01 lintrans
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Chapter 1 of a linear algebra course, apparently from Matthews' notes (file is in a folder named matthews linear algebra). It defines linear transformations, kernel and image, and proves the rank plus nullity theorem and the dimension theorem for subspaces via direct sums. It also covers the matrix of a transformation relative to bases, composition, invertible maps, and isomorphisms, with examples such as T(X)=AX-XA.
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1 Linear Transformations
We will study mainly nite-dimensional vector spaces over an arbitrary eld
F|i.e. vector spaces with a basis. (Recall that the dimension of a vector
spaceV(dimV) is the number of elements in a basis of V.)
DEFINITION 1.1
(Linear transformation )
Given vector spaces UandV,T:U7!Vis a linear transformation (LT)
if
T(u+v) =T(u) +T(v)
for all;2F, andu;v2U. ThenT(u+v) =T(u)+T(v); T(u) =T(u)
and
T nX
k=1kuk!
=nX
k=1kT(uk):
EXAMPLES 1.1
Consider the linear transformation
T=TA:Vn(F)7!Vm(F)
whereA= [aij] ismn, de ned by TA(X) =AX.
Note thatVn(F) = the set of all n-dimensional column vectors2
64x1
...
xn3
75of
F|sometimes written Fn.
Note that if T:Vn(F)7!Vm(F) is a linear transformation, then T=TA,
whereA= [T(E1)jjT(En)] and
E1=2
66641
0
...
03
7775;:::;En=2
66640
...
0
13
7775
Note:
v2Vn(F); v =2
64x1
...
xn3
75=x1E1++xnEn
1
IfVis a vector space of all in nitely di erentiable functions on R, then
T(f) =a0Dnf+a1Dn 1f++an 1Df+anf
de nes a linear transformation T:V7!V.
The set offsuch thatT(f) = 0 (i.e. the kernel of T) is important.
LetT:U7!Vbe a linear transformation. Then we have the following
de nition:
DEFINITIONS 1.1
(Kernel of a linear transformation )
KerT=fu2UjT(u) = 0g
(Image ofT)
ImT=fv2Vj9u2Usuch thatT(u) =vg
Note : KerTis a subspace of U. Recall that Wis a subspace of Uif
1. 02W,
2.Wis closed under addition, and
3.Wis closed under scalar multiplication.
PROOF. that Ker Tis a subspace of U:
1.T(0) + 0 = T(0) =T(0 + 0) = T(0) +T(0). Thus T(0) = 0, so
02KerT.
2. Letu;v2KerT; thenT(u) = 0 and T(v) = 0. So T(u+v) =
T(u) +T(v) = 0 + 0 = 0 and u+v2KerT.
3. Letu2KerTand2F. ThenT(u) =T(u) =0 = 0. So
u2KerT.
EXAMPLE 1.1
KerTA=N(A);the null space of A
=fX2Vn(F)jAX= 0g
and ImTA=C(A);the column space of A
=hA1;:::;Ani
2
Generally, if U=hu1;:::;uni, then ImT=hT(u1);:::;T (un)i.
Note: Even ifu1;:::;unform a basis for U,T(u1);:::;T (un) may not
form a basis for Im T. I.e. it may happen that T(u1);:::;T (un) are linearly
dependent.
1.1 Rank + Nullity Theorems (for Linear Maps)
THEOREM 1.1 (General rank + nullity theorem)
IfT:U7!Vis a linear transformation then
rankT+ nullityT= dimU:
PROOF.
1. KerT=f0g.
Then nullity T= 0.
We rst show that the vectors T(u1);:::;T (un), whereu1;:::;unare
a basis forU, are LI (linearly independent):
Supposex1T(u1) ++xnT(un) = 0 where x1;:::;xn2F.
Then
T(x1u1++xnun) = 0 (by linearity)
x1u1++xnun= 0 (since Ker T=f0g)
x1= 0;:::;xn= 0 (since uiare LI)
Hence ImT=hT(u1);:::;T (un)iso
rankT+ nullityT= dim ImT+ 0 =n= dimV:
2. KerT=U.
So nullityT= dimU.
Hence ImT=f0g) rankT= 0
)rankT+ nullityT= 0 + dim U
= dimU:
3. 0<nullityT <dimU.
Letu1;:::;urbe a basis for Ker Tandn= dimU, sor= nullityT
andr<n .
Extend the basis u1;:::;urto form a basis u1;:::;ur;ur+1;:::;unof
3
U(refer to last year's notes to show that this can be done).
ThenT(ur+1);:::;T (un) span ImT. For
ImT=hT(u1);:::;T (ur);T(ur+1);:::;T (un)i
=h0;:::; 0;T(ur+1);:::;T (un)i
=hT(ur+1);:::;T (un)i
So assume
x1T(ur+1) ++xn rT(un) = 0
)T(x1ur+1++xn run) = 0
)x1ur+1++xn run2KerT
)x1ur+1++xn run=y1u1++yrur
for somey1;:::;yr
)( y1)u1++ ( yr)ur+x1ur+1++xn run= 0
and sinceu1;:::;unis a basis for U, all coecients vanish.
Thus
rankT+ nullityT= (n r) +r
=n
= dimU:
We now apply this theorem to prove the following result:
THEOREM 1.2 (Dimension theorem for subspaces)
dim(U\V) + dim(U+V) = dimU+ dimV
whereUandVare subspaces of a vector space W.
(Recall that U+V=fu+vju2U;v2Vg.)
For the proof we need the following de nition:
DEFINITION 1.2
IfUandVare any two vector spaces, then the direct sum is
UV=f(u;v)ju2U;v2Vg
(i.e. the cartesian product of UandV) made into a vector space by the
component-wise de nitions:
4
1.(u1;v1) + (u2;v2) = (u1+u2;v1+v2),
2.(u;v) = (u;v ), and
3.(0;0)is an identity for UVand( u; v)is an additive inverse for
(u;v).
We need the following result:
THEOREM 1.3
dim(UV) = dimU+ dimV
PROOF.
Case 1:U=f0g
Case 2:V=f0g
Proof of cases 1 and 2 are left as an exercise.
Case 3:U6=f0gandV6=f0g
Letu1;:::;umbe a basis for U, and
v1;:::;vnbe a basis for V.
We assert that ( u1;0);:::; (um;0);(0;v1);:::; (0;vn) form a basis for UV.
Firstly, spanning:
Let (u;v)2UV, sayu=x1u1++xmumandv=y1v1++ynvn.
Then
(u;v) = (u;0) + (0;v)
= (x1u1++xmum;0) + (0;y1v1++ynvn)
=x1(u1;0) ++xm(um;0) +y1(0;v1) ++yn(0;vn)
SoUV=h(u1;0);:::; (um;0);(0;v1);:::; (0;vn)i
Secondly, independence: assume x1(u1;0) ++xm(um;0) +y1(0;v1) +
+yn(0;vn) = (0;0). Then
(x1u1++xmum;y1v1++ynvn) = 0
)x1u1++xmum= 0
andy1v1++ynvn= 0
)xi= 0;8i
andyi= 0;8i
5
Hence the assertion is true and the result follows.
PROOF.
LetT:UV7!U+VwhereUandVare subspaces of some W, such
thatT(u;v) =u+v.
Thus ImT=U+V, and
KerT=f(u;v)ju2U;v2V;andu+v= 0g
=f(t; t)jt2U\Vg
Clearly then, dim Ker T= dim(U\V)1and so
rankT+ nullityT= dim(UV)
) dim(U+V) + dim(U\V) = dimU+ dimV:
1.2 Matrix of a Linear Transformation
DEFINITION 1.3
LetT:U7!Vbe a LT with bases :u1;:::;unand
:v1;:::;vmfor
UandVrespectively.
Then
T(uj) =a1jv1
+
a2jv2
+
...
+
amjvmfor somea1j
...
amj2F:
Themnmatrix
A= [aij]
is called the matrix of Trelative to the bases and
and is also written
A= [T]
Note: Thej-th column of Ais the co-ordinate vector of T(uj), where
ujis thej-th vector of the basis .
Also ifu=x1u1++xnun, the co-ordinate vector2
64x1
...
xn3
75is denoted by
[u] .
1True ifU\V=f0g; if not, let S= KerTandu1;:::;urbe a basis for U\V. Then
(u1; u1);:::; (ur; ur) form a basis for S and hence dim Ker T= dimS.
6
EXAMPLE 1.2
LetA=a b
c d
2M22(F)and letT:M22(F)7!M22(F)be
de ned by
T(X) =AX XA:
ThenTis linear2, and KerTconsists of all 22matricesAwhereAX=
XA.
Take to be the basis E11,E12,E21, andE22, de ned by
E11=1 0
0 0
;E12=0 1
0 0
;E21=0 0
1 0
;E22=0 0
0 1
(so we can de ne a matrix for the transformation, consider these henceforth
to be column vectors of four elements).
Calculate [T]
=B:
T(E11) =AE11 E11A
=a b
c d1 0
0 0
1 0
0 0a b
c d
=0 b
c0
= 0E11 bE12+cE21+ 0E22
and similar calculations for the image of other basis vectors show that
B=2
6640 c b 0
b a d 0b
c 0d a c
0c b 03
775
Exercise: Prove that rankB= 2 ifAis not a scalar matrix (i.e. if
A6=tIn).
Later, we will show that rankB= rankT. Hence
nullityT= 4 2 = 2
2
T(X+Y) =A(X+Y) (X+Y)A
=(AX XA) +(AY YA)
=T(X) +T(Y)
7
Note:I2;A2KerTwhich has dimension 2. Hence ifAis not a scalar
matrix, since I2andAare LI they form a basis for KerT. Hence
AX=XA)X= I2+ A:
DEFINITIONS 1.2
LetT1andT2be LT's mapping U to V.
ThenT1+T2:U7!Vis de ned by
(T1+T2)(x) =T1(x) +T2(x);8x2U
ForTa LT and2F, de neT:U7!Vby
(T)(x) =T(x)8x2U
Now . . .
[T1+T2]
= [T1]
+ [T2]
[T]
=[T]
DEFINITION 1.4
Hom (U;V) =fTjT:U7!Vis a LTg:
Hom (U;V)is sometimes written L(U;V).
The zero transformation 0 : U7!Vis such that 0( x) = 0,8x.
IfT2Hom (U;V), then ( T)2Hom (U;V) is de ned by
( T)(x) = (T(x))8x2U:
Clearly, Hom ( U;V) is a vector space.
Also
[0]
= 0
and [ T]
= [T]
The following result reduces the computation of T(u) to matrix multi-
plication:
THEOREM 1.4
[T(u)]
= [T]
[u]
8
PROOF.
LetA= [T]
, where is the basis u1;:::;un,
is the basis v1;:::;vm,
and
T(uj) =mX
i=1aijvi:
Also let [u] =2
64x1
...
xn3
75.
Thenu=Pn
j=1xjuj, so
T(u) =nX
j=1xjT(uj)
=nX
j=1xjmX
i=1aijvi
=mX
i=10
@nX
j=1aijxj1
Avi
)[T(u)]
=2
64a11x1++a1nxn
...
am1x1++amnxn3
75
=A[u]
DEFINITION 1.5
(Composition of LTs)
IfT1:U7!VandT2:V7!Ware LTs, then T2T1:U7!Wde ned by
(T2T1)(x) =T2(T1(x))8x2U
is a LT.
THEOREM 1.5
If ,
andare bases for U,VandW, then
[T2T1]
= [T2]
[T1]
9
PROOF. Let u2U. Then
[T2T1(u)]= [T2T1]
[u]
and = [T2(T1(u))]
= [T2]
[T1(u)]
Hence
[T2T1]
[u] = [T2]
[T1]
[u] (1)
(note that we can't just \cancel o " the [ u] to obtain the desired result!)
Finally, if isu1;:::;un, note that [ uj] =Ej(sinceuj= 0u1++
0uj 1+ 1uj+ 0uj+1++ 0un) then for an appropriately sized matrix B,
BEj=Bj;thejth column of B.
Then (1) shows that the matrices
[T2T1]
and [T2]
[T1]
have their rst, second, . . . , nth columns respectively equal.
EXAMPLE 1.3
IfAismnandBisnp, then
TATB=TAB:
DEFINITION 1.6
(theidentity transformation )
LetUbe a vector space. Then the identity transformation IU:U7!U
de ned by
IU(x) =x8x2U
is a linear transformation, and
[IU]
=Inifn= dimU.
Also note that IVn(F)=TIn.
THEOREM 1.6
LetT:U7!Vbe a LT. Then
IVT=TIU=T:
10
Then
TImTA=TImA=TA=TATAIn=TAIn
and consequently we have the familiar result
ImA=A=AIn:
DEFINITION 1.7
(Invertible LTs )
LetT:U7!Vbe a LT.
If9S:V7!Usuch thatSis linear and satis es
ST=IUandTS=IV
then we say that Tisinvertible and thatSis an inverse ofT.
Such inverses are unique and we thus denote SbyT 1.
Explicitly,
S(T(x)) =x8x2UandT(S(y)) =y8y2V
There is a corresponding de nition of an invertible matrix :A2Mmn(F)
is called invertible if 9B2Mnm(F)such that
AB=ImandBA=In
Evidently
THEOREM 1.7
TAis invertible i Ais invertible (i.e. if A 1exists). Then,
(TA) 1=TA 1
THEOREM 1.8
Ifu1;:::;unis a basis for Uandv1;:::;vnare vectors in V, then there
is one and only one linear transformation T:U!Vsatisfying
T(u1) =v1;:::;T (un) =vn;
namelyT(x1u1++xnun) =x1v1++xnvn.
(In words, a linear transformation is determined by its action on a basis.)
11
1.3 Isomorphisms
DEFINITION 1.8
A linear map T:U7!Vis called an isomorphism ifTis 1-1 and onto,
i.e.
1.T(x) =T(y))x=y8x;y2U, and
2.ImT=V, that is, if v2V,9u2Usuch thatT(u) =v.
Lemma: A linear map Tis 1-1 i Ker T=f0g.
Proof:
1. ()) SupposeTis 1-1 and let x2KerT.
We haveT(x) = 0 =T(0), and so x= 0.
2. (() Assume Ker T=f0gandT(x) =T(y) for somex;y2U.
Then
T(x y) =T(x) T(y) = 0
)x y2KerT
)x y= 0)x=y
THEOREM 1.9
LetA2Mmn(F). ThenTA:Vn(F)!Vm(F)is
(a) onto:,dimC(A) =m,the rows of Aare LI;
(b) 1{1:,dimN(A) = 0,rankA=n,the columns of Aare LI.
EXAMPLE 1.4
LetTA:Vn(F)7!Vn(F)withAinvertible; so TA(X) =AX.
We will show this to be an isomorphism.
1. LetX2KerTA, i.e.AX= 0. Then
A 1(AX) =A 10
)InX= 0
)X= 0
)KerT=f0g
,Tis 1-1.
12
2. LetY2Vn(F): then,
T(A 1Y) =A(A 1Y)
=InY=Y
soImTA=Vn(F)
THEOREM 1.10
IfTis an isomorphism between UandV, then
dimU= dimV
PROOF.
Letu1;:::;unbe a basis for U. Then
T(u1);:::;T (un)
is a basis for V(i.e.huii=UandhT(ui)i=V, withui,viindependent
families), so
dimU=n= dimV
THEOREM 1.11
: Hom (U;V)7!Mmn(F)de ned by (T) = [T]
is an isomorphism.
Here dimU=n,dimV=m, and and
are bases for UandV, re-
spectively.
THEOREM 1.12
T:U7!Vis invertible
,Tis an isomorphism between UandV.
PROOF.
)AssumeTis invertible. Then
T 1T=IU
andTT 1=IV
)T 1(T(x)) =x8x2U
andT(T 1(y)) =y8y2V
13
1. We prove Ker T=f0g.
LetT(x) = 0. Then
T 1(T(x)) =T 1(0) = 0 =x
SoTis 1-1.
2. We show Im T=V.
Lety2V. NowT(T 1(y)) =y, so taking x=T 1(y) gives
T(x) =y:
Hence ImT=V.
(Assume T is an isomorphism, and let S be the inverse map of T
S:V7!U
ThenST=IUandTS=IV. It remains to show that Sis linear.
We note that
x=S(y),y=T(x)
And thus, using linearity of Tonly, for any y1;y22V,x1=S(y1), and
x2=S(y2) we obtain
S(y1+y2) =S(T(x1) +T(x2))
=S(T(x1+x2))
=x1+x2
=S(y1) +S(y2)
COROLLARY 1.1
IfA2Mmn(F)is invertible, then m=n.
PROOF.
SupposeAis invertible. Then TAis invertible and thus an isomorphism
betweenVn(F) andVm(F).
Hence dimVn(F) = dimVm(F) and hence m=n.
THEOREM 1.13
IfdimU= dimVandT:U7!Vis a LT, then
Tis 1-1 (injective) ,Tis onto (surjective)
(,Tis an isomorphism )
14
PROOF.
)SupposeTis 1-1.
Then KerT=f0gand we have to show that Im T=V.
rankT+ nullityT= dimU
)rankT+ 0 = dim V
i.e. dim( Im T) = dimV
)ImT=VasTV.
(SupposeTis onto.
Then ImT=Vand we must show that Ker T=f0g. The above
argument is reversible:
ImT=V
rankT= dimV
= dimU
= rankT+ nullityT
)nullityT= 0
or KerT=f0g
COROLLARY 1.2
LetA;B2Mnn(F). Then
AB=In)BA=In:
PROOF Suppose AB=In. Then Ker TB=f0g. For
BX= 0)A(BX) =A0 = 0
)InX= 0)X= 0:
But dimU= dimV=n, soTBis an isomorphism and hence invertible.
Thus9C2Mnn(F) such that
TBTC=IVn(F)=TCTB
)BC =In =CB;
noting that IVn(F)=TIn.
15
Now, knowing AB=In,
)A(BC) =A
(AB)C=A
InC=A
)C=A
)BA =In
DEFINITION 1.9
Another standard isomorphism: Let dimV=m, with basis
=v1;:::;vm.
Then
:V7!Vm(F)is the isomorphism de ned by
(v) = [v]
THEOREM 1.14
rankT= rank [T]
PROOF
UT !V
# #
Vn(F) !
TAVm(F)
With
:u1;:::;un
:v1;:::;vma basis forU
V;
letA= [T]
:Then the commutative diagram is an abbreviation for the
equation
T=TA : (2)
Equivalently
T(u) =TA (u)8u2U
or
[T(u)]
=A[u]
which we saw in Theorem 1.4.
But rank (ST) = rankTifSis invertible and rank ( TR) = rankTifR
is invertible. Hence, since and
are both invertible,
(2))rankT= rankTA= rankA
16
and the result is proven.
Note:
Observe that
(T(uj)) =Aj, thejth column of A. So ImTis mapped
under
intoC(A). Also Ker Tis mapped by intoN(A). Consequently
we get bases for Im Tand KerTfrom bases for C(A) andN(A), respectively.
(u2KerT,T(u) = 0,
(T(u)) = 0
,TA (u) = 0
, (u)2N(A):)
THEOREM 1.15
Let and
be bases for some vector space V. Then, with n= dimV,
[IV]
is non-singular and its inverse
n
[IV]
o 1
= [IV]
:
PROOF
IVIV=IV
)[IVIV]
= [IV]
=In
= [IV]
[IV]
:
The matrix P= [IV]
= [pij] is called the change of basis matrix . For if
:u1;:::;unand
:v1;:::;vnthen
uj=IV(uj)
=p1jv1++pnjvnforj= 1;:::;n .
It is also called the change of co-ordinate matrix , since
[v]
= [IV(v)]
[v]
i.e. if
v=x1u1++xnun
=y1v1++ynvn
17
then 2
64y1
...
yn3
75=P2
64x1
...
xn3
75;
or, more explicitly,
y1=p11x1++p1nxn
...
yn=pn1x1++p1nxn:
THEOREM 1.16 (E ect of changing basis on matrices of LTs)
LetT:V7!Vbe a LT with bases and
. Then
[T]
=P 1[T]
P
where
P= [IV]
as above.
PROOF
IVT=T=TIV
)[IVT]
= [TIV]
)[IV]
[T]
= [T]
[IV]
DEFINITION 1.10
(Similar matrices )
IfAandBare two matrices in Mmn(F), then if there exists a non-
singular matrix Psuch that
B=P 1AP
we say that AandBaresimilar overF.
1.4 Change of Basis Theorem for TA
In the MP274 course we are often proving results about linear transforma-
tionsT:V7!Vwhich state that a basis can be found for Vso that
[T]
=B, whereBhas some special property. If we apply the result to
the linear transformation TA:Vn(F)7!Vn(F), the change of basis theorem
applied toTAtells us that Ais similar to B. More explicitly, we have the
following:
18
THEOREM 1.17
LetA2Mnn(F)and suppose that v1;:::;vn2Vn(F)form a basis
forVn(F). Then ifP= [v1jjvn]we have
P 1AP= [TA]
:
PROOF. Let
be the standard basis for Vn(F) consisting of the unit vectors
E1;:::;Enand let :v1;:::;vnbe a basis for Vn(F). Then the change of
basis theorem applied to T=TAgives
[TA]
=P 1[TA]
P;
whereP= [IV]
is the change of coordinate matrix.
Now the de nition of Pgives
v1=IV(v1) =p11E1++pn1En
...
vn=IV(vn) =p1nE1++pnnEn;
or, more explicitly,
v1=2
64p11
...
pn13
75; :::;2
64p1n
...
pnn3
75:
In other words, P= [v1jjvn], the matrix whose columns are v1;:::;vn
respectively.
Finally, we observe that [ TA]
=A.
19