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Chapter 3 of a linear algebra course text (folder name suggests Matthews), covering direct sums, T-invariant subspaces, T-cyclic subspaces and companion matrices. It includes the Jordan and hypercompanion bases, a proof of the Cayley-Hamilton theorem, an algorithm for finding the minimum polynomial, and the primary decomposition theorem. It ends with simultaneous diagonalization of commuting maps and Fitting's lemma.
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3 Invariant subspaces
DEFINITIONS 3.1
SubspacesV1;:::;VtofVare called independent if
v1++vt= 0)v1= 0;:::;vt= 0
8v12V1;:::;vt2Vt:
We say that Vis the ( internal )direct sum of the subspaces V1;:::;Vtif
(a)V1;:::;Vtare independent and
(b)V=V1++Vt.
I.e. every element v2Visuniquely expressible as
v=v1++vt
withvi2Vi.
ThenVis isomorphic to the (external) direct sum V1Vtunder
the isomorphism v7!(v1;:::;vt)and we write V=V1Vt:
THEOREM 3.1
IfV=V1Vt(an internal direct sum) and 1;:::; tare bases for
V1;:::;Vtrespectively, then
= 1[[ t;
the sequence formed by juxtaposing the separate bases, is a basis for V.
Also
dimV= dimV1++ dimVt:
Proof: Left as an exercise.
DEFINITION. Let T:V7!Vbe a LT and Wa subspace of V. Then if
w2W)T(w)2W;
we sayWis aT-invariant subspace ofV. We can then consider the linear
transformation TW:W!Wde ned by
TW(w) =T(w)8w2W:
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If 0is a basis for W,f0gWV, and is an extension to a basis of
V, then
[T]
="
[TW] 0
0B1
0B2#
:
A situation of great interest is when we have T-invariant subspaces
W1;:::;WtandV=W1Wt. For if = 1[[ t, where iis
a basis forWi, we see that
[T]
= [TW1] 1
1 [TWt] t
t:
There are two important examples of T{invariant subspaces that arise in
our study of Jordan and rational canonical forms - Ker pt(T) andT{cyclic
subspaces.
3.1T{cyclic subspaces
DEFINITION 3.1
The unique monic polynomial finF[x]of least degree satisfying
f(T)(v) = 0
is called the minimum polynomial of the vector v2Vrelative to the
transformation T:V7!Vand is denoted mT;v.
Thenf(T)(v) = 0)mT;vjf, somT;vjmT. AlsomT;v= 1,v= 0 and
so ifv6= 0, degmT;v1.
EXAMPLE. Let T=TA, whereA=0 0
1 0
. Also let
v1=1
0
andv2=0
1
:
ThenAv1=0
1
6=c0v1, somT;v16=x c0. NextA2v1=0
0
, so
mT;v1=x2. AlsoAv2=0
0
, somT;v2=x.
DEFINITION 3.2
(T-cyclic subspace generated by v.)
Ifv2V, the set of all vectors of the form f(T)(v); f2F[x];forms a
subspace of Vcalled theT-cyclic subspace generated by v. It is denoted by
CT;v.
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PROOF. Exercise.
Also,CT;vis aT-invariant subspace of V. For
w2CT;v)w=f(T)(v)
)T(w) =T(f(T)(v)) = (Tf(T))(v) = ((xf)(T))(v)2CT;v:
We see that v= 0 if and only if CT;v=f0g.
THEOREM 3.2
Letv6= 0; v2V. ThenCT;vhas the basis
v;T(v);T2(v);:::;Tk 1(v)
wherek= degmT;v. ( is called the T-cyclic basis generated by v. ) Note
that dimCT;v= degmT;v.
Finally,
TCT;v
=C(mT;v);
the companion matrix of the minimum polynomial of v.
PROOF.
