma√sto−→[T(v)]/tildewideB=/tildewideA[v]/tildewideB, (2.11)
where now/tildewideAis the matrix of the transformation Twith respect to the basis /tildewideB.
We want to find now the relation between Aand/tildewideA. LetL:=L/tildewideBBbe the matrix
of the change of basis from Bto/tildewideB. Then, for any v∈V,
[v]/tildewideB=L−1[v]B. (2.12)
In particular the above equation holds for the vector T(v), that is
[T(v)]/tildewideB=L−1[T(v)]B. (2.13)
Using (2.12), (2.11), (2.13) and (2.10) in this order, we have
/tildewideAL−1[v]B=/tildewideA[v]/tildewideB= [T(v)]/tildewideB=L−1[T(v)]B=L−1A[v]B
for every vector v∈V. If follows that /tildewideAL−1=L−1Aor equivalently
/tildewideA=L−1AL, (2.14)
which in Einstein notation reads
/tildewideAi
j= Λi
kAk
mLm
j.
We say that the linear transformation Tis atensor of type (1,1).
Example 2.19.LetV=R2and letBand/tildewideBbe the bases in Example 2.16. The
matrices corresponding to the change of coordinates are
L:=L/tildewideBB/bracketleftbigg
1 1
1−1/bracketrightbigg
andL−1=1
−2/bracketleftbigg
−1−1
−1 1/bracketrightbigg
=/bracketleftbigg1
21
21
2−1
2/bracketrightbigg
,
where in the last equality we used the formula for the inverse of a mat rix in (2.7).
LetT:R2→R2be the linear transformation that in the basis Btakes the form
A=/bracketleftbigg
1 3
2 4/bracketrightbigg
.
Thenaccordingto(2.14)thematrix /tildewideAofthelineartransformation Twithrespect
to the basis/tildewideBis
/tildewideA=L−1AL=/bracketleftbigg1
21
21
2−1
2/bracketrightbigg/bracketleftbigg
1 3
2 4/bracketrightbigg/bracketleftbigg
1 1
1−1/bracketrightbigg
=/bracketleftbigg
5−2
−1 0/bracketrightbigg
.
/square
16 2. REVIEW OF LINEAR ALGEBRA
Example 2.20.We now look for the standard matrix of T, that is the matrix M
that represents Twith respect to the standard basis of R2, which we denote by
E:=/braceleftbigg/bracketleftbigg
1
0/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
e1,/bracketleftbigg
0
1/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
e2/bracerightbigg
.
We want to applyagainthe formula (2.14) andhence we first need tofi nd the matrix
S:=LBEof the change of basis from EtoB. Recall that the columns of Sare the
coordinates of bjwith respect to the basis E, that is
S=/bracketleftbigg
1 2
0 1/bracketrightbigg
.
According to (2.14),
A=S−1MS,
from which, using again (2.7), we obtain
M=SAS−1=/bracketleftbigg
1 2
0 1/bracketrightbigg/bracketleftbigg
1 3
2 4/bracketrightbigg/bracketleftbigg
1−2
0 1/bracketrightbigg
=/bracketleftbigg
1 2
0 1/bracketrightbigg/bracketleftbigg
1 1
2 0/bracketrightbigg
=/bracketleftbigg
5 1
2 0/bracketrightbigg
.
/square
Example 2.21.LetT:R3→R3be theorthogonal projection onto the planePof
equation
2x+y−z= 0.
This means that the transformation Tis characterized by the fact that
– it does not change vectors in the plane P, and
– it takes to zero vectors perpendicular to P.
We want to find the standard matrix for T.
Idea:First compute the matrix of Twith respect to a basis BofR3well adapted
to the problem, then use (2.14) after having found the matrix LBEof the change of
basis.
To this purpose, we choose two linearly independent vectors in the p lanePand
a third vector perpendicular to P. For instance, we set
B:=/braceleftbigg
1
0
2
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
b1,
0
1
1
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
b2,
2
1
−1
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
b3/bracerightbigg
,
where the coordinates of b1andb2satisfy the equation of the plane, while the
coordinates of b3are the coefficients of the equation describing P. LetEbe the
standard basis of R3.
2.4. LINEAR TRANSFORMATIONS 17
Since
T(b1) =b1, T(b2) =b2andT(b3) = 0,
the matrix of Twith respect toBis
A=
1 0 0
0 1 0
0 0 0
, (2.15)
where we recall that the j-th column are the coordinates [ T(bj)]Bof the vector T(bj)
with respect to the basis B.
The matrix of the change of basis from EtoBis
L=
1 0 2
0 1 1
2 1−1
,
hence, by Gauss–Jordan elimination,
L−1=
1
3−1
31
3
−1
35
61
61
31
6−1
6
.
Therefore
M=LAL−1=···=
1
3−1
31
3
−1
35
61
61
31
65
6
.
/square
Example 2.22.LetV:=R[x]2be the vector space of polynomials of degree ≤2,
and letT:R[x]2→R[x]2be the linear transformation given by differentiating a
polynomial and then multiplying the derivative by x,
T(p(x)) :=xp′(x),
so thatT(a+bx+cx2) =x(b+2cx) =bx+2cx2. Let
B:={1,x,x2}and/tildewideB:={x,x−1,x2−1}
be two bases of R[x]2. Since
T(1) = 0 = 0·1+0·x+0·x2
T(x) =x= 0·1+1·x+0·x2
T(x2) = 2x2= 0·1+0·x+2·x2
and
T(x) =x= 1·x+0·(x−1)+0·(x2−1)
T(x−1) =x= 1·x+0·(x−1)+0·(x2−1)
T(x2−1) = 2x2= 2·x−2·(x−1)+2·(x2−1),
18 2. REVIEW OF LINEAR ALGEBRA
then
A=
0 0 0
0 1 0
0 0 2
and/tildewideA=
1 1 2
0 0−2
0 0 2
.
One can check that L=
0−1−1
1 1 0
0 0 1
and that indeed AL=L/tildewideAor, equivalently
/tildewideA=L−1AL. /square
2.4.1. Similar matrices. The above calculations can be summarized by the
commutativity of the following diagram. Here the vertical arrows correspond to t he
operation of the change of basis from Bto/tildewideB(recall that the coordinate vectors are
contravariant tensors, that is they transform as [ v]/tildewideB=L−1[v]B) and the horizontal
arrows to the operation of applying the transformation Tin the two different basis
[v]B✤A/d47/d47
❴
L−1
/d15/d15[T(v)]B❴
L−1
/d15/d15
[v]/tildewideB✤
/tildewideA/d47/d47[T(v)]/tildewideB.
Sayingthethediagramiscommutative isexactlythesamethingassay ing thatifone
startsfromtheupperlefthandcorner, reaching thelower righth andcornerfollowing
either one of the two paths has exactly the same effect. In other w ords, changing
coordinates first then applying the transformation Tyields exactly the same affect
as applying first the transformation Tand then the change of coordinates, that is
AL−1=L−1M
or equivalently
A=L−1ML.
In this case we say that Aand/tildewideAaresimilarmatrices. This means that Aand
/tildewideArepresent the same transformation with respect to different bas es.
Definition 2.23.We say that two matrices Aand/tildewideAaresimilarif there exists and
invertible matrix Lsuch that/tildewideA=L−1AL.
Examples 2.24.(1) The matrices in Example 2.19 and Example 2.20
A=/bracketleftbigg
1 3
2 4/bracketrightbigg
M=/bracketleftbigg
5 1
2 0/bracketrightbigg
and/tildewideA=/bracketleftbigg
5−2
−1 0/bracketrightbigg
are similar.
(2) The matrices Ain (1) and A′=/bracketleftbigg
1 2
2 4/bracketrightbigg
are not similar. In fact, Ais
invertible, as det A=−2/ne}ationslash= 0, while det A′= 0, so that A′is not invertible.
/square
2.5. EIGENBASES 19
We collect here few facts about similar matrices. Recall that the eigenvalues
of a matrix Aare the roots of the characteristic polynomial
pA(λ) := det(A−λI).
Moreover
(1) thedeterminant of a matrix is the product of its eigenvalues, and
(2) thetraceof a matrix is the sum of its eigenvalues.
Let us assume that Aand/tildewideAare similar matrices, that is /tildewideA=L−1ALfor some
invertible matrix L. Then
p/tildewideA(λ) = det(/tildewideA−λI) = det(L−1AL−λL−1IL)
= det(L−1(A−λI)L)
=✘✘✘✘✘(detL−1) det(A−λI)✘✘✘✘(detL) =pA(λ),(2.16)
which means that any two similar matrices have the same characteris tic polynomial.
Facts about similar matrices: From (2.16) if follows immediately that if the
matricesAand/tildewideAare similar, then:
•Aand/tildewideAhave the same size;
•the eigenvalues of A(as well as their multiplicity) are the same as those of
/tildewideA;
•detA= det/tildewideA;
•trA= tr/tildewideA;
•Ais invertible if and only if /tildewideAis invertible.
2.5. Eigenbases
The possibility of choosing different bases is very important and ofte n simplifies
the calculations. Example 2.21 is such an example, where we choose an appropriate
basis according to the specific problem. Other times a basis can be ch osen according
to the symmetries and, completely at the opposite side, sometime th ere is just not
a basis that is a preferred one. One basis that is particularly importa nt, when it
exists, is an eigenbasis with respect to some linear transformation AofV.
Recall that an eigenvector of a linear transformation Acorresponding to an
eigenvalue λis a non-zero vector v∈Eλ:= ker(A−λI). Aneigenbasis of a
vector space Vis a basis consisting of eigenvectors of a linear transformation Aof
V. The point of having an eigenbasis is that, with respect to this eigenb asis, the
linear transformation is as simple as possible, that is is as close as poss ible to be
diagonal. This diagonal matrix similar to Ais called the Jordan canonical form
ofA.
Given a linear transformation T:V→V, in order to find an eigenbasis of T,
we need to perform the following steps:
(1) Compute the eigenvalues
(2) Compute the eigenspaces
20 2. REVIEW OF LINEAR ALGEBRA
(3) Find a eigenbasis.
We will do this in the following example.
Example 2.25.LetT:R2→R2be the linear transformation given by the matrix
A=/bracketleftbigg
3−4
−4−3/bracketrightbigg
with respect to the standard basis of R2.
(1) The eigenvalues are the roots of the characteristic polynomial pλ(A). Since
pA(λ) = det(A−λI) = det/bracketleftbigg
3−λ−4
−4−3−λ/bracketrightbigg
= (3−λ)(−3−λ)−16 =λ2−25 = (λ−5)(λ+5),
henceλ=±5 are the eigenvalues of A.
(2) Ifλis an eigenvalue of A, the eigenspace corresponding to λis given by
Eλ= ker(A−λI). Note that
v∈Eλ⇐⇒Av=λv.
With our choice of Aand with the resulting eigenvalues, we have
E5= ker(A−5I) = ker/bracketleftbigg
−2−4
−4−8/bracketrightbigg
= span/bracketleftbigg
2
−1/bracketrightbigg
E−5= ker(A+5I) = ker/bracketleftbigg
−8−4
−4−2/bracketrightbigg
= span/bracketleftbigg
1
2/bracketrightbigg
.
(3) The following is an eigenbasis of R2
/tildewideB=/braceleftbigg
˜b1=/bracketleftbigg
2
1/bracketrightbigg
,˜b2=/bracketleftbigg
1
2/bracketrightbigg/bracerightbigg
and
T(˜b1) = 5˜b1= 5·˜b1+0·˜b2
T(˜b2) =−5˜b2= 0·˜b1−5·˜b2,
so thatA=/bracketleftbigg
5 0
0−5/bracketrightbigg
.
Notice that the eigenspace E5consists of vectors on the line x+2y= 0
and these vectors get scaled by the transformation Tby a factor of 5. On
the other hand, the eigenspace E−5consists of vectors perpendicular to the
linex+2y= 0 and these vectors get flipped by the transformation Tand
then also scaled by a factor of 5. Hence Tis just the reflection across the
linex+2y= 0 followed by multiplication by 5.
/square
Summarizing, in Examples 2.19 and 2.20 we looked at how the matrix of a
transformation changes with respect to two different basis that w e were given. In
Example 2.21 we looked for a particular basis appropriate to the tran sformation at
2.5. EIGENBASES 21
hand. InExample2.25welookedforaneigenbasiswithrespect tothe giventransfor-
mation. Example 2.21 in this respect fits into the same framework as E xample 2.25,
but the orthogonal projection has a zero eigenvalue (see (2.15)) .
CHAPTER 3
Multilinear Forms
3.1. Linear Forms
3.1.1. Definition, Examples, Dual and Dual Basis.
Definition 3.1.LetVbe a vector space. A linear form onVis a mapα:V→R
such that for every a,b∈Rand for every v,w∈V
α(av+bw) =aα(v)+bα(w).
Alternative terminologies for “linear form” are tensor of type (0,1),1-form,
linear functional andcovector .
Exercise 3.2.IfV=R3, which of the following is a linear form?
(1)α(x,y,z) :=xy+z;
(2)α(x,y,z) :=x+y+z+1;
(3)α(x,y,z) :=πx−7
2z.
Exercise 3.3.IfVis the infinite dimensional vector space of continuous functions
f:R→R, which of the following is a linear form?
(1)α(f) :=f(7)−f(0);
(2)α(f) :=/integraltext33
0exf(x)dx;
(3)α(f) :=ef(x).
Example 3.4.[Coordinate forms] This is the most important example of linear
form. LetB:={b1,...,b n}be a basis of Vand letv=vibi∈Vbe a generic vector.
Defineβi:V→Rby
βi(v) :=vi, (3.1)
that isβiwill extract the i-th coordinate of a vector with respect to the basis B.
The linear form βiis called coordinate form . Notice that
βi(bj) =δi
j, (3.2)
since thei-th coordinate of the basis vector bjwith respect to the basis Bis equal
to 1 ifi=jand 0 otherwise. /square
23
24 3. MULTILINEAR FORMS
Example 3.5.LetV=R3and letEbe its standard basis. The three coordinate
forms are defined by
β1
x
y
z
:=x, β2
x
y
z
:=y, β3
x
y
z
:=z.
/square
Example 3.6.LetV=R2and letB:=/braceleftbigg/bracketleftbigg
1
1/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
b1,/bracketleftbigg
1
−1/bracketrightbigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
b2/bracerightbigg
. We want to describe the
elements ofB∗:={β1,β2}, in other words we want to find
β1(v) and β2(v)
for a generic vector v∈V.
To this purpose we need to find [ v]B. Recall that ifEdenotes the standard basis
ofR2andL:=LBEthe matrix of the change of coordinate from EtoB, then
[v]B=L−1[v]E=L−1/parenleftbigg
v1
v2/parenrightbigg
.
Since
L=/bracketleftbigg
1 1
1−1/bracketrightbigg
and hence
L−1=1
2/bracketleftbigg
1 1
1−1/bracketrightbigg
,
then
[v]B=/parenleftbigg1
2(v1+v2)
1
2(v1−v2)/parenrightbigg
.
Thus, according to (3.1), we deduce that
β1(v) =1
2(v1+v2) andβ2(v) =1
2(v1−v2).
/square
Let us define
V∗:={all linear forms α:V→R}.
Exercise 3.7.Check that V∗is a vector space whose null vector is the linear form
identically equal to zero.
We calledV∗thedualofV.
3.1. LINEAR FORMS 25
Proposition 3.8.LetB:={b1,...,b n}be a basis of Vandβ1,...,βnare the
corresponding coordinate forms. Then B∗:={β1,...,βn}is a basis of V∗. As a
consequence
dimV= dimV∗.
Proof. According to Definition 2.6 , we need to check that the linear forms in
B∗
(1) spanVand
(2) are linearly independent.
(1) To check that B∗spansVwe need to verify that any α∈V∗is a linear combi-
nation ofβ1,...,βn, that is that
α=αiβi(3.3)
for someαi∈R. Because of (3.2), if we apply both sides of (3.3) to the j-th basis
vectorbi, we obtain
α(bj) =αiβi(bj) =αiδi
j=αj, (3.4)
which identifies the coefficients in (3.3).
Now letv=vibi∈Vbe an arbitrary vector. Then
α(v) =α(vibi) =viα(bi) =viαi,
where the second equality follows form the definition of linear form an d the third
from (3.4).
On the other hand
αiβi(v) =αiβi(vjbj) =αivjβi(bj) =αivjδi
j=αivi.
Thus (3.3) is verified.
(2) We need to check that the only linear combination of β1,...,βnthat gives
the zero linear form is the trivial linear combination. Let ciβi= 0 be a linear
combination of the βi. Then for every basis vector bj, withj= 1,...,n,
0 = (ciβi)(bj) =ci(βi(bj)) =ciδi
j=cj,
thus showing the linear independence. /square
The basisB∗ofV∗is called the basis ofVdual toB. We emphasize that the
coordinates (or components) of a linear form αwith respect toB∗are exactly the
values ofαon the elements of B,
αi=α(bi).
Example 3.9.LetV=R[x]2be the vector space of polynomials of degree ≤2, let
α:V→Rbe the linear form given by
α(p(x)) :=p(2)−p′(2) (3.5)
and letB:={1,x,x2}be a basis of V. We want to:
26 3. MULTILINEAR FORMS
(1) find the components of αwith respect toB∗;
(2) describe the basis B∗={β1,β2,β3};
(1) Since
α1=α(b1) =α(1) = 1−0 = 1
α2=α(b2) =α(x) = 2−1 = 1
α3=α(b3) =α(x2) = 4−4 = 0,
then
[α]B∗=/parenleftbig1 1 0/parenrightbig
. (3.6)
(2) The generic element p(x)∈R[x]2written as combination of basis elements 1 ,x
andx2is
p(x) =a+bx+cx2.
HenceB∗={β1,β2,β3}, is given by
β1(a+bx+cx2) =a
β2(a+bx+cx2) =b
β3(a+bx+cx2) =c.(3.7)
/square
Remark 3.10.Note that it does not make sense to talk about a “dual basis” of V∗,
as for every basis BofVthere is going to be a basis B∗ofV∗dual to the basis B.
In the next section we are going to see how the dual basis transfor m with a change
of basis.
3.1.2. Transformation of Linear Forms under a Change of Basi s.We
want to study how a linear form α:V→Rbehaves with respect to a change a
basis inV. To this purpose, let
B:={b1,...,b n}and/tildewideB:={˜b1,...,˜bn}
be two bases of Vand let
B∗:={β1,...,βn}and/tildewideB∗:={˜β1,...,˜βn}
the corresponding dual bases. Let
[α]B∗=/parenleftbigα1... α n/parenrightbig
and [α]/tildewideB∗=/parenleftbig/tildewideα1.../tildewideαn/parenrightbig
be the coordinate vectors of αwith respect toB∗and/tildewideB∗, that is
α(bi) =αiandα(˜bi) =/tildewideαi.
LetL:=L/tildewideBBbe the matrix of the change of basis in (2.3)
˜bj=Li
jbi.
3.1. LINEAR FORMS 27
Then
/tildewideαj=α(˜bj) =α(Li
jbi) =Li
jα(bi) =Li
jαi=αiLi
j, (3.8)
so that
/tildewideαj=αiLi
j. (3.9)
Exercise 3.11.Verify that (3.9) is equivalent to saying that
[α]/tildewideB∗= [α]B∗L. (3.10)
Note that we have exchanged the order of αiandLi
jin the last equation in (3.8) to
respect the order in which the matrix multiplication in (3.10) has to be p erformed.
This was possible because both αiandLi
jare real numbers.
We say that the component of a linear form αarecovariant1because they
change byLwhen the basis changes by L. A linear form αis hence a covariant
tensoror atensor of type (1,0).
Example 3.12.We continue with Example 3.9. We consider the bases as in Exam-
ple 2.22, that is
B:={1,x,x2}and/tildewideB:={x,x−1,x2−1}
and the linear form α:V→Ras in (3.5). We will:
(1) find the components of αwith respect toB∗;
(2) describe the basis B∗={β1,β2,β3};
(3) find the components of αwith respect to /tildewideB∗;
(4) describe the basis /tildewideB∗={˜β1,˜β2,˜β3};
(5) find the matrix of change of basis L:=L/tildewideBBand compute Λ = L−1;
(6) check the covariance of α;
(7) check the contravariance of B∗.
(1) This is done in (3.6).
(2) This is done in (3.7).
(3) We proceed as in (3.6). Namely,
α1=α(˜b1) =α(x) = 2−1 = 1
α2=α(˜b2) =α(x−1) = 1−1 = 0
α3=α(˜b3) =α(x2−1) = 3−4 =−1,
so that
[α]/tildewideB∗=/parenleftbig1−1−1/parenrightbig
.
1“co” is a prefix that in Latin means “joint”.
28 3. MULTILINEAR FORMS
(4) Since ˜βi(v) = ˜vi, to proceed as in (3.7) we first need to write the generic polyno-
mialp(x) =a+bx+cx2as a linear combination of elements in /tildewideB, namely we need
to find ˜a,˜band ˜csuch that
p(x) =a+bx+cx2= ˜ax+˜b(x−1)+˜c(x2−1).
By multiplying and collecting the terms, we obtain that
−˜b−˜c=a
˜a+˜b=b
˜c=cthat is
˜a=a+b+c
˜b=−a−c
˜c=c.
Hence
p(x) =a+bx+cx2= (a+b+c)x+(−a−c)(x−1)+c(x2−1),
so that it follows that
β1(p(x)) =a+b+c
β2(p(x)) =−a−c
β3(p(x)) =c,
(5) The matrix of the change of bases is given by
L:=L/tildewideBB=
0−1−1
1 1 0
0 0 1
,
since for example ˜b3can be written as a linear combination with respect to Bas
˜b3=x2−1 =−1b1+0b2+1b3, and hence its coordinates form the third column of
L.
To compute Λ = L−1we can use the Gauss–Jordan elimination process
0−1−11 0 0
1 1 0 0 1 0
0 0 1 0 0 1
/squiggleleftright.../squiggleleftright
1 0 0 1 1 1
0 1 0−1 0−1
0 0 1 0 0 1
Hence
Λ =
1 1 1
−1 0−1
0 0 1
(6) The linear form αiscovariant since
/parenleftbig/tildewideα1/tildewideα2/tildewideα3/parenrightbig
=/parenleftbig1 0−1/parenrightbig
=/parenleftbig1 1 0/parenrightbig
0−1−1
1 1 0
0 0 1
=/parenleftbigα1α2α3/parenrightbig
L
(7) The dual basis B∗iscontravariant since
3.1. LINEAR FORMS 29
Table 1. Covariance and Contravariance
Thecovariance Thecontravariace
of a tensor of a tensor
is characterized by lowerindices upperindices
vectors are indicated as rowvectors columnvectors
the tensor transforms w.r.t.
a change of basis B→/tildewideBby
multiplication by Lon therightL−1on theleft
(for later use) if a tensor
is of type (p,q) (p,q) (p,q)
˜β1
˜β2
˜β3
= Λ
β1
β2
β3
,
as it can be verified by looking at an arbitrary vector p(x) =a+bx+cx2
a+b+c
−a−c
c
=
1 1 1
−1 0−1
0 0 1
a
b
c
.
/square
In fact, the statement in Example 3.9(7) holds in general, namely:
Claim3.13.Dual bases are contravariant .
Proof. We will check that when bases Band/tildewideBare related by
˜bj=Li
jbi
the corresponding dual bases B∗and/tildewideB∗ofV∗are related by
˜βj= Λj
iβi. (3.11)
It is enough to check that the Λj
iβiaredualof theLi
jbi. In fact, since Λ L=I, then
(Λk
ℓβℓ)(Li
jbi) = Λk
ℓLi
jβℓ(bi) = Λk
ℓLi
jδℓ
i= Λk
iLi
j=δk
j=βj(˜bj).
/square
In Table 1 you will find a summary of the properties that characteriz e covariance
and contravariance, while in Table 2 you can find a summary of the pro perties that
bases and dual bases, coordinate vectors and coordinates of line ar forms satisfy
with respect to a change of coordinates and hence whether they a re covariant or
contravariant.
30 3. MULTILINEAR FORMS
Table 2. Summary
Vreal vector space V∗={α:V→R}=linear forms
with dimV=n = dual vector space
B:={b1,...,b n} B∗={β1,...,βn}
basis ofV dual basis of V∗w.r.tB
/tildewideB:={˜b1,...,˜bn} /tildewideB∗={˜β1,...,˜βn}
another basis of V dual basis of V∗w.r.t/tildewideB∗
L:=L/tildewideBB=matrix of the change Λ =L−1=matrix of the change
of basis fromBto/tildewideB of basis from /tildewideBtoB
˜bj=Li
jbii.e./parenleftbig˜b1...˜bn/parenrightbig
=/parenleftbigb1... b n/parenrightbig
L˜βi= Λi
jβji.e.
˜β1
...
˜βn
=L−1
β1
...
βn
covariance of a basis contravariance of the dual basis
Ifvis any vector in V Ifαis any linear form in V∗
thenv=vibi= ˜vi˜bi thenα=αjβj=/tildewideαj˜βj
where where
˜vi= Λi
jvji.e. [v]/tildewideB=L−1[v]B /tildewideαj=Li
jαii.e. [α]/tildewideB= [α]BL
or
˜v1
...
˜vn
=L−1
v1
...
vn
or/parenleftbig˜α1...˜αn/parenrightbig
=/parenleftbigα1... α n/parenrightbig
L
contravariance of the coordinate vectors covariance of linear forms
3.2. Bilinear Forms
3.2.1. Definition, Examples and Basis.
Definition 3.14.Abilinear form onVis a function ϕ:V×V→Rthat is linear
in each variable, that is
ϕ(u,λv+µw) =λϕ(u,v)+µϕ(u,w)
ϕ(λv+µw,u) =λϕ(v,u)+µϕ(w,u),
for everyλ,µ∈Rand for every u,v,w∈V.
Examples 3.15.LetV=R3.
(1) Thescalar product
ϕ(v,w) :=v•w=|v||w|cosθ,
whereθis the angle between vandwis a bilinear form. It can be defined
also forn>3.
3.2. BILINEAR FORMS 31
(2) Choose a vector u∈R3and for any two vectors v,w∈R3, denote by v×w
theircross product . Thescalar triple product
ϕu(v,w) :=u•(v×w) = det
u
v
w
(3.12)
is a bilinear form in vandw, where
u
v
w
denotes the matrix with rows u,v
andw. The quantity ϕu(v,w) calculates the signed volume of the paral-
lelepiped spannedby u,v,w: thesignof ϕu(v,w)depends ontheorientation
of the triple u,v,w.
Since the cross product is defined only in R3, contrary to the scalar
product, the scalar triple product cannot be defined in Rnwithn >3
(although there is a formula for an ndimensional parallelediped involving
some “generalization” of it).
/square
Exercise 3.16.Verify the equality in (3.12) using the Leibniz formula for the de-
terminant of a 3×3 matrix. Recall that
det
a11a12a13
a21a22a23
a31a32a33
=a11a22a33−a11a23a32+a12a23a31
−a12a21a33+a13a21a32−a13a22a31
=/summationdisplay
σ∈S3sign(σ)a1σ(1)a2σ(2)a3σ(3),
where
σ= (σ(1),σ(2),σ(3))∈S3:={permutations of 3 elements }
={(1,2,3),(1,3,2),(2,3,1),(2,1,3),(3,1,2),(3,2,1)}.
Examples 3.17.LetV=R[x]2.
(1) Letp,q∈R[x]2. The function ϕ(p,q) :=p(π)q(33) is a bilinear form.
(2) Likewise,
ϕ(p,q) :=p′(0)q(4)−5p′(3)q′′(1
2)
is a bilinear form.
/square
Exercise 3.18.Are the following functions bilinear forms?
(1)V=R2andϕ(u,v) := det/bracketleftbigg
u
v/bracketrightbigg
;
(2)V=R[x]2andϕ(p,q) :=/integraltext1
0p(x)q(x)dx;
32 3. MULTILINEAR FORMS
(3)V=M2×2(R), the space of real 2 ×2 matrices, and ϕ(L,M) :=L1
1trM,
whereL1
1it the (1,1)-entry of Land trMis the trace of M;
(4)V=R3andϕ(v,w) :=v×w;
(5)V=R2andϕ(v,w) is the area of the parallelogram spanned by vandw.
3.2.2. Tensor product of two linear forms on V.Letα,β∈V∗be two
linear forms, α,β:V→R, and define ϕ:V×V→R, by
ϕ(v,w) :=α(v)β(w).
Thenϕis bilinear, is called the tensor product ofαandβand is denoted by
ϕ=α⊗β.
Note3.19.In generalα⊗β/ne}ationslash=β⊗α, as there could be vectors vandwsuch that
α(v)β(w)/ne}ationslash=β(v)α(w).
Example 3.20.LetV=R[x]2, letα(p) =p(2)−p′(2) andβ(p) =/integraltext4
3p(x)dxbe two
linear forms. Then
(α⊗β)(p,q) = (p(2)−p′(2))/integraldisplay4
3q(x)dx
is a bilinear form. /square
Example 3.21.Letϕ:R×R→Rbe a function:
(1)ϕ(x,y) := 2x−yis alinear form in (x,y)∈R2;
(2)ϕ(x,y) := 2xyisbilinear, hencelinearinx∈Randlineariny∈R, but it
isnot linear in (x,y)∈R2.
/square
Let
Bil(V×V,R) :={all bilinear forms ϕ:V×V→R}.
Exercise 3.22.Check that Bil( V×V,R) is a vector space with the zero element
equal to the bilinear form identically equal to zero.
Hint:It is enough to check that if ϕ,ψ∈Bil(V×V,R), andλ,µ∈R, then
λϕ+µψ∈Bil(V×V,R). Why? (Recall Example 2.3(4).)
Assuming Exercise 3.22, we are going to find a basis of Bil( V×V,R) and deter-
mine its dimension. Let B:={b1,...,b n}be a basis of Vand letB∗={β1,...,βn}
be the dual basis of V∗(that isβi(bj) =δi
j).
Proposition 3.23.The bilinear forms β1⊗βj,i,j= 1,...,nform a basis of
Bil(V×V,R). As a consequence dimBil(V×V,R) =n2.
Notation. We denote
Bil(V×V,R) =V∗⊗V∗
thetensor product ofV∗andV∗.
3.2. BILINEAR FORMS 33
Proof of Proposition 3.23. The proof will be similar to the one of Proposi-
tion 3.8 for linear forms. We first check that the set of bilinear forms {β1⊗βj, i,j=
1,...,n}span Bil(V×V,R) and than that it consists of linearly independent ele-
ments.
To check that span {βi⊗βj, i,j= 1,...,n}= Bil(V×V,R), we need to check
that ifϕ∈Bil(V×V,R), there exists Bij∈Rsuch that
ϕ=Bijβi⊗βj.
Because of (3.2), we obtain
ϕ(bk,bℓ) =Bijβi(bk)βj(bℓ) =Bijδi
kδj
ℓ=Bkℓ,
for every pair ( bk,bℓ)∈V×V. Hence we are forced to choose Bkℓ:=ϕ(bk,bℓ). Now
we have to check that with this choice of Bkℓwe have indeed
ϕ(v,w) =Bijβi(v)βj(w)
for arbitrary v=vkbk∈Vandw=wℓbℓ∈V.
On the one hand we have that
ϕ(v,w) =ϕ(vkbk,wℓbℓ) =vkwℓϕ(bk,bℓ) =vkwℓBkℓ,
where the next to the last equality follows from the bilinearity of ϕand the last one
from the definition of Bkℓ.
On the other hand,
Bijβi(v)βj(w) =Bijβi(vkbk)βj(vℓbℓ)
=Bijvkβi(bk)vℓβj(bℓ)
=Bijvkvℓδi
kδj
ℓ
=Bkℓvkwℓ,
where the second equality follows from the bilinearity of βiand the next to the last
from (3.2).
Now we need to check that the only linear combination of the βi⊗βjthat gives
the zero bilinear form is the trivial linear combination. Let cijβi⊗βj= 0 be a
linear combination of the βi⊗βj. Then for all pairs of basis vectors ( bk,bℓ), with
k,ℓ= 1,...,n, we have
0 =cijβi⊗βj(bk,bℓ) =cijδi
kδj
ℓ=ckℓ,
thus showing the linear independence. /square
3.2.3. Transformation of Bilinear Forms under a Change of Ba sis.If
we summarize what we have done so far, we see that once we choose a basisB:=
{b1,...,b n}ofV, we automatically have a basis B∗={β1,...,βn}ofV∗and a basis
{β1⊗βj, i,j= 1,...,n}ofV∗⊗V∗.
That is, any bilinear form ϕ:V×V→Rcan be represented by its components
Bij=ϕ(bi,bj) (3.13)
34 3. MULTILINEAR FORMS
and these components can be arranged in a matrix
B:=
B11... B 1n
......
Bn1... B nn
called the matrix of the bilinear form ϕwith respect to the chosen basis B.
The natural question of course is: how does the matrix Bchange when we choose a
different basis of V?
So, let us choose a different basis /tildewideB:={˜b1,...,˜bn}and corresponding bases
/tildewideB∗={˜β1,...,˜βn}ofV∗and{˜βi⊗˜βj, i,j= 1,...,n}ofV∗⊗V∗, with respect to
whichϕwill be represented by a matrix /tildewideB, whose entries are /tildewideBij=ϕ(˜bi,˜bj).
To see the relation between Band/tildewideB, due to the change of basis from Bto/tildewideB,
we start with the matrix of the change of basis L:=L/tildewideBB, according to which
˜bj=Li
jbi. (3.14)
Then
/tildewideBij=ϕ(˜bi,˜bj) =ϕ(Lk
ibk,Lℓ
jbℓ) =Lk
iLℓ
jϕ(bk,bℓ) =Lk
iLℓ
jBkℓ,
where the first and the last equality follow from (3.13), the second f rom (3.14) (
after having renamed the dummy indices to avoid conflicts) and the r emaining one
from the bilinearity.
Exercise 3.24.Show that the formula of the transformation of the component of
a bilinear form in terms of the matrices of the change of coordinates is
/tildewideB=tLBL, (3.15)
wheretLdenotes the transpose of the matrix L.
We hence say that a bilinear form ϕis acovariant 2-tensor or atensor of
type(0,2).
3.3. Multilinear forms
We saw in§3.1.2 that linear forms are covariant 1-tensors – or tensor of type
(0,1) – and in§3.2.3 that bilinear forms are covariant 2-tensors – or tensors of ty pe
(0,2).
Completely analogously to what was done until now, one can define trilinear
forms, that is functions T:V×V×V→Rthat are linear in each of the three
variables. The space of trilinear forms is denoted by V∗⊗V∗⊗V∗, has basis
{βj⊗βj⊗βk, i,j,k= 1,...,n}and hence dimension n3.
Since the components of a trilinear form T:V×V×V→Rsatisfy the following
transformation with respect to a change of basis
/tildewideTijk=Lℓ
iLp
jLq
kTℓpq,
a trilinear form is a covariant 3-tensoror atensor of type (0,3).
3.4. EXAMPLES 35
In fact, there is nothing special about k= 1,2 or 3.
Definition 3.25.Amultilinear form if a function f:V×···×V→Rfrom
k-copies ofVintoR, that is linear with respect to each variable.
A multilinear form is a covariant k-tensor or atensor of type (0,k). The
vectors space of multilinear forms V∗⊗···⊗V∗has basisβi1⊗βi2×···⊗βik,
i1,...,i k:= 1,...,nand hence dim( V∗⊗···⊗V∗) =nk.
3.4. Examples
3.4.1. A Bilinear Form.
Example 3.26.We continue with the study of the scalar triple product , that was
defined in Example 3.15. We want to find the components Bijofϕuwith respect
to the standard basis of R3. Letu=
u1
u2
u3
be the fixed vector. Recall the cross
product in R3is defined as
ei×ej:=
0 ifi=j
ekif (i,j,k) is a cyclic permutation of (1 ,2,3)
−ekif (i,j,k) is a non-cyclic permutation of (1 ,2,3),
that is
cyclic
e1×e2=e3
e2×e3=e1
e3×e1=e2and
non-cyclic
e2×e1=−e3
e3×e2=−e1
e1×e3=−e2
Sinceu•ek=uk, then
Bij=ϕu(ei,ej) =u•(ei×ej) =
0 ifi=j
ukif (i,j,k) is a cyclic permutation of (1 ,2,3)
−ukif (i,j,k) is a non-cyclic permutation of (1 ,2,3)
Thus
B12=u3=−B21
B23=u1=−B32
BII= 0 (that is the diagonal components are zero) ,
which can be written as a matrix
B=
0u3−u2
−u30u1
u2−u10
.
36 3. MULTILINEAR FORMS
We look now for the matrix of the scalar tripe product with respect t o the basis
/tildewideB:=/braceleftigg
0
2
0
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
˜b1,
1
0
1
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
˜b2,
0
0
1
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
˜b3/bracerightigg
.
The matrix of the change of coordinates from the standard basis t o/tildewideBis
L=
0 1 0
1 0 0
0 1 1
,
so that
/tildewideB=
0 1 0
1 0 1
0 0 1
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
t
0u3−u2
−u30u1
u2−u10
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
B
0 1 0
1 0 0
0 1 1
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
L
=
0 1 0
1 0 1
0 0 1
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
tL
u3−u2−u2
0u1−u3u1
−u1u20
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
BL=
0u1−u3u1
u3−u10−u2
−u1u20
.
It is easy to check that /tildewideBis antisymmetric just like Bis, and to check that the
components of /tildewideBare correct by using the formula for ϕ. In fact
/tildewideB12=ϕ(˜b1,˜b2) =u•(e2×(e1+e3)) =u1−u3
/tildewideB13=ϕ(˜b2,˜b3) =u•((e2)×e3) =u1
/tildewideB23=ϕ(˜b2,˜b3) =u•((e1+e3)×e3) =−u2
/tildewideB11=ϕ(˜b1,b1) =u•(e2×e2) = 0
/tildewideB22=ϕ(˜b2,b2) =u•((e1+e3)×(e1+e3)) = 0
/tildewideB33=ϕ(˜b3,b3) =u•(e3×e3) = 0
/square
3.4.2. A Trilinear Form.
Example 3.27.If in the definition of the scalar triple product instead of fixing a
vectora∈R, we let the vector vary, we have a function ϕ:R3×R3×R3→R,
defined by
ϕ(u,v,w) :=u•(v×w) = det
u
v
w
.
3.5. BASIC OPERATION ON MULTILINEAR FORMS 37
One can verify that such function is trilinear, that is linear in each of t he three
variables separately.
3.5. Basic Operation on Multilinear Forms
LetT:V×···×V/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
ktimes→RandU:V×···×V/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
ℓtimes→Rbe respectively a k-linear
and anℓ-linear form. Then the tensor product ofTandU
T⊗U:V×···×V/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
k+ℓtimes→R,
defined by
T⊗U(v1,...,v k+ℓ) :=T(v1,...,v k)U(vk+1,...,v k+ℓ)
is a (k+ℓ)-linear form.
Likewise, one can take the tensor product of a tensor of type (0 ,k) and a tensor
of type (0,ℓ) to obtain a tensor of type (0 ,k+ℓ).
CHAPTER 4
Inner Products
4.1. Definitions and First Properties
Inner products are a special case of bilinear forms. They add an imp ortant
structure to a vector space, as for example they allow to compute the length of a
vector. Moreover, they provide a canonical identification betwee n the vector space
Vand its dual V∗.
Definition 4.1.Aninner product g:V×V→Ronavector space Vis abilinear
formonVthat is
(1)symmetric , that isg(v,w) =g(w,v) for allv,w∈Vand
(2)positive definite , that isg(v,v)≥0 for allv∈V, andg(v) = 0 if and only
ifv= 0.
Exercise 4.2.LetV=R3. Determine whether the following bilinear forms are
inner products, by verifying whether they are symmetric and posit ive definite:
(1) the scalar or dot product ϕ(v,w) :=v•w, defined as
v•w=viwi,
wherev=
v1
v2
v3
andw=
w1
w2
w3
;
(2)ϕ(v,w) :=−v•w, for allv,w∈V;
(3)ϕ(v,w) =v•w+2v1w2, forv,w∈V;
(4)ϕ(v,w) =v•3w, forv,w∈V.
Exercise 4.3.LetV:=R[x]2be the vector space of polynomials of degree ≤2.
Determine whether the following bilinear forms are inner products, b y verifying
whether they are symmetric and positive definite:
(1)ϕ(p,q) =/integraltext1
0p(x)q(x)dx;
(2)ϕ(p,q) =/integraltext1
0p′(x)q′(x)dx;
(3)ϕ(p,q) =/integraltextπ
3exp(x)q(x)dx;
(4)ϕ(p,q) =p(1)q(1)+p(2)q(2);
(5)ϕ(p,q) =p(1)q(1)+p(2)q(2)+p(3)q(3).
Definition 4.4.Letg:V×V→Rbe an inner product on V.
39
40 4. INNER PRODUCTS
(1) Thenorm/bardblv/bardblof a vector v∈Vis defined as
/bardblv/bardbl:=/radicalbig
g(v,v).
(2) A vector v∈Visunit vector if/bardblv/bardbl= 1;
(3) Two vectors v,w∈Vareorthogonal (that isperpendicular or v⊥w),
ifg(v,w) = 0;
(4) Two vectors v,w∈Vareorthonormal if they are orthogonal and /bardblv/bardbl=
/bardblw/bardbl= 1;
(5) A basisBofVis anorthonormal basis ifb1,...,b nare pairwise orthonor-
mal vectors, that is
g(bi,bj) =δij:=/braceleftigg
1 ifi=j
0 ifi/ne}ationslash=j,(4.1)
for alli,j= 1...,n. The condition for i=jimplies that an orthonormal
basis consists of unit vectors, while the one for i/ne}ationslash=jimplies that it consists
of pairwise orthogonal vectors.
Example 4.5.(1) LetV=R[x]2andgthe standard inner product. The
standard basisB={e1,...,e n}is an orthonormal basis with respect to the
standard inner product.
(2) LetV=R[x]2and letg(p,q) :=/integraltext1
−1p(x)q(x)dx. Check that the basis
B={p1,p2,p3},
where
p1(x) :=1√
2, p2(x) :=/radicalbigg
3
2x, p3(x) :=/radicalbigg
5
8(3x2−1),
is an orthonormal basis with respect to the inner product g.
Remark 4.6.p1,p2,p3are the first three Legendre polynomials up to
scaling.
An inner product gon a vector space Vis also called a metriconV.
4.1.1. Correspondence Between Inner Products and Symmetri c Pos-
itive Definite Matrices. Recall that a matrix S∈Mn×n(R) issymmetric if
S=tS, that is if
S=
∗A B...
A∗C...
B C∗...
......∗
.
Moreover if Sis symmetric, then
4.1. DEFINITIONS AND FIRST PROPERTIES 41
(1)Sispositive definite iftvSv>0 for allv∈Rn;
(2)Sisnegative definite iftvSv<0 for allv∈Rn;
(3)Sispositive semidefinite iftvSv≥0 for allv∈Rn;
(4)Sispositive semidefinite iftvSv≤0 for allv∈Rn;
(5)Sisindefinite ifvtSvtakes both positive and negative values for different
v∈Rn.
Definition 4.7.Aquadratic form Q:Rn→Ris a homogeneous quadratic
polynomial in nvariables.
Any symmetric matrix Scorrespond to a quadratic form as follow:
S/ma√sto→QS,
whereQS:Rn→Ris defined by
QS(v) =tvSv=/bracketleftbigv1... vn/bracketrightbig
S
v1
...
vn
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
matrix notationSij=vivjSij/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
Einstein notation. (4.2)
Note thatQisnotlinear inv.
LetSbe a symmetric matrix and QSbe the corresponding quadratic form.
The notion of positive definiteness, etc. for Scan be translated into corresponding
properties for QS, namely:
(1)Qispositive definite ifQ(v)>0 for allv∈V;
(2)Qisnegative definite ifQ(v)<0 for allv∈V;
(3)Qispositive semidefinite ifQ(v)≥0 for allv∈V;
(4)Qisnegative semidefinite ifQ(v)≤0 for allv∈V;
(5)Qisindefinite ifQ(v) takes both positive and negative values.
To find out the type of a symmetric matrix S(or, equivalently of a quadratic
formQS) it is enough to look at the eigenvalues of S, namely:
(1)SandQSarepositive definite if all eigenvalues of Sare positive:
(2)SandQSarenegative definite if all eigenvalues of Sare negative;
(3)SandQSarepositive semidefinite if all eigenvalues of Sare non-negative;
(4)SandQSarenegative semidefinite if all eigenvalues of Sare non-positive;
(5)SandQSareindefinite ifShas both positive and negative eigenvalues.
The reason this makes sense is the same reason for which we need to restrict our
attention to symmetric matrices and lies in the so-called Spectral Th eorem:
Theorem 4.8.[Spectral Theorem] Any symmetric matrix Shas the following prop-
erties:
(1)it has only realeigenvalues;
(2)it isdiagonalizable ;
42 4. INNER PRODUCTS
(3)it admits an orthonormal eigenbasis , that is a basis{b1,...,b n}such that
thebjare orthonormal and are eigenvectors of S.
4.1.1.1.From Inner Products to Symmetric Positive Definite Matrices .LetB:=
{b1,...,b n}be a basis of V. Thecomponents of gwith respect to Bare
gij=g(bi,bj). (4.3)
LetGbe the matrix with entries gij
G=
g11... g 1n
.........
gn1... g nn
. (4.4)
We claim that Gis symmetric and positive definite. In fact:
(1) Sincegissymmetric , then for 1≤i,j≤n,
gij=g(bi,bj) =g(bj,bi) =gji⇒Gis asymmetric matrix;
(2) Sincegispositive definite , thenGispositive definite asa symmetric matrix.
In fact, let v=vibi,w=wjbj∈Vbe two vectors. Then, using the
bilinearity of gin (1), (4.3) and with the Einstein notation, we have:
g(v,w) =g(vibi,wjbj)(1)=viwjg(bi,bj)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
gij=viwjgij
or, in matrix notation,
g(v,w) =t[v]BG[w]B=/bracketleftbigv1... vn/bracketrightbig
G
w1
...
wn
.
4.1.1.2.From Symmetric Positive Definite Matrices to Inner Products .IfSis a
symmetric positive definite matrix, then the assignment
(v,w)/ma√sto→tvSv
defines a map that is easily seen to be bilinear, symmetric and positive d efinite and
is hence an inner product.
4.1.2. Orthonormal Basis. Suppose that there in a basis B:={b1,...,b n}of
Vconsisting of orthonormal vectors with respect to g, so that
gij=δij,
because of Definition 4.4(5) and of (4.3). In other words the symme tric matrix
corresponding to the inner product gin the basis consisting of orthonormal vectors
is the identity matrix. Moreover
g(v,w) =viwjgij=viwjδij=viwi,
4.1. DEFINITIONS AND FIRST PROPERTIES 43
so that, ifv=w,
/bardblv/bardbl2=g(v,v) =vivi= (v1)2+···+(vn)2.
We deduce the following important
Fact4.9.Any inner product gcan be expressed in the standard form
g(v,w) =viwi,
as long as [v]B=
v1
...
vn
and[w]B=
w1
...
wn
are the coordinates of vandwwith
respect to an orthonormal basis Bforg.
Example 4.10.Letgbe an inner product of R3with respect to which
/tildewideB:=/braceleftigg
1
0
0
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
˜b1,
1
1
0
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
˜b2,
1
1
1
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
˜b3/bracerightigg
is an orthonormal basis. We want to express gwith respect to the standard basis E
ofR3
E:=/braceleftigg
1
0
0
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
e1,
0
1
0
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
e2,
0
0
1
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
e3/bracerightigg
.
The matrices of the change of basis are
L:=L/tildewideBE=
1 1 1
0 1 1
0 0 1
and Λ =L−1=
1−1 0
0 1−1
0 0 1
.
Sincegis a bilinear form, we saw in (3.15) that its matrices with respect to a ch ange
of basis are related by the formula
/tildewideG=tLGL.
Since the basis /tildewideBis orthonormal with respect to g, the associated matrix /tildewideGis the
identity matrix, so that
G=tΛ/tildewideGΛ =tΛΛ
=
1 0 0
−1 1 0
0−1 1
1−1 0
0 1−1
0 0 1
=
1−1 0
−1 2−1
0−1 2
.(4.5)
44 4. INNER PRODUCTS
It follows that, with respect to the standard basis, gis given by
g(v,w) =/parenleftbigv1v2v3/parenrightbig
1−1 0
−1 2−1
0−1 2
w1
w2
w3
=v1w1−v1w2−v2w1+2v2w2
−v2w3−w3v2+2v3w3.(4.6)
/square
Exercise 4.11.Verify the formula (4.6) for the inner product gin the coordinates
of the basis/tildewideBby applying the matrix of the change of coordinate directly on the
coordinates vectors [ v]E.
Remark 4.12.Norm and inner product of vectors depend onlyon the choice of g,
butnoton the choice of basis: different coordinate expressions yield the sa me result.
Example 4.13.We verify the assertion of the previous remark with the inner prod-
uct in Example 4.10. Let v,w∈R3such that
[v]E=
v1
v2
v3
=
3
2
1
and [v]/tildewideB=
˜v1
˜v2
˜v3
=L−1
v1
v2
v3
=
3
1
1
and
[w]E=
w1
w2
w3
=
1
2
3
and [w]/tildewideB=
˜w1
˜w2
˜w3
=L−1
w1
w2
w3
=
−1
−1
3
.
Then with respect to the basis /tildewideBwe have that
g(v,w) = 1·(−1)+1·(−1)+1·3 = 1,
and also with respect to the basis E
g(v,w) = 3·1−3·2−2·1+2·2·2−2·3−1·2+2·1·3 = 1.
/square
Exercise 4.14.Verifythat/bardblv/bardbl=√
3and/bardblw/bardbl=√
11, whencomputedwithrespect
of both bases.
LetB:={b1,...,b n}be an orthonormal basis and let v=vibibe a vector in V.
We want to compute the coordinates viofvwith respect of the metric gand of the
elements of the basis. In fact
g(v,bj) =g(vibi,bj) =vig(bi,bj) =viδij=vj,
that is the coordinates of a vector with respect to an orthonorma l basis are the inner
product of the vector with the basis vectors. This is particularly nic e, so that we
have to make sure that we remember how to construct an orthono rmal basis from a
given arbitrary basis.
4.1. DEFINITIONS AND FIRST PROPERTIES 45
Recall (Gram–Schmidt orthogonalization process) .The Gram–Schmidt orthogo-
nalization process is a recursive process that allows us to obtain an o rthonormal
basis starting from an arbitrary one. Let B:={b1,...,b n}be an arbitrary basis, let
g:V×V→Rbe an inner product and /bardbl·/bardblthe corresponding norm.
We start by defining
u1:=1
/bardblb1/bardblb1.
Next, observe that g(b2,u1)u1is the projection of the vector b2in the direction of
u1. It follows that
b⊥
2:=b2−g(b2,u1)u1
is a vector orthogonal to u1, but not necessarily of unit norm. Hence we set
u2:=1
/bardblb2/bardblb⊥
2.
Likewiseg(b3,u1)u1+(b3,u2)u2is the projection of b3on the plane generated by u1
andu2, so that
b⊥
3:=b3−g(b3,u1)u1−g(b3,u2)u2
is orthogonal both to u1and tou2. Set
u3:=1
/bardblb3/bardblb⊥
3.
Continuing until we have exhausted all elements of the basis B, we obtain an or-
thonormal basis{u1,...,u n}.
u3
b1 b1b2 b2b3 b3 b3
u1 u1u2 u2
Example 4.15.LetVbe the subspace of R4spanned by
b1=
1
1
−1
−1
b2=
2
2
0
0
b3=
1
1
1
0
.
(One can check that b1,b2,b3are linearly independent and hence form a basis of V.)
We look for an orthonormal basis of Vwith respect to the standard inner product
/an}bracketle{t·,·/an}bracketri}ht. Since
/bardblb1/bardbl= (11+12+(−1)2+(−1)2)1/2= 2⇒u1:=1
2b1.
46 4. INNER PRODUCTS
Moreover
/an}bracketle{tb2,u1/an}bracketri}ht=1
2(1+1) = 2 =⇒b⊥
2:=b2−/an}bracketle{tb2,u1/an}bracketri}htu1=
1
1
1
1
,
so that
/bardblb2/bardbl= 2 and u2=1
2
1
1
1
1
.
Finally,
/an}bracketle{tb3,u1/an}bracketri}ht=1
2(1+1−1) =1
2and/an}bracketle{tb3,u2/an}bracketri}ht=1
2(1+1+1) =3
2
implies that
b⊥
3:=b3−/an}bracketle{tb3,u1/an}bracketri}htu1−/an}bracketle{tb3,u2/an}bracketri}htu2=
0
0
1
2
−1
2
.
Since
/bardblb⊥
3/bardbl=√
2
2=⇒u3:=√
2
2
0
0
1
−1
.
4.2. Reciprocal Basis
Letg:V×V→Rbe an inner product and B:={b1,...,b n}any basis of V.
FromgandBwe can define another basis of V, denoted by
Bg={b1,...,bn}
and satisfying
g(bi,bj) =δi
j. (4.7)
The basisBgis called the reciprocal basis ofVwith respect to gandB.
Note that, strictly speaking, we are very imprecise here. In fact, while it is
certainly possible to define a set of n= dimVvectors as in (4.7), we should justify
the fact that we call it a basis. This will be done in Claim 4.18.
Remark 4.16.In generalBg/ne}ationslash=Band in fact, because of Definition 4.4(5),
B=Bg⇐⇒Bis an orthonormal basis.
4.2. RECIPROCAL BASIS 47
Example 4.17.Letgbe the inner product in (4.6) in Example 4.10 and let Ethe
standard basis of R3. We want to find the reciprocal basis Eg, that is we want to
findEg:={b1,b2,b3}such that
g(bi,ej) =δi
j.
IfGis the matrix of the inner product in (4.5), using the matrix notation a nd
considering bjas a row vector and eias a column vector for i,j= 1,2,3,
/bracketleftbig––tbi––/bracketrightbig
G
|
ej
|
=δi
j.
Lettingiandjvary from 1 to 3, we obtain
––tb1––
––tb2––
––tb3––
1−1 0
−1 2−1
0−1 2
| | |
e1e2e3
| | |
=
1 0 0
0 1 0
0 0 1
,
from which we conclude that
––tb1––
––tb2––
––tb3––
=
| | |
e1e2e3
| | |
−1
1−1 0
−1 2−1
0−1 2
−1
=
| | |
e1e2e3
| | |
3 2 1
2 2 1
1 1 1
=
3 2 1
2 2 1
1 1 1
.
Hence
b1=
3
2
1
, b2=
2
2
1
, b3=
1
1
1
. (4.8)
Observethatinordertocompute G−1weusedtheGauss–Jordaneliminationmethod
1−1 01 0 0
−1 2−10 1 0
0−1 20 0 1
/squiggleleftright
1−1 01 0 0
0 1−11 1 0
0−1 20 0 1
/squiggleleftright
1 0−12 1 0
0 1−11 1 0
0 0 1 1 1 1
/squiggleleftright
1 0 0 3 2 1
0 1 0 2 2 1
0 0 1 1 1 1
48 4. INNER PRODUCTS
4.2.1. Properties of Reciprocal Bases.
Claim4.18.Givenavectorspace VwithabasisBandaninnerproduct g:V×V→
R, a reciprocal basis existsand isunique.
As we pointed out right after the definition of reciprocal basis, wha t this claim
really says is that there is a set of vectors {b1,...,bn}inVthat satisfy (4.7), that
form a basis and that this basis is unique.
Proof. LetB:={b1,...,b n}be the given basis. Any other basis {b1,...,bn}
is related toBby the relation
bi=Mijbj (4.9)
for some invertible matrixM. We want to show that there exists a uniquematrix
Msuch that, when (4.9) is plugged into g(bi,bj), we have
g(bi,bj) =δi
j. (4.10)
From (4.9) and (4.10) we obtain
δi
j=g(bi,bj) =g(Mikbk,bj) =Mikg(bk,bj) =Mikgkj,
which, in matrix notation becomes
I=MG,
whereGis the matrix of gwith respect toBwhose entries are gijas in (4.4). Since
Gis invertible because it is positive definite, then M=G−1exists and is unique. /square
Remark 4.19.Note that in the course of the proof we have found that, since M=
LBgB, then
G= (LBgB)−1=LBBg.
We denotewith gijtheentries of M=G−1. Fromtheabove discussion, it follows
that with this notation
gikgkj=δi
j (4.11)
as well as
bi=gijbj, (4.12)
or
/parenleftbigb1... bn/parenrightbig
=/parenleftbigb1... b n/parenrightbig
G−1. (4.13)
(check for example the dimensions and the indices to understand wh yG−1has to
be multiplied on the right). We can now compute g(bi,bj)
g(bi,bj)(4.12)=g(gikbk,gjℓbℓ) =gikgjℓg(bk,bℓ)
(4.1)=gjℓδi
ℓ(4.11)=gji=gij,
4.2. RECIPROCAL BASIS 49
wherewe usedinthesecondequality thebilinearity of gandinthelast itssymmetry.
Thus, similarly to (4.1), we have
gij=g(bi,bj). (4.14)
Given that we just proved that reciprocal basis are unique, we can talk about
thereciprocal basis (of a fixed vector space Vassociated to a basis and an inner
product).
Claim4.20.The reciprocal basis is contravariant .
Proof. LetBand/tildewideBbe two bases of VandL:=L/tildewideBBbe the corresponding
matrix of the change of bases, with Λ = L−1. Recall that this means that
˜bi=Lj
ibj.
We have to check that if Bg={b1,...,bn}is a reciprocal basis for B, then the
basis{˜b1,...,˜bn}defined by
˜bi= Λi
kbk(4.15)
is a reciprocal basis for /tildewideB. Then the assertion will be proven, since {˜b1,...,˜bn}is
contravariant by construction.
To check that{˜b1,...,˜bn}is the reciprocal basis, we need with check that with
the choice of ˜bias in (4.15), the property (4.1) of the reciprocal basis is verified,
namely that
g(˜bi,˜bj) =δi
j.
But in fact,
g(˜bi,˜bj)(4.15)=g(Λi
kbk,Lℓ
jbℓ) = Λi
kLℓ
jg(bk,bℓ)(4.10)= Λi
kLℓ
jδk
ℓ= Λi
kLk
j=δi
j,
wherethesecondequalitycomesfromthebilinearityof g,thethirdfromtheproperty
(4.7) defining reciprocal basis and the last from the fact that Λ = L−1./square
Suppose now that Vis a vector space with a basis Band thatBgis the reciprocal
basis ofVwith respect toBand to a fixed inner product g:V×V→R. Then
there are two ways of writing a vector v∈V, namely
v=vibi/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
with respect to B=vjbj
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
with respect to Bg.
Recall that the (ordinary) coordinates of vwith respect toBarecontravariant
(see Example 1.2).
Claim4.21.Vector coordinates with respect to the reciprocal basis are covariant .
50 4. INNER PRODUCTS
Proof. This will follow from the fact that the reciprocal basis is contravaria nt
and the idea of the proof is the same as in Claim 4.20.
Namely, letB,/tildewideBbe two bases of V,L:=L/tildewideBBthe matrix of the change of basis
and Λ =L−1. LetBgand/tildewideBgbe the corresponding reciprocal bases and v=vjbja
vector with respect to Bg.
It is enough to check that the numbers
˜vi:=Lj
ivj
are the coordinates of vwith respect to /tildewideBg, because in fact these coordinates are
covariant by definition. But in fact, using this and (4.15), we obtain
˜vi˜bi= (Lj
ivj)(Λi
kbk) =Lj
iΛi
k/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
δj
kvjbk=vjbj=v
/square
Definition 4.22.Thecoordinates viofavectorv∈Vwithrespect tothereciprocal
basisBgare called the covariant coordinates ofv.
4.2.2. Change of basis from a basis Bto its reciprocal basis Bg.Wewant
to look now at the direct relationship between the covariant and the contravariant
coordinates of a vector v. Recall that we can write
vibi/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
with respect to B=v=vjbj
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
with respect to Bg.
from which we obtain
(vigij)bj=vi(gijbj) =vibi=v=vjbj,
and hence
vj=vigijort[v]Bg=G[v]B,. (4.16)
Likewise, from
vibi=v=vjbj=vj(gjibi) = (vjgji)bi
it follows that
vi=vjgjior [v]B=G−1t[v]Bg. (4.17)
Example 4.23.LetB={e1,e2,e3}be the standard basis of R3and let
G=
1−1 0
−1 2−1
0−1 2
4.2. RECIPROCAL BASIS 51
be the matrix of gwith respect toB. In (4.8) Example 4.17 we saw that
Bg=/braceleftigg
b1=
3
2
1
,b2=
2
2
1
,b3=
1
1
1
/bracerightigg
is the reciprocal basis. We find the covariant coordinates of v=
4
5
6
with respect
toBgusing (4.16), namely
[v]Bg=G[v]B=
1−1 0
−1 2−1
0−1 2
4
5
6
=/parenleftbig−1 0 7/parenrightbig
.
In fact,
vibi= (−1)
3
2
1
+0
2
2
1
+7
1
1
1
=
4
5
6
.
/square
Example 4.24.LetV:=R[x]1be the vector space of polynomials of degree ≤1
(that is “linear” polynomial, or of the form a+bx). Letg:V×V→Rbe defined
by
g(p,q) :=/integraldisplay1
0p(x)q(x)dx,
and letB:={1,x}be a basis of V. Determine:
(1) the matrix G;
(2) the matrix G−1;
(3) the reciprocal basis Bg;
(4) the contravariant coordinates of p(x) = 6x(that is the coordinates of p(x)
with respect toB);
(5) the covariant coordinates of p(x) = 6x(that is the coordinates of p(x) with
respect toB).
52 4. INNER PRODUCTS
(1) The matrix Ghas entries gij=g(bi,bi), that is
g11=g(b1,b1) =/integraldisplay1
0(b1)2dx=/integraldisplay1
0dx= 1
g12=g(b1,b2) =/integraldisplay1
0b1b2dx=/integraldisplay1
0x=1
2
g21=g(b2,b1) =/integraldisplay1
0b2b1dx=1
2
g22=g(b2,b2) =/integraldisplay1
0(b2)2dx=/integraldisplay1
0x2dx=1
3,
so that
G=/parenleftbigg
11
21
21
3/parenrightbigg
.
(2) Since det G= 1·1
3−1
2·1
2=1
12, then by using (2.7), we get
G−1=/parenleftbigg
4−6
−6 12/parenrightbigg
,.
(3) Using (4.13), we obtain that
/parenleftbigb1b2/parenrightbig
=/parenleftbig1x/parenrightbig
G−1=/parenleftbig1x/parenrightbig/parenleftbigg
4−6
−6 12/parenrightbigg
=/parenleftbig4−6x−6+12x/parenrightbig
,
so thatBg={4−6x,−6+12x}.
(4)p(x) = 6x= 0·1+6·x, so thatp(x) hascontravariant coordinates[ p(x)]B=/parenleftbigg
0
6/parenrightbigg
.
(5) From (4.16) it follows that if v=p(x), then
/parenleftbigg
v1
v2/parenrightbigg
=G/parenleftbigg
v1
v2/parenrightbigg
=/parenleftbigg
11
21
21
3/parenrightbigg/parenleftbigg
0
6/parenrightbigg
=/parenleftbigg
3
2/parenrightbigg
.
In fact, one can easily check that
v1b1+v2b2= 3·(4−6x)+2·(−6+12x) = 6x.
4.2.3. Isomorphisms Between a Vector Space and its Dual. We saw
already in Proposition 3.8 that If Vis a vector space and V∗is its dual, then
dimV= dimV∗. In particular this means that VandV∗can be identified, once we
choose a basisBofVand a basisB∗ofV∗. In fact, the basis B∗ofV∗is given once
we choose the basis BofV, as the dual basis of V∗with respect toB. Then there
is the following correspondence:
v∈V/squiggleleftrightα∈V∗,
4.2. RECIPROCAL BASIS 53
Table 1. Summary of covariance and contravariance of vector coordinate s
B:={b1,...,b n} | Bg={b1,...,bn}
basis | reciprocal basis
they are related by g(bi,bj) =δi
j
v=vibi| v=vibi
contravariant | covariant
coordinates | coordinates
the matrices of gare gij=g(bi,bj)|gij=g(bi,bj)
the matrices are inverse
of each other, that is gikgkj=δi
j
the relation between the basis |
and the reciprocal basis is bi=gijbj| bi=gijbj
the relation between covariant |
coordinates and contravariant vi=gijvj| vi=gijvj
coordinates is |
exactly when vandαhave the same coordinates, respectively with respect to Band
B∗, However this correspondence depends on the choice of the basis Band hence
not canonical .
If however Vis endowed with an inner product, then there is a canonical
identification ofVwithV∗(that is an identification that does not depend on
the basisBovV). In fact, let g:V×V→Rbe an inner product and let v∈V.
Then
g(v,·) :V−→R
w/ma√sto−→g(v,w)
is a linear form and hence we have the following canonical identification given by
the metric
V←→V∗
v←→v∗:=g(v,·).(4.18)
Note that the isomorphism sends the zero vector to the linear form identically equal
to zero, since g(v,·)≡0 if and only if v= 0, sincegis positive definite.
So far, we have two bases of the vector space V, namely the basis Band the
reciprocal basisBgand we have also the dual basis of the dual vector space V∗. In
fact, under the isomorphism (4.18), the reciprocal basis of Vand the dual basis of
V∗correspond to each other. This is easily seen because, under the is omorphism
(4.18) an element of the reciprocal basis bicorrespond to the linear form g(bi,·)
bi←→g(bi,·)
54 4. INNER PRODUCTS
and the linear form g(bi,·) :V→Rhas the property that
g(bi,bj) =δi
j.
Hence
g(bi,·)≡βi,
and under the canonical identification between VandV∗the reciprocal basis of V
corresponds to the dual basis of V∗.
4.2.4. Geometric Interpretation. Letg:V×V→Rbe an inner product
andB:={b1,...,b n}a basis ofV. The orthogonal projection of a vector ∈Vonto
bkis defined as
projbkv=g(v,bk)
g(bk,bk)bk. (4.19)
bkv
projbkv
In fact, projbkvis obviously parallel to bkand the following exercises shows that the
component v−projbkvis orthogonal to bk.
Exercise 4.25.With projbkvdefined as in , we have
v−projbkv⊥bk,
where the orthogonality is meant with respect to the inner product g.
Now letv=vibi∈Vbe a vector written in terms of its covariant coordinates
(that is the coordinates with respect to the reciprocal basis). Th en
g(v,bk) =g(vibi,bk) =vig(bi,bk)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
δi
k=vk,
so that (4.19) becomes
projbkv=vk
g(bk,bk)bk.
If we assume that the elements of the basis B:={b1,...,b n}are unit vectors,
then (4.19) further simplifies to give
projbkv=vkbk. (4.20)
This equation shows the following:
Fact4.26.The covariant coordinates of vgive the orthogonal projection of vonto
b1,...,b n.
Likewise, the following holds basically by definition:
4.2. RECIPROCAL BASIS 55
Fact4.27.The contravariant coordinates of vgive the “parallel” projection of v
ontob1,...,b n.
b1b2
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
/bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipuprightv2v2
v1
v1v
CHAPTER 5
Tensors
5.1. Generalities
LetVbe a vector space. Up to now we saw several objects related to V, that we
said were “tensors”. We summarize them in Table 1. So, we seem to ha ve a good
candidate for the definition of a tensor of type (0 ,q) for allq∈N, but we cannot
say the same for a tensor of type ( p,0) for allp∈N. The next discussion will lead
us to that point, and in the meantime we will discuss an important point .
5.1.1. Canonical isomorphism between Vand(V∗)∗.We saw in§4.2.3
that any vector space is isomorphic to its dual, but the the isomorph ism is not
canonical (that is, it depends on the choice of basis). We also saw th at if there is
an inner product on V, then there is a canonical isomorphism. The point of this
section is to show that, even without an inner product, there is a alw ays a canonical
isomorphism between Vand itsbidual(V∗)∗, that is the dual of its dual.
To see this, let us observe first of all that
dimV= dim(V∗)∗. (5.1)
If fact, for any vector space W, we saw in Proposition 3.8 that dim W= dimW∗. If
we apply this equality both to W=Vand toW=V∗, we obtain
dimV= dimV∗and dim V∗= dim(V∗)∗,
from which (5.1) follows immediately. From (4.2.3) we deduce immediately thatV
and(V∗)∗areisomorphic, andweonlyhavetoseethattheisomorphism iscanon ical.
Table 1. Covariance and Contravariance
Tensor Components Behavior under a change of basis Type
vectors inV vicontravariant tensor (1,0)
linear forms V→R αj covariant tensor (0,1)
linear transformations V→VAi
jmixed:/braceleftigg
contravariant
covarianttensor (1,1)
bilinear forms1V×V→RBij covariant 2-tensor (0,2)
k-linear forms V×···×V/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
k→RFi1i2...ik covariantk-tensor (0,k)
57
58 5. TENSORS
To this end, observe that a vector v∈Vgives rise to a linear form on V∗defined
by
ϕv:V∗−→R
α/ma√sto−→α(v).
Then we can define a linear map as follows:
Φ :V−→(V∗)∗
v/ma√sto→ϕv{α/ma√sto→α(v)}(5.2)
Since for any linear map T:W→Wbetween vector spaces
dimW= dimRange( T)+dimker( T),
it will be enough to show that kerΦ = {0}, because then
dim(V∗)∗= dimV= dimRange(Φ) ,
that is Φ is an isomorphism. Notice that the important fact is that we h avenot
chosen a basis to define the isomorphism Φ.
To see that kerΦ = {0}, observe that the kernel consists of all vectors v∈V
such thatα(v) = 0 for all α∈V∗. We want to see that the only vector v∈Vfor
which this happens is the zero vector. In fact, if 0 /ne}ationslash=v∈VandB:={b1,...,b n}
isanybasis ofV, then we can write v=vibi, where at least one vj/ne}ationslash= 0. But
then, ifB∗={β1,...,β n}is the dual basis, 0 /ne}ationslash=vj=βj(v). Notice again that the
dimension of the kernel of a linear map is invariant under a change of b asis, and
hence kerΦ ={0}no matter what the basis Bhere was.
We record this fact as follows:
Fact5.1.LetVbe a vector space and V∗its dual. The dual (V∗)∗ofV∗is
canonically isomorphic to V.
5.1.2. Towards general tensors. Recall that
V∗:={α:V→R:αis a linear form}={(0,1)-tensors}.
Applying this formula to the vector space V∗we obtain
(V∗)∗:={α:V∗→R:αis a linear form}.
Using the isomorphism ( V∗)∗∼=Vand the fact that coordinate vectors are con-
travariant, we conclude that
{α:V∗→R:αis a linear form}= (V∗)∗=V={(1,0)-tensors}.
So changing the vector space from Vto its dualV∗seems to have had the effect
to transform a covariant tensor of type (0 ,1) into a contravariant one of type ( 1,0).
We are going to apply this procedure to try to transform a covarian t of type (0 ,2)
into a contravariant one of type ( 2,0).
Recall that
{ϕ:V×V→R: bilinear}={(0,2)-tensors}.
5.1. GENERALITIES 59
and consider
{ϕ:V∗×V∗→R: bilinear}.
We can hence give the following definition:
Definition 5.2.Atensor of type (2,0) is a bilinear form on V∗, that is a bilinear
functionσ:V∗×V∗→R.
Let
Bil(V∗×V∗,R) ={σ:V∗×V∗→R:σis a bilinear form}={(2,0)-tensors}.
Exercise 5.3.Check that Bil( V∗×V∗,R) is a vector space, that is that if σ,τ∈
Bil(V∗×V∗,R) andλ,µ∈R, thenλσ+µτ∈Bil(V∗×V∗,R).
5.1.3. Tensor product of (1,0)-tensors on V∗.Ifv,w∈Vare two vectors
(that is two (1 ,0)-tensors), we define
σv,w:V∗×V∗→R
by
σv,w(α,β) :=α(v)β(w).
Sinceαandβare two linear forms, then σ
σv,w=:v⊗w
isbilinearand called the tensor product ofvandw. Henceσis a (2,0)-tensor.
Note5.4.In general
v⊗w/ne}ationslash=w⊗v,
as there can be linear forms α,βsuch thatα(v)β(w)/ne}ationslash=α(w)β(v).
Similar to what we saw in §3.2.2, we can define a basis for the space of (2 ,0)-
tensors by considering the (2 ,0)-tensors defined by bi⊗bj, whereB:={b1,...,b n}
is a basis of V.
Proposition 5.5.The elements bj⊗bj,i,j= 1,...,nform a basis of Bil(V∗×
V∗,R). ThusdimBil(V∗×V∗,R) =n2.
We will not prove the proposition here, as the proof is be completely a nalogous
to the one of Proposition 3.23.
Notation. We write
Bil(V∗×V∗,R) =V⊗V.
60 5. TENSORS
5.1.4. Components of a (2,0)-tensor and their contravariance. Letσ:
V∗×V∗→Rbe a bilinear form on V∗, that is a (2 ,0)-tensor, as we just saw. We
want to verify that it behaves as we expect with respect to a chang e of basis. After
choosing a basisB:={b1,...,b n}ofV, we have the dual basis B∗={β1,...,βn}
ofV∗and the basis{bi⊗bj:i,j= 1,...,n}of the space of (2 ,0)-tensors.
The (2,0)-tensorσis represented by its components
Sij=σ(βi,βj),
that is
σ=Sijbj⊗bj,
and the components Sijcan be arranged into a matrix
S=
S11... S1n
.........
Sn1... Snn
called the matrix of the (2,0)-tensor with respect to the chosen basis of V.
We look now at how the components of a (2 ,0)-tensor change with a change
of basis. LetB:={b1,...,b n}and/tildewideB:={˜b1,...,˜bn}be two basis of Vand let
B∗:={β1,...,βn}and/tildewideB∗:={˜β1,...,˜βn}be the corresponding dual basis of V∗.
Letσ:V∗×V∗→Rbe a (2,0)-tensor with components
Sij=σ(βi,βj) and/tildewideSij=σ(˜βi,˜βj)
with respect toB∗and/tildewideB∗respectively. Let L:=L/tildewideBBbe the matrix of the change
of basis fromBto/tildewideB, and let Λ = L−1. Then, as seen in (1.2) and (3.11) we have
that
˜bj=Li
jbiand˜βi= Λi
jβj.
It follows that
/tildewideSij=σ(˜βi,˜βj) =σ(Λi
kβk,Λj
ℓβℓ) = Λi
kΛj
ℓσ(βk,βℓ) = Λi
kΛj
ℓSkℓ,
where the first and the last equality follow from the definition of /tildewideSijand ofSkℓ
respectively, the second from the change of bases and the third f rom the bilinearity
ofσ. We conclude that
/tildewideSij= Λi
kΛj
ℓSkl. (5.3)
Henceσis acontravariant 2-tensor..
Exercise 5.6.Verify that in terms of matrices (5.3) translates into
/tildewideS=tΛSΛ.
(Compare with (3.15).)
5.3. TENSOR PRODUCT 61
5.2. Tensors of type (p,q)
Definition 5.7.Atensor of type (p,q) or (p,q)-tensoris a multilinear form
T:V∗×...V∗
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
p×V×···×V/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
q−→R.
LetTbe a (p,q)-tensor,B:={b1,...,b n}a basis ofVandB∗={β1,...,β n}the
corresponding dual basis of V∗. The components of Twith respect to these bases
are
Ti1,...,ip
j1,...,jq=T(βi1,...,βip,bj1,...,b jq).
If moreover/tildewideB:={˜b1,...,˜bn}is another basis, /tildewideB∗={˜β1,...,˜βn}is the coresponding
dual basis of V∗andL:=L/tildewideBBis the matrix of the change of basis with inverse
Λ :=L−1, then the components of Twith respect to these new bases are
/tildewideTi1,...,ip
j1,...,jq= Λi1
k1...Λip
kpLℓ1
j1...Lℓq
jqTk1,...,lp
ℓ1,...,ℓq.
5.3. Tensor product
We saw already in §3.2.2 and§3.5 the tensor product of two multilinear forms.
Since multilinear forms are covariant tensors, we said that this corr esponds to the
tensor product of two covariant tensors. More generally, we can define the tensor
product of any two tensors as follows:
Definition 5.8.Let
T:V∗×...V∗
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
p×V×···×V/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
q−→R
be a (p,q)-tensor and
U:V∗×...V∗
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
k×V×···×V/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
ℓ−→R
a (k,ℓ) tensor. The tensor product T⊗UofTandUis a (p+k,q+ℓ)-tensor
T⊗U:V∗×...V∗
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
p+k×V×···×V/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
q+ℓ−→R
defined by
(T⊗U)(α1,...,α p+k,v1,...,v q+ℓ) :=
T(α1,...,α p,v1,...,v q)U(αp+1,...,α p+k,vq+1,...,v q+ℓ).
Note that both T⊗UandU⊗Tare tensors of the same type ( p+k,q+ℓ), but
in general
T⊗U/ne}ationslash=U⊗T.
62 5. TENSORS
The set of all tensors of type ( p,q) on a vector space Vis denoted by
Tp
q(V) :={all (p,q)-tensors on V}.
Analogously to how we proceeded in the case of (0 ,2)-tensors, we compute the
dimension ofTp
q(V). IfB:={b1,...,b n}is a basis of VandB∗:={β1,...,βn}is
the corresponding dual basis of V∗. Just like we saw in Proposition 5.5 in the case
of (0,2)-tensors, a basis of Tp
q(V) is
{bi1⊗bi2⊗···⊗bip⊗βj1⊗βj2⊗···⊗βjq: 1≤i1,...,i p≤n,1≤j1,...,j q≤n}.
Since there are n×···×n/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
p×n×···×n/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
q=np+qelements in this basis (correspond-
ing to the possible choices of bikandβjℓ), we deduce that
dimTp
q(V) =np+q.
We now define the tensor product of two vector spaces:
Definition 5.9.LetVandWbetwofinitedimensionalvectorspaces, withdim V=
nand dimW=m. Choose{b1,...,b n}a basis ofVand{a1,...,a m}a basis ofW.
Then the tensor product V⊗WofVandWis an (n·m)-dimensional vector
space with basis
{bi⊗aj: 1≤i≤n,1≤j≤m}.
We remark that there is no reason to restrict oneself to the tenso r product of
only two factors. One can equally define the tensor product V1⊗···⊗Vk, and obtain
a vector space of dimension dim V1×···×dimVk.
There is hence an identification
Tp
q(V)∼=V⊗···⊗V/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
p⊗V∗⊗···⊗V∗
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
q,
that follows from the fact that both spaces have the same basis. T he following
proposition gives other useful identifications.
Proposition 5.10.LetVandWbe two finite dimensional vector spaces, with
dimV=nanddimW=mand let us denote by
Lin(V,W∗) :={linear maps V→W∗}.
Then
Bil(V×W,R)∼=Lin(V,W∗)
∼=Lin(W,V∗)
∼=V∗⊗W∗
∼=(V⊗W)∗
∼=Lin(V⊗W,R).
5.3. TENSOR PRODUCT 63
Proof. Here is the idea behind this chain of identifications. Let f∈Bil(V×
W,R), that is a bilinear function f:V×W→R. This means that ftakes two
vectors,v∈Vandw∈W, as input and gives back a real number f(v,w)∈R. If
on the other hand we only feed fone vectorv∈V, then there is a remaining spot
waiting for a vector w∈Wto produce a real number. Since fis linear on Vand
onW, the mapf(v,·) :W→Ris a linear form, so f(v,·)∈W∗. In other words,
we can view f∈Lin(V,W∗). It follows that there is a linear map
Bil(V×W,R)−→Lin(V,W∗)
f/ma√sto−→Tf,
where
Tf(v)(w) :=f(v,w).
Conversely, any T∈Lin(V,W∗) can be identified with a bilinear map fT∈Bil(V×
W,R) defined by
fT(v,w) :=T(v)(w).
SincefTf=fandTfT=T, we have proven thefirst identification intheproposition.
Analogously, iftheinputisonlyavector w∈W, thenf(·,w) :V→Risalinear
map and hence f∈Bil(V×W,R) defines a linear map Tf∈Lin(W,V∗). The same
reasoning as in the previous paragraph, shows that Bil( V×W,R)∼=Lin(W,V∗).
To proceed with the identifications, observe that, because of our definition of
V∗⊗W∗, we have
Bil(V×W,R)∼=V∗⊗W∗,
since these spaces both have basis2
{βi⊗αj: 1≤i≤n,1≤j≤m},
where{b1,...,b n}is a basis of Vwith corresponding dual basis {β1,...,βn}ofV∗,
and{a1,...,a n}is a basis of Wwith corresponding dual basis {α1,...,αn}ofW∗.
Finally, an element Dijβi⊗αj∈V∗⊗W∗may be viewed as a linear map
V⊗W→R, that is as an element of ( V⊗W)∗by
V⊗W−→ R
Ckℓbk⊗aℓ/ma√sto−→DijCkℓβi(bk)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
δi
kαj(aℓ)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
δj
ℓ=DijCkℓ.
/square
Because of the identification Bil( V×W,R)∼=Lin(V⊗W,R), sometimes one says
that “the tensor product linearizes what was bilinear or multilinear”.
2There is a way of defining the tensor product of vector spaces with out involving bases, but
we will not do it here.
CHAPTER 6
Applications
6.1. Inertia tensor
6.1.1. Moment of inertia with respect to the axis determined by the
angular velocity. LetMbe a rigid body fixed at a point O. The motion of this
rigid body at time tis byrotationby an angle θwithangular velocity ωabout some
axis through O. The angular velocity has magnitude |·|or/bardbl·/bardbl
|ω|=/vextendsingle/vextendsingle/vextendsingle/vextendsingledθ
dt/vextendsingle/vextendsingle/vextendsingle/vextendsingle,
direction given by the axis of rotation and orientation by the right-h and rule. The
position vector of a pointPin the body Mrelative to the origin Ois
x=−→OP
while the linear velocity of a pointPis
v=ω×x.
The linear velocity vhas magnitude
|v|=|ω|/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
|dθ
dt||x|sinα/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
r
and direction tangent at Pto the circle of radius rperpendicular to the axis of
rotation.
65
66 6. APPLICATIONS
..α
rO
Px
vω
Thekinetic energy of an infinitesimal region dMofMaroundPis
dE=1
2v2dm,
wherev2=v•vanddmis the mass of dM. Thetotal kinetic energy ofMis
E=1
2/integraldisplay
Mv2dm=1
2/integraldisplay
M(ω×x)2dm.
Note that, depending on the type of the rigid body, we might take he re the sum
of the integral, and what type of integral depends again from the k ind of rigid body
we have. More precisely:
(1) IfMis a solid in 3-dimensional space, then
E=1
2/integraldisplay/integraldisplay/integraldisplay
M(ωP×xP)2ρPdx1dx2dx3,
where (ω×xP)2ρPis a function of P(x1,x2,x3).
(2) IfMis a flat sheet in 3-dimensional space, then
E=1
2/integraldisplay/integraldisplay
M(ωP×xP)2ρPdx1dx2.
(3) IfMis a surface in 3-dimensional space, then
E=1
2/integraldisplay/integraldisplay/integraldisplay
M(ωP×xP)2ρPdσ,
wheredσis the infinitesimal element of the surface for a surface integral.
6.1. INERTIA TENSOR 67
(4) IfMis a wire in 3-dimensional space, then
E=1
2/integraldisplay
M(ωP×xP)2ρPds,
wheredsis the infinitesimal element of length for a line integral.
(5) IfMis a finite set of point masses with rigid relative positions, then
E=1
2N/summationdisplay
i=1(ωP×xi)2mi.
In any case we need to work out the quantity
(ω×x)2
for vectors ωandxiin 3-dimensional space.
To this purpose we use the Lagrange identity1, according to which
(a×b)·(c×d) = det/bracketleftbigg
a·c a·d
b·c b·d/bracketrightbigg
. (6.1)
Applying (6.1) with a=c=ωandb=d=x, we obtain
(ω×x)2= (ω×x)·(ω×x) = det/bracketleftbigg
ω·ω ω·x
x·ωx·x/bracketrightbigg
=ω2x2−(ω·x)2.
Let nowB={e1,e2,e3}be an orthonormal2basis ofR, so that
ω=ωieiandx=xiei.
Then
ω2=ω·ω=δijωiωj=ω1ω1+ω2ω2+ω3ω3
x2=x·x=δkℓxkxℓ=x1x1+x2x2+x3x3
ω·x=δikωixk
so that
(ω×x)2=ω2x2−(ω·x)2
= (δijωiωj)(δkℓxkxℓ)−(δikωixk)(δjlωjxℓ)
= (δijδkℓ−δikδjℓ)ωiωjxkxℓ.
Therefore the total kinetic energy is
E=1
2(δijδkℓ−δikδjℓ)ωiωj/integraldisplay
Mxkxℓdm
and it depends only on ω1,ω2,ω3(since we have integrated over the x1,x2,x3).
1The Lagrange identity can easily be proven in coordinates.
2We could use any basis of R3. Then, instead of the δij, the formula would have involved
the metric tensor gij. However computations with orthonormal bases are simpler. In ad dition
in this case we will see that the metric tensor is symmetric, and hence it admits an orthonormal
eigenbasis.
68 6. APPLICATIONS
Definition 6.1.Theinertia tensor is the tensor whose components with respect
to an orthonormal basis Bare
Iij= (δijδkℓ−δikδjℓ)/integraldisplay
Mxkxℓdm.
Then the kinetic energy of the rotating rigid body is
E=1
2Iijωiωj
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
Einstein
notation=1
2ω·Iω
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
matrix
notation.
One can check that
I11=/integraldisplay
M(x2x2+x3x3)dm
I22=/integraldisplay
M(x1x1+x3x3)dm
I33=/integraldisplay
M(x1x1+x2x2)dm
I23=I32=−/integraldisplay
Mx2x3dm
I31=I13=−/integraldisplay
Mx1x3dm
I12=I21=−/integraldisplay
Mx1x2dm,
sothatwithrespect tothebasis B, themetrictensorisrepresented bythesymmetric
matrix
I=
I11I12I13
I21I22I23
I31I32I33
.
We check only the formula for I11. In fact,
I11= (δ11δkℓ/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
=0
unless
k=ℓ−δ1kδ1ℓ/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
=0
unless
k=ℓ=1)/integraldisplay
Mxkxℓdm
Ifk=ℓ= 1, thenδ11δ11−δ11δ11= 0, sothatthenon-vanishingtermshave k=ℓ/ne}ationslash= 1.
SoI11,I22,I33are themoments of inertia of the rigid body Mwith respect to
the axesOx1,Ox2,Ox3respectively; I12,I23,I31are thepolar moments of inertia
or theproducts of inertia of the rigid body M.
Example 6.2.Findtheinertia tensor ofahomogeneous rectangular platewithside s
aandband total mass m, assuming that the center of rotation Ocoincides with
6.1. INERTIA TENSOR 69
the center of inertia. We choose an orthonormal basis with e1aligned with the side
of lengtha,e2aligned with the side of length bande3perpendicular to the plate.
.e1e2
ab
Since the plate is assumed to be homogeneous, it has a constant mass density
equal to
ρ=total mass
area=m
ab.
Denote byx,yandzthe coordinates. Then
I11/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
Ixx=/integraldisplaya
2
−a
2/integraldisplayb
2
−b
2(y2+z2/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
=0)ρ/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
m
abdydx
=m
aba/integraldisplayb
2
−b
2y2dy
=m
b/bracketleftbiggy3
3/bracketrightbiggb
2
−b
2
=m
12b2.
Similarly
I22/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
Iyy=m
12a2,
and
I33/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
Izz=/integraldisplaya
2
−a
2/integraldisplayb
2
−b
2(x2+y2)ρdydx=m
12(a2+b2),
which turns out to be just the sum of I11andI22.
70 6. APPLICATIONS
Furthermore,
I23=I32=−/integraldisplaya
2
−a
2/integraldisplayb
2
−b
2y z/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
=0ρdydx= 0,
and similarly I31=I13= 0. Finally
I21=I21=−/integraldisplaya
2
−a
2/integraldisplayb
2
−b
2xyρdydx =−m
ab/parenleftigg/integraldisplaya
2
−a
2xdx/parenrightigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
=0/parenleftigg/integraldisplayb
2
−b
2ydy/parenrightigg
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
=0/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
because the integral of an odd function
on a symmetric interval is 0.
We conclude that the inertia tensor is given by the matrix
m
12
b20 0
0a20
0 0a2+b2
./square
Exercise 6.3.Compute the inertia tensor of the same plate but now with center
of rotation Ocoinciding with a vertex of the rectangular plate.
6.1.2. Moment of inertia about any axis through the fixed poin t.We
compute the moment of inertia of the body Mabout an axis through O. Letpbe
a unit vector defining an axis through O.
..α
rOPxp
Themoment of inertia of an infinitesimal region of MaroundPis
dI=r2/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
ris the distance
fromPto the axisdm/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
infinitesimal
mass=/bardblp×x/bardbl2dm,
6.1. INERTIA TENSOR 71
where the last equality follows from the fact that /bardblp×x/bardbl=/bardblp/bardbl/bardblx/bardblsinα=r, since
pis a unit vector. Hence the total moment of inertia ofMwith respect to the
axis given by pis
I=/integraldisplay
M/bardblp×x/bardbl2dm≥0.
(This is very similar to the total kinetic energy E: just replace ωbypand omit the
factor1
2.) By the earlier computations, we conclude that
I=Iijpipj,
whereIijis the inertia tensor. This formula shows that the total moment of inertia
of the rigid body Mwith respect to an arbitrary axis passing through the point Ois
determined only by the inertia tensor of the rigid body.
Example 6.4.For the rectangular plate in Example 6.2, compute the moment of
inertia with respect to the diagonal of the plate.
.O
ab
Choosep=1√
a2+b2(ae1+be2) (the other possibility is the negative of this vector).
So
p1=a√
a2+b2, p2=b√
a2+b2p3= 0.
The moment of inertia is
I=Iijpipj
=/parenleftig
a√
a2+b2b√
a2+b20/parenrightig
m
12b20 0
0m
12a20
0 0m
12(a2+b2)
a√
a2+b2
b√
a2+b2
0
=m
6a2b2
a2+b2./square
72 6. APPLICATIONS
6.1.3. Moment of inertia with respect to an eigenbasis of the inertia
tensor. Observe that the inertia tensor is symmetric and recall the spectral the-
oremfor symmetric matrices that we saw in Theorem 4.8. Let (˜ e1,˜e2,˜e3) be an
orthonormal basis for the inertia tensor of a rigid body M. LetI1,I2,I3be the
corresponding eigenvalues of the inertia tensor. The matrix repre senting the inertia
tensor with respect to this eigenbasis is
I10 0
0I20
0 0I3
.
The axes of the eigenvectors ˜ e1,˜e2,˜e3are called the principal axes of inertia
of the rigid body M. The eigenvalues Iiare called the principal moments of
inertia.
Theprincipal moments of inertia are the moments if inertia with respect to the
principal axes of inertia , hence they are non-negative
I1, ,I2, ,I3≥0.
A rigid body is called
(1) anasymmetric top ifI1/ne}ationslash=I2/ne}ationslash=I3/ne}ationslash=I1;
(2) asymmetric top ifI1=I2/ne}ationslash=I3: any axis passing through the plane
determined by e1ande2is a principal axis of inertia;
(3) aspherical top ifI1=I2=I3: any axis passing through Ois a principal
axis of inertia.
With respect to the eigenbasis {˜e1,˜e2,˜e3}thekinetic energy is
E=1
2(I1(˜ω1)2+I2(˜ω2)2+I3(˜ω3)2),
whereω= ˜ωi˜ei, with ˜ωithe components of the angular velocity with respect to the
basis{˜e1,˜e2,˜e3}.
The surface determined by the equation (with respect to the coor dinatesx,y,z)
I1x2+I2y2+I3z2= 1
is called the ellipsoid of inertia . The symmetry axes of the ellipsoid coincide with
the principal axes of inertia. Note that for a spherical top, the ellip soid of inertia is
actually a sphere.
The ellipsoid of inertia gives the moment of inertia with respect to any a xis
as follows: Consider an axis given by the unit factor pand letqbe a vector of
intersection of the axis with the ellipsoid of inertia.
q=cp
wherecis the (signed) distance to Oof the intersection of the axis with the ellipsoid
of inertia.
6.1. INERTIA TENSOR 73
Oqpaxis
The moment of inertia with respect to this axis if
I=Iijpipj=1
c2Iijqiqj=1
c2,
where the last equality follows from the fact that, since qis on the ellipsoid, then
Iijqiqj= 1.
6.1.4. Angular momentum. LetMbe a body rotating with angular velocity
ωabout an axis through the point O. Letx=−→OPbe the position vector of a point
Pandv=ω×xthe linear velocity of P.
..αO
Px
vω
74 6. APPLICATIONS
Then the angular momentum of an infinitesimal region of MaroundPis
dL= (x×v)dm,
so that the total angular momentum ofMis
L=/integraldisplay
M(x×(ω×x))dm.
We need to work out x×(ω×x) for vectors xandωin three dimensional space.
It is easy to prove, using coordinates3, the equality
x×(ω×x) =ω(x•x)−x(ω•x). (6.2)
LetB={e1,e2,e3}be an orthonormal basis of R3. Then, replacing the following
equalities
ω=ωiei=δi
jωjei (6.3)
x=xiei=δi
kxkei (6.4)
x•x=δkℓxkxℓ(6.5)
ω•x=δjℓωjxℓ(6.6)
into (6.2), we obtain
x×(ω×x) =δi
jωjei/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
(6.3)(δkℓxkxℓ
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
(6.5))−δi
kxkei/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
(6.4)(δjℓωjxℓ
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
(6.6)) = (δi
jδkℓ−δi
kδjℓ)ωjxkxℓei.
Therefore the total angular momentum is
L=Liei,
where the components Liare
Li= (δi
jδkℓ−δi
kδjℓ)ωj/integraldisplay
Mxkxℓdm.
Since we restrict to orthonormal bases, we have always
δi
j=δij=δij=/braceleftigg
1 ifi=j
0 ifi/ne}ationslash=j.
Hence the above expression for Lican be written in terms of the inertia tensor Iij
as
Li=Iijωj.
3Consider only the case in which xis a basis vector and use the linearity in ω.
6.2. STRESS TENSOR (SPANNUNG) 75
Example 6.5.Suppose the rectangular plate in the previous examples is rotating
about an axis through Owith angular velocity
ω=e1+2e2+3e3,or
ω1
ω2
ω3
=
1
2
3
.
Compute its angular momentum.
The inertia tensor is given by the matrix Iij
m
12b20 0
0m
12a20
0 0m
12(a2+b2)
.
The total angular momentum has components given by
m
12b20 0
0m
12a20
0 0m
12(a2+b2)
1
2
3
=
m
12b2
m
6a2
m
4(a2+b2)
=
L1
L2
L3
,
so that
L=m
12b2e1+m
6a2e2+m
4(a2+b2)e3.
6.2. Stress tensor (Spannung)
It was the concept of stressin mechanics that originally led to the invention of
tensors
tenseur
/d38/d38▲▲▲▲▲▲▲▲▲▲
stress/d57/d57rrrrrrrrrr
/d37/d37▲▲▲▲▲▲▲▲▲▲ tensor
tension/d56/d56rrrrrrrrrr
Let us consider a rigid body Macted upon by external forces but in static
equilibrium , and let us consider an infinitesimal region dMaround a point P. There
are two types of external forces:
(1) The body forces , that is forces whose magnitude is proportional to the
volume/mass of the region. For instance, gravity,attractive force or the
centrifugal force .
(2) Thesurface forces , that is forces exerted on the surface of the element by
the material surrounding it. They are forces whose magnitude is pr opor-
tional to the area of the region in consideration.
76 6. APPLICATIONS
Thesurface force per unit area is called the stress. We will concentrate on homo-
geneous stress , that is stress that does not depend on the location of the element
in the body, but depends only on the orientation of the surface/pla ne. Moreover,
we assume that the body in consideration is in static equilibrium .
Choose an orthonormal basis {e1,e2,e3}and the plane πthroughPparallel to
thee2e3coordinate plane. The normal to the plane is the vector e1. Let ∆A1be
the area of the slice of the infinitesimal region around Pcut by the plane and let
∆Fbe the force acting on that slice. We write ∆ Fin terms of its components
∆F= ∆F1e1+∆F2ee+∆F3e3
and, since we defined the stress to be the surface force per unit a rea, we define, for
j= 1,2,3,
σ1j:= lim
∆A1→0∆Fj
∆A1.
Similarly we can consider planes parallel to the other coordinate plane s
e1e2e3
PPPπ1π2
π3
and define
σij:= lim
∆Ai→0∆Fj
∆Ai.
It turns out that the resulting nine numbers σijform a contravariant 2-tensor called
thestress tensor . To see this, we compute the stress tensor across otherslices
throughP, that is other planes with other normal vectors. Let Π be a plane pa ssing
throughP,nthe unit vector through Pperpendicular to the plane π, ∆s=π∩dM
the area of a small element of the plane Π containing Pand ∆Fthe force acting on
that element.
Π∆sn
P
Claim6.6.The stress at Pacross the surface perpendicular to nis
σ(n)def= lim
∆s→0∆F
∆s=σij(n•ei)ej.
6.2. STRESS TENSOR (SPANNUNG) 77
It follows from the claim that the stress σis a vector valued function that depends
linearly on the normal nto the surface element, and we will see in §6.2.2 that the
matrixσijof this linear vector valued function forms a second-order tensor .
Proof. Consider the tetrahedron OA1A2A3formed by the triangular slice on
the plane Π having area ∆ sand three triangles on planes parallel to the coordinate
planes
e1e2e3
On
α1α2α3
Consider all forces acting on this tetrahedron as a volume element o f the rigid body.
There can be two types of forces:
(1)Body forces =f•∆v, wherefis the force per unit of volume and ∆ vis the
volume of the tetrahedron. We actually do not know these forces, but we
will see later that this is not relevant.
(2)Surface forces , that is the sum of the forces on each of the four sides of the
tetrahedron.
We want to assess each of the four surface contributions due to t he surface forces.
If ∆sis the area of the slice on the plane Π, the contribution of that slice is
σ(n)∆s.
If ∆s1is the area of the slice on the plane with normal −e1, the contribution of that
slice is
−σ1jej∆s1,
and similarly the contributions of the other two slices are
−σ2jej∆s2and−σ3jej∆s3.
78 6. APPLICATIONS
−e1
−e2
−e3O
O OOn
A1 A1A1 A1
A2A2 A2
A2A3 A3
A3 A3
Note that the minus sign comes from the fact that we use everywhe re outside point-
ing normals.
So the total surface force is
σ(n)∆s−σ1jej∆s1−σ2jej∆s2−σ3jej∆s3.
Since there is static equilibrium the sum of all (body and surface) for ces must be
zero
f∆v+σ(n)∆s−σijej∆si= 0.
The termf∆vcan be neglected when ∆ sis small, as it contains terms of higher
order (in fact ∆ v→0 faster than ∆ s→0). We conclude that
σ(n)∆s=σijej∆si.
It remains to relate ∆ sto ∆s1,∆s2,∆s3. The side with area ∆ siis the orthogonal
projection of the side with area ∆ sonto the plane with normal ei. The scaling
factor for the area under projection is cos αi, whereαiis the convex angle between
the plane normal vectors
ein
αi
∆si
∆s= cosαi= cosαi/bardbln/bardbl/bardblei/bardbl=n•ei.
Therefore
σ(n)∆s=σijej(n•ei)∆s
6.2. STRESS TENSOR (SPANNUNG) 79
or, equivalently,
σ(n) =σij(n•ei)ej.
/square
Remark 6.7.(1) Forhomogeneousstress, thestresstensor σijdoesnotdepend
on the point P. However, when we flip the orientation of the normal to the
plane, the stress tensor changes sign. In other words, if σ(n) is the stress
across a surface with normal n, then
σ(−n) =−σ(n).
The stress considers orientation as if the forces on each side of th e surface
have to balance each other in static equilibrium.
σ(n)σ(−n)
−nn
(2) In the formula σ(n) =σij(n•ei)ej, the quantities n•eiare the coordinates
ofnwith respect to the orthonormal basis {e1,e2,e3}, namely
n= (n•e1)e1+(n•e2)e2+(n•e3)e3=n1e1+n2e2+n3e3.
Claim6.8.The stress tensor is a symmetric tensor , that isσij=σji.
In fact, let us consider an infinitesimal cube of side ∆ ℓsurrounding Pand with
faces parallel to the coordinate planes.
e1e2e3
A BC DA′B′C′D′
The force acting on each of the six faces of the cube are:
•σ1j∆A1ejand−σ1j∆A1ej, respectively for the front and the back faces,
ABB′A′andDCC′D′;
80 6. APPLICATIONS
•σ2j∆A2ejand−σ2j∆A2ej, respectively for the right and the left faces
BCC′B′andADD′A′;
•σ3j∆A3ejand−σ3j∆A3ej, respectively for the top and the bottom faces
ABCDandA′B′C′D′,
where ∆A1= ∆A2= ∆A3= ∆s= (∆ℓ)2is the common face area. We compute
now the torque µ, assuming the forces are applied at the center of the faces (whos e
distance is1
2∆ℓto the center point P). Recall that the torque is the tendency of a
force to twist or rotate an object.
µ=∆ℓ
2e1×σ1j∆sej+/parenleftbigg
−∆ℓ
2e1/parenrightbigg
×(−σ1j∆sej)
+∆ℓ
2e2×σ2j∆sej+/parenleftbigg
−∆ℓ
2e2/parenrightbigg
×(−σ2j∆sej)
+∆ℓ
2e2×σ3j∆sej+/parenleftbigg
−∆ℓ
2e3/parenrightbigg
×(−σ3j∆sej)
=∆ℓ∆s(ei×σijej) =
=∆ℓ∆s/parenleftbig
(σ23−σ32)e1+(σ31−σ13)e2+(σ12−σ21)e3/parenrightbig
.
Since the equilibrium is static, then L= 0, so that σij=σji.
We can hence write
σ=
σ11σ12σ13
σ12σ22σ23
σ13σ23σ33
,
where thediagonalentries σ11,σ22andσ33arethenormal components , thatisthe
components of the forces perpendicular to the coordinate planes and the remaining
entriesσ12,σ13andσ23are theshear components , that is the components of the
forces parallel to the coordinate planes.
Since the stress tensor is symmetric, it can be orthogonally diagona lized, that is
σ=
σ10 0
0σ20
0 0σ3
,
where now σ1,σ2andσ3are theprincipal stresses , that is the eigenvalues of
σ. The eigenspaces of σare theprincipal directions and the shear components
disappear for the principal planes .
6.2.1. Special forms of the stress tensor (written with resp ect to an
orthonormal eigenbasis or another special basis).
6.2. STRESS TENSOR (SPANNUNG) 81
•Uniaxial stress with stress tensor given by
σ0 0
0 0 0
0 0 0
Example 6.9.This is the stress tensor in a long vertical rod loaded by
hanging a weight on the end.
•Plane stressed state orbiaxial stress with stress tensor given by
σ10 0
0σ20
0 0 0
Example 6.10.This isthestress tensor inplateonwhich forcesareapplied
as in the picture.
•Pure shear with stress tensor given by
−σ0 0
0σ0
0 0 0
or
0σ0
σ0 0
0 0 0
. (6.7)
This is special case of the biaxial stress, in the case in which σ1=σ2. In
(6.7)thefirst isthestress tensor writtenwithrespect toaneigen basis, while
the second is the stress tensor written with respect to an orthon ormal basis
obtained by rotating an eigenbasis by 45◦about the third axis. In fact
0σ0
σ0 0
0 0 0
=
√
2
2√
2
20
−√
2
2√
2
20
0 0 1
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
tL
−σ0 0
0σ0
0 0 0
√
2
2−√
2
20√
2
2√
2
20
0 0 1
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
L
whereLis the matrix of the change of coordinates.
•Shear deformation with stress tensor given by
0σ12σ13
σ120σ23
σ13σ230
82 6. APPLICATIONS
with respect to some orthonormal basis.
Fact6.11.The stress tensor σis a shear deformation if and only if its trace
is zero.
Example 6.12.The stress tensor
2−4 0
−4 0 4
0 4−2
represents a shear deformation. In fact one can check that
√
2
20√
2
2
0 1 0
−√
2
20√
2
2
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
tL
2−4 0
−4 0 4
0 4−2
√
2
20−√
2
2
0 1 0√
2
20√
2
2
/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
L=
0 0−2
0 0 4√
2
−2 4√
2 0
•Hydrostatic pressure with stress tensor given by
−p0 0
0−p0
0 0−p
,
wherep/ne}ationslash= 0 is the pressure. Here all eigenvalues are equal to −p.
Example 6.13.Pressure of a fluid on a bubble.
Exercise 6.14.Anystresstensorcanbewrittenasthesumofahydrostaticpres sure
and a shear deformation. Hint:look at the trace.
6.2.2. Contravariance of the stress tensor. LetB={e1,e2,e3}and/tildewideB=
{˜e1,˜e2,˜e3}be two basis, and let
˜ei=Lj
iejandei= Λj
i˜ej, (6.8)
whereL:=L/tildewideBBis the matrix of the change of basis and Λ = L−1is the inverse. Let
nbe a given unit vector and Sthe stress across a surface perpendicular to n. Then
Scan be expressed in two way, respectively with respect to Band to/tildewideB
S=σij(n•ei)ej (6.9)
S= ˜σij(n•˜ei)˜ej, (6.10)
and we want to relate σijto ˜σij. We start with the expression for Sin (6.9) and
rename the indices for later convenience.
S=σkm(n•ek)em=σkm(n•Λi
k˜ei)(Λj
m˜ej) =σkmΛi
kΛj
m(n•˜ei)˜ej,
where in the second equality we used (6.8), and in the third we used line arity.
Comparing the last expression with the expression in (6.10) we obtain
˜σij=σkmΛi
kΛj
m,
thus showing that σis acontravariant 2-tensoror atensor of type (0,2).
6.3. STRAIN TENSOR (VERZERRUNG) 83
6.3. Strain tensor (Verzerrung)
Consider a slightly deformation of a body
..
..
/squiggleright
PP
P1P1
/tildewideP/tildewideP/tildewideP1
/tildewideP1
∆x∆˜x
displacement of P1displacement of P
∆u
We have
∆˜x= ∆x+∆u,
where ∆xis the old relative position of PandP1, ∆˜xis their new relative position
and ∆uis the difference of the displacement, which hence measures the def ormation.
Assume that we have a small homogeneous deformation, that is
∆u=f(∆x);
in other words fis a small linear function independent of the point P. If we write
the components of ∆ uand ∆xwith respect to an orthonormal basis {e1,e2,e3}, the
functionfwill be represented by a matrix with entries that we denote by fij,
∆ui=fij∆xj.
The matrix ( fij) can be written as a sum of a symmetric and an antisymmetric
matrix as follows:
fij=ǫij+ωij,
where
ǫij=1
2(fij+fji)
84 6. APPLICATIONS
is a symmetric matrix and is called the strain tensor ordeformation tensor and
ωij=1
2(fij−fji)
is an antisymmetric matrix called the rotation tensor . We try to understand now
where these names come from.
Remark 6.15.First we verify that a (small) antisymmetric 3 ×3 matrix represents
a (small) rotation in 3-dimensional space.
Fact6.16.LetVbe a vector space with orthonormal basis B={e1,e2,e3}, and let
ω=
a
b
c
. The matrix of the linear map V→Vdefined byv/ma√sto→ω×vwith respect
to the basisBis
0−c b
c0−a
−b a0
.
In fact
ω×v=
a
b
c
×
x
y
z
= det
e1e2e3
a b c
x y z
=
bz−cy
cx−az
ay−bx
=
0−c b
c0−a
−b a0
x
y
z
.
Note that the matrix ( ωij) =
0ω12−ω13
−ω120ω23
ω13−ω230
corresponds to the cross
product with the vector ω=
−ω23
−ω13
−ω12
. /square
The antisymmetric case. Suppose that the matrix ( fij) was already antisymmet-
ric, so that
ωij=fijandǫij= 0.
By the Fact 6.16, the relation
∆ui=fij∆xj(6.11)
is equivalent to
∆u=ω×∆x,
so that
∆˜x= ∆x+∆u= ∆x+ω×∆x.
6.3. STRAIN TENSOR (VERZERRUNG) 85
Whenωis small, this represents an infinitesimal rotation of an angle /bardblω/bardblabout the
axisOω.
∆x ∆˜x
αr
Oω
In fact, since ω×∆xis orthogonal to the plane determined by ωand by ∆x, it
is tangent to the circle with center along the axis Oωand radius determined by ∆ x.
Moreover,
/bardbl∆u/bardbl=/bardblω×∆x/bardbl=/bardblω/bardbl/bardbl∆x/bardblsinα/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright
r,
and hence, since the length of an arc of a circle of radius rcorresponding to an angle
θisrθ, infinitesimally this represents a rotation by an angle /bardblω
.
The symmetric case. The opposite extreme case is when the matrix fijwas
alreadysymmetric , so that
ǫij=fijandωij= 0.
We will see that it is ǫijthat encodes the changes in the distances: in fact,
/bardbl∆˜x/bardbl2= ∆˜x•∆˜x= (∆x+∆u)•(∆x+∆u)
= ∆x•∆x+2∆x•∆u+∆u•∆u
≃/bardbl∆x/bardbl2+2ǫij∆xi∆xj,(6.12)
where in the last step we neglected the term /bardbl∆u/bardbl2since it is small compared to
∆uwhen ∆u→0 and used (6.11).
Remark 6.17.Even when fijis not purely symmetric, only the symmetric part ǫij
is relevant for the distortion of the distances. In fact, if ωijis antisymmetric, the
86 6. APPLICATIONS
term 2ωij∆xi∆xj= 0, so that
/bardbl∆˜x/bardbl2≃/bardbl∆x/bardbl2+2fij∆xi∆xj=/bardbl∆x/bardbl2+2ǫij∆xi∆xj./square
Recall that a metric tensor (or inner product) encodes the distan ces among
points. It follows that a deformation changes the metric tensor. L et us denote by
gthe metric before the deformation and by ˜ gthe metric after the deformation. By
definition we have
/bardbl∆˜x/bardbl2def= ˜g(∆˜x,∆˜x) = ˜gij∆˜xi∆˜xj= ˜gij(∆xi+∆ui)(∆xj+∆uj) (6.13)
and
/bardbl∆x/bardbl2def=g(∆x,∆x) =gij∆xi∆xj. (6.14)
For infinitesimal deformations (that is if ∆ u∼0), (6.13) becomes
/bardbl∆˜x/bardbl2= ˜gij∆xi∆xj.
This, together with (6.14) and (6.12), leads to
˜gij∆xi∆xj≃gij∆xi∆xj+2ǫij∆xi∆xj
and hence
ǫij≃1
2(˜gij−gij),
that isǫijmeasures the change in the metric.
By definition the strain tensor ǫijis symmetric
E=
ǫ11ǫ12ǫ13
ǫ12ǫ22ǫ23
ǫ13ǫ23ǫ33
,
where the terms on the diagonal (in green) determine the elongation or the contrac-
tion of the body along the coordinate directions e1,e2,e3, and the terms above the
diagonal (in orange) are the shear components of the strain tensor; that is ǫijis
the movement of a line element parallel to OejtowardsOei. Since it is a symmetric
tensor it can be orthogonally diagonalized
ǫ10 0
0ǫ20
0 0ǫ3
,
The eigenvalues of Eare theprincipal coefficients of the deformation and the
eigenspaces are the principal directions of the deformation.
6.4. ELASTICITY TENSOR 87
6.3.1. Special forms of the strain tensor.
(1)Shear deformation whenEis traceless
trE=ǫ11+ǫ22+ǫ33= 0.
(2)Uniform compression when the principal coefficients of Eare equal (and
nonzero)
k0 0
0k0
0 0k
Exercise 6.18.Any strain tensor can be written as the sum of a uniform
compression and a shear deformation.
6.4. Elasticity tensor
The stress tensor represents an external exertion onthematerial, while thestrain
tensor represents the material reaction to that exertion. In crystallography these are
calledfield tensors because they represent imposed conditions, opposed to matter
tensors, that represents material properties.
Hooke’s law says that, for small deformations, stress is related to strain by a
matter tensor called elasticity tensor orstiffness tensor
σij=Eijkℓǫkl,
while the tensor relating strain to stress is the compliance tensor
ǫkℓ=Sijkℓσij.
The elasticity tensor has rank 4, and hence in 3-dimensional space it has 34= 81
components. However symmetry reduces the number of independ ent components
forEijkℓ.
(1)Minor symmetries: The symmetry of the stress tensor
σij=σji
implies that
Eijkℓ=Ejikℓfor eachk,ℓ;
it follows that for each k,ℓfixed there are only 6 independent components
Eijkℓ
E11kℓE12kℓE13kℓ
E12kℓE22kℓE23kℓ
E13kℓE23kℓE33kℓ.
Having taken this in consideration, the number of independent comp onents
decreases to 6×32at the most. However the symmetry also of the strain
tensor
ǫkℓ=ǫℓk
88 6. APPLICATIONS
implies that
Eijkℓ=Eijℓkfor eachi,j.
This means that for each i,jfixed there are also only 6 independent com-
ponentsEijkℓ, so thatEijkℓhas at most 62= 36 independent components.
(2)Major symmetries: Since (under appropriate conditions) partial derivatives
commute, if follows from the existence of a strain energy density functional
Usatisfying
∂2U
∂ǫij∂ǫkℓ=Eijkℓ
that
Eijkℓ=Ekℓij,
that means the matrix with rows labelled by ( i,j) and columns labelled by
(k,ℓ) is symmetric. Since from (1) that there are only 6 entries ( i,j) for
a fixed (k,ℓ),Eijkℓcan be written in a 6 ×6 matrix with rows labelled by
(i,j) and columns labelled by ( k,ℓ)
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗
∗ ∗
∗
so thatEijkℓhas in fact only 6+5+4+3+2+1 = 21 components.
6.5. Conductivity tensor
Consider a homogeneous continuous crystal. Its properties can b e divided into
two classes:
•Properties that do not depend on a direction, and are hence described by
scalars. Examples are density and heat capacity.
•Properties that dependson a direction, and are hence described by tensors.
Examples are elasticity ,electrical conductivity andheat conductiv-
ity. We say that a crystal is anisotropic when it has such “tensorial”
properties.
6.5.1. Electrical conductivity. LetEbe theelectric field andJtheelectrical
current density . We assume that these are constant throughout the crystal. At each
point of the crystal:
(1)Egives the electric force (in Volts/m) that would be exerted on a positive
test charge (of 1 Coulomb) placed at the point;
6.5. CONDUCTIVITY TENSOR 89
(2)J(in Amperes/m2) gives the direction the charge carriers move and the
rate of electric current across an infinitesimal surface perpendicular to
that direction.
Jis a function of E,
J=f(E).
Consider asmall increment ∆ JinJcausedby asmall increment ∆ EinE, andwrite
these increments in terms of their components with respect to a ch osen orthonormal
basis{e1,e2,e3}.
∆J= ∆Jieiand ∆E= ∆Eiei.
The increments are related by
∆Ji=∂fi
∂Ej∆Ej+higher order terms in (∆ Ej)2,(∆Ej)2,...
If the quantities ∆ Ejare small, we can assume that
∆Ji=∂fi
∂Ej∆Ej(6.15)
If we assume that∂fi
∂Ejis independent of the point of the crystal,
∂fi
∂Ej=σi
j∈R
we obtain the relation
∆Ji=σi
j∆Ej
or simply
∆J=σ∆E,
whereσif theelectrical conductivity tensor . This is a (1 ,1)-tensor and may
depend4on the initial value of E, that is the electrical conductivity may be different
for small and large electric forces. If initially E= 0 andσ0is the corresponding
electrical conductivity tensor, we obtain the relation
J=σ0E
that is called the generalized Ohm law . This is always under the assumption that
∆Eand ∆Jare small and that the relation is linear.
Theelectrical resistivity tensor is
ρ=K−1,
that is, it is the (1 ,1)-tensor such that
ρj
iKℓ
j=δℓ
i.
4Typically if the dependence between EandJis linear for any value, and not only for small
ones, the tensor ail not depend on the initial value of E.
90 6. APPLICATIONS
The electrical conductivity measures the material’s ability to conduc t an electrical
current, while theelectrical resistivity quantifies theability of them aterial to oppose
the flow of the electrical current.
For anisotropic crystal, all directions are equivalent and these tensors are spher -
ical
σj
i=kδj
iandρj
i=1
kδj
i, (6.16)
wherekis a scalar, called the electrical conductivity of the crystal. Equa-
tion (6.16) can also be written as
k0 0
0k0
0 0k
and
1
k0 0
01
k0
0 01
k
.
In general, σj
iis neither symmetric nor antisymmetric (and actually symmetry does
not even make sense for a (1 ,1) tensor unless a metric is fixed, since it does require
a canonical identification of VwithV∗).
6.5.2. Heat conductivity. LetTbe thetemperature andHtheheat flux vec-
tor. For a homogeneous crystal and constant Hand for a constant gradient of T,
Fourier heat conduction law says that
H=−KgradT. (6.17)
At each point of the crystal:
(1) gradTpoints in the direction of the highest ascent of the temperature an d
measures therateofincreaseof Tinthatdirection. Theminussignin(6.17)
comes from the fact that the heat flows in the direction of the decr easing
temperature.
(2)Hmeasure the amount of heat passing per unit area perpendicular to its
direction per unit time.
HereKis theheat conductivity tensor orthermal conductivity tensor . In
terms of components with respect to a chosen orthonormal basis
Hi=−Kij(gradT)j.
Exercise 6.19.Verify that the gradient of a real function is a covariant 1-tensor.
The heat conductivity tensor is a contravariant 2-tensor and exp eriments show
that it is symmetric and hence can be orthogonally diagonalized. The heat resis-
tivity tensor is
r=K−1,
6.5. CONDUCTIVITY TENSOR 91
andhence isalso symmetric. Withrespect toanorthonormalbasis, Kisrepresented
by
K10 0
0K20
0 0K3
,
where the eigenvalues of Kare called the principal coefficients of heat conduc-
tivity.
Physical considerations (that is the fact that heat flows always in t he direction
of decreasing temperature) show that the eigenvalues are positiv e
Ki>0.
The eigenspaces of Kare called the principal directions of heat conductivity.
CHAPTER 7
Solutions
Exercise 2.4 (1) yes; (2) no, (3) no, (4) yes, (5) no, (6) yes.
Exercise 2.14
(1) The vectors in BspanVsince
/bracketleftbigg
a b
c−a/bracketrightbigg
=a/bracketleftbigg
1 0
0−1/bracketrightbigg
+b/bracketleftbigg
0 1
0 0/bracketrightbigg
+c/bracketleftbigg
0 0
1 0/bracketrightbigg
.
Moreover they are linearly independent since
a/bracketleftbigg
1 0
0−1/bracketrightbigg
+b/bracketleftbigg
0 1
0 0/bracketrightbigg
+c/bracketleftbigg
0 0
1 0/bracketrightbigg
=/bracketleftbigg
0 0
0 0/bracketrightbigg
if and only if
/bracketleftbigg
a b
c−a/bracketrightbigg
=/bracketleftbigg
0 0
0 0/bracketrightbigg
,
that is if and only if a=b=c= 0.
(2)Bis a basis of V, hence dim V= 3. Since/tildewideBhas three elements, it is enough
to check either that it spans Vor that it consists of linearly independent
vectors. We will check this last condition. In fact
a/bracketleftbigg
1 0
0−1/bracketrightbigg
+b/bracketleftbigg
0−1
1 0/bracketrightbigg
+c/bracketleftbigg
0 1
1 0/bracketrightbigg
=/bracketleftbigg
0 0
0 0/bracketrightbigg
⇐⇒/bracketleftbigg
a c−b
b+c−a/bracketrightbigg
=/bracketleftbigg
0 0
0 0/bracketrightbigg
that is
a= 0
b+c= 0
c−b= 0⇐⇒
a= 0
b= 0
c= 0
(3)
/bracketleftbigg
2 1
7−2/bracketrightbigg
= 2/bracketleftbigg
1 0
0−1/bracketrightbigg
+1/bracketleftbigg
0 1
0 0/bracketrightbigg
+7/bracketleftbigg
0 0
1 0/bracketrightbigg
,
therefore
[v]B=
2
1
7
.
93
94 7. SOLUTIONS
To computethecoordinatesof vwithrespect to /tildewideBweneedtofind a,b,c∈R
such that/bracketleftbigg
2 1
7−2/bracketrightbigg
=a/bracketleftbigg
1 0
0−1/bracketrightbigg
+b/bracketleftbigg
0−1
1 0/bracketrightbigg
+c/bracketleftbigg
0 1
1 0/bracketrightbigg
.
Similar calculations to the above ones yield
[v]/tildewideB=
2
3
4
.
Exercise 3.2: (1) no; (2) no; (3) yes.
Exercise 3.3: (1) yes; (2) yes; (3) no.
Exercise 3.11:
Exercise 3.18: (1) yes; (2) yes; (3) yes; (4) no, because v×wis not a real number;
(5) yes.
Exercise 4.2: (1) yes, this is the standard inner product ; (2) no, as ϕisnega-
tive definite , that isϕ(v,v)<0 ifv∈V,v/ne}ationslash= 0; (3) no, as ϕis not symmetric; (4)
yes.
Exercise 4.3:
(1) Yes, in fact:
(a)/integraltext1
0p(x)q(x)dx=/integraltext1
0q(x)p(x)dxbecausep(x)q(x) =q(x)p(x);
(b)/integraltext1
0(p(x))2dx≥0forallp∈R[x]2because(p(x))2≥0,and/integraltext1
0(p(x))2dx=
0 only when p(x) = 0 for all x∈[0,1], that is only if p≡0.
(2) No, since/integraltext1
0(p′(x))2dx= 0 implies that p′(x) = 0 for all x∈[0,1], butpis
not necessarily the zero polynomial.
(3) Yes
(4) No. Is there p∈R[x]2,p/ne}ationslash= 0 such that ( p(1))2+(p(2))2= 0?
(5) Yes. Is there a non-zero polynomial of degree 2 with 3 distinct z eros?
Exercise 4.11. We write
[v]/tildewideB=
˜v1
˜v2
˜v3
and [w]/tildewideB=
˜w1
˜w2
˜w3
and we know that gwith respect to the basis /tildewideBhas the standard form g(v,w) = ˜vi˜wi
and we want to verify (4.6) using the matrix of the change of coordin atesL−1= Λ.
If
[v]B=
v1
v2
v3
and [w]B=
w1
w2
w3
7. SOLUTIONS 95
then we have that
˜v1
˜v2
˜v3
= Λ
v1
v2
v3
=
v1−v2
v2−v3
v3
and
˜w1
˜w2
˜w3
= Λ
w1
w2
w3
=
w1−w2
w2−w3
w3
It follows that
g(v,w) = ˜vi˜wi= (v1−v2)(w1−w2)+(v2−v3)(w2−w3)+v3w3
=v1w1−v1w2−v2w1+2v2w2−v2w3−w3v2+2v3w3.
Exercise 4.14 With respect to /tildewideB, we have
/bardblv/bardbl= (12+12+12)1/2=√
3
/bardblw/bardbl= ((−1)2+(−1)2+32)1/2=√
11
and with respect to E
/bardblv/bardbl= (3·3−3·2−2·3+2·2·2−2·1−1·2+2·1·1)1/2=√
3
/bardblw/bardbl= (1·1−1·2−2·1+2·2·2−2·3−3·2+2·3·3)1/2=√
11.
Exercise: 4.25. Saying that the orthogonality is meant with respect to g, means
that we have to show that g(v−projbkv,bk) = 0. In fact,
g(v−projbkv,bk) =g(v−g(v,bk)
g(bk,bk)bk,bk) =g(v,bk)−g(v,bk)
✘✘✘✘✘g(bk,bk)✘✘✘✘✘g(bk,bk) = 0