1. TheT-cyclic basis is a basis for CT;v:
Spanning:
Letw2hv;T(v);:::;Tk 1(v)i, so
w=w0v+w1T(v) ++wk 1Tk 1(v)
= (w0IV++wk 1Tk 1)(v)
=g(T)(v);
whereg=w0++wk 1xk 1, sow2CT;v. Hence
hv;T(v);:::;Tk 1(v)iCT;v:
Conversely, suppose that w2CT;vso
w=f(T)(v)
and
f=qmT;v+r
wherer=a0+a1x++ak 1xk 1anda0;:::;ak 12F. So
f(T)(v) =q(T)mT;v(T)(v) +r(T)(v)
=q(T)mT;v(T)(v) +a0v+a1T(v) ++ak 1Tk 1(v)
=a0v+a1T(v) ++ak 1Tk 1(v)
2 hv;T(v);:::;Tk 1(v)i:
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Independence:
Assume
a0v+a1T(v) ++ak 1Tk 1(v) = 0;
wherea0;:::;ak 12F; that is,f(T)(v) = 0 where
f=a0+a1x++ak 1xk 1:
HencemT;vjfand since
degf=k 1<k= degmT;v;
we havef= 0 and thus ai= 08i.
2. [TCT;v]
=C(mT;v):
LetL=TCT;v, the restriction of TtoCT;v.
We want to nd [ L]
. So
L(v) =T(v) = 0v+ 1T(v) + 0T2(v) ++ 0Tk 1(v)
L(T(v)) =T2(v) = 0v+ 0T(v) + 1T2(v) ++ 0Tk 1(v)
...
L(Tk 2(v)) =Tk 1(v) = 0v+ 0T(v) + 0T2(v) ++ 1Tk 1(v)
Finally, to calculate L(Tk 1(v)) =Tk(v), we let
mT;v=a0+a1x++ak 1xk 1+xk:
ThenmT;v(T)(v) = 0 and hence
L(Tk 1(v)) =Tk(v) = a0v a1T(v) ak 1Tk 1(v):
Hence
[L]
=2
66640 0:::0 a0
1 0 a1
.........
0 01 ak 13
7775=C(mT;v);
as required.
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THEOREM 3.3
Suppose that mT;v= (x c)k. Then the vectors
v;(T cIV)(v);:::; (T cIV)k 1(v)
form a basis forW=CT;vwhich we call the elementary Jordan basis .
Also
[TW]
=Jk(c):
More generally suppose mT;v=pk, wherepis a monic irreducible polynomial
inF[x], withn= degp. Then the vectors
v; T (v); :::; Tn 1(v)
p(T)(v); Tp (T)(v); :::; Tn 1p(T)(v)
............
pk 1(T)(v); Tpk 1(T)(v); :::; Tn 1pk 1(T)(v);
form a basis for W=CT;v, which reduces to the elementary Jordan basis
whenp=x c. Also
[TW]
=H(pk);
whereH(pk)is ahypercompanion matrix, which reduces to the elemen-
tary Jordan matrix Jk(c)whenp=x c:
H(pk) =2
666664C(p) 0 0
N C (p) 0
0N 0
............
0N C (p)3
777775;
where there are kblocks on the diagonal and Nis a square matrix of same
size asC(p)which is everywhere zero, except in the top right{hand corner,
where there is a 1. The overall e ect is an unbroken subdiagonal of 10s.
3.1.1 A nice proof of the Cayley-Hamilton theorem
(From Insel, Friedberg and Spence.)
Letf= chT, for someT:V7!V. We must show that f(T) = 0V|i.e.
thatf(T)(v) = 08v2V.
This is immediate if v= 0, so assume v6= 0 and let W=CT;v. Let 0
be a basis of Wand be an extension of 0to a basis of V. Then
[T]
="
[TW] 0
0B1
0B2#
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and chT= chTWchB2. So chTWjchTand since we know that
chTW=mT;v;
we havemT;vjchT.
Hence ch T=gmT;vand
chT(T)(v) = (g(T)mT;v(T))(v) =g(T)(mT;v(T)(v)) =g(T)(0) = 0:
3.2 An Algorithm for Finding mT
We use the factorization of ch Tinto monic irreducibles.
THEOREM 3.4
SupposeT:V7!V,
mT=pb1
1:::pbt
t
whereb1;:::;bt1, andp1;:::;ptare distinct monic irreducibles.
Then fori= 1;:::;t we have
(a)
VImpi(T) Impbi 1
i(T)Impbi
i(T) =
(b)
f0gKerpi(T) Kerpbi 1
i(T)Kerpbi
i(T) =:
Note : In terms of nullities, conclusion (b) says that
0<(pi(T))<<(pbi 1
i(T))<(pbi
i(T)) =
so this gives us a method of calculating bi.
Presently we'll show that if ch T=pa1
1:::pat
t;then
nullity (pbi
i(T)) =aidegpi:
Hencebiis also characterised as the smallest integer hsuch that
nullity (ph
i(T)) =aidegpi:
Also note that (a) and (b) are equivalent, and it is the latter that we
prove.
A notational simpli cation| the leftF[x]-module notation .
Iff2F[x] andv2V, we de ne
fv=f(T)(v):
It is easy to verify that
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1.
(f+g)v=fv+gv8f;g2F[x];v2V;
2.
f(v+w) =fv+fw8f2F[x];v;w2V;
3.
(fg)v=f(gv)8f;g2F[x];v2V;
4.
1v=v8v2V:
These axioms, together with the four axioms for addition on V, turnVinto
what is called a \left F[x]-module". (So there are deeper considerations
lurking in the background|ideas of greater generality which make the algo-
rithm we unravel for the rational canonical form also apply to other things
such as the theorem that any nite abelian group is a direct product of cyclic
prime power subgroups.)
(i) We rst prove that f0gKerpi(T). We write p=pi,b=bifor
brevity; no confusion should arise since iis xed.
PROOF.mT=pf; f2F[x] andf(T)6= 0V. Hence9v2Vsuch that
fv6= 0. Then
p(fv) = (pf)v=mTv= 0;
sofv2Kerp(T).
(ii) We next prove that
Kerpb(T) = Kerpb+1(T):
The containment
Kerpb(T)Kerpb+1(T)
is obvious, so we need only show that
Kerpb(T)Kerpb+1(T):
Letw2Kerpb+1(T), i.e.pb+1w= 0. Now if mT=pbq, then gcd(pb;q) = 1.
So9u;v2F[x] such that 1 = upb+vq. Hence
pb=up2b+vmT:
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Hence
pb(T) = (up2b)(T) +v(T)mT(T)
= (up2b)(T)
and thus
pbw= (pb 1)(pb+1w) =pb 10 = 0
andw2Kerpb, as required.
(iii)
Kerph(T) = Kerph+1(T))Kerph+1(T) = Kerph+2(T);
i.e. Ker ph(T)Kerph+1(T))Kerph+1(T)Kerph+2(T):
PROOF. Suppose that Ker ph(T)Kerph+1(T). Then
v2Kerph+2(T))ph+2v= 0
)ph+1(pv) = 0)pv2Kerph+1(T)
)pv2Kerph(T))ph(pv) = 0
)ph+1v= 0)v2Kerph+1(T):
So it follows by induction from (ii) that
Kerpbi
i(T) = Kerpbi+1
i(T) =:
(iv)
Kerpb 1(T)Kerpb(T)
and this forces a chain of proper inclusions:
f0gKerp(T) Kerpb 1(T)Kerpb(T) =
which is the desired result. For
pb 1q(T)6= 0V, so9vsuch thatpb 1qv6= 0. Then
qv62Kerpb 1(T);
butqv2Kerpb(T) as
pbqv=mTv= 0:
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3.3 Primary Decomposition Theorem
THEOREM 3.5 (Primary Decomposition)
IfT:V7!Vis a LT with mT=pb1
1:::pbt
t, wherep1;:::;ptare monic
irreducibles, then
V= Kerpb1
1(T) Kerpbt
t(T);
a direct sum of T-invariant subspaces. Moreover for 1it,
(pi(T)bi) =aidegpi;
wherechT=pa1
1pat
t.
REMARK. The same proof gives a slightly more general result:
Ifp=pb1
1pbt
t, then
Kerp(T) = Kerpb1
1(T) Kerpbt
t(T):
We subsequently give an application of the decomposition theorem in
this form to the solution of the n{th order linear di erential equations with
constant coecients. (See Ho man and Kunze, pages 184{185.)
PROOF. Let mT=pbi
iqi8i= 1;:::;t . Then
(qiqj)(T) = 0Vifi6=jasmTjqiqjifi6=j.
Now gcd(q1;:::;qt) = 1, so9f1;:::;ft2F[x] such that
1 =f1q1++ftqt
and withTi= (fiqi)(T) we have
IV=T1++Tt: (6)
Also
TiTj= (fiqi)(T)(fjqj)(T)
= (fifj)(T)(qiqj)(T)
= 0V ifi6=j,
Then
V=tM
i=1ImTi:
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ForT2
i=Ti(T1++Tt) =TiIV=Ti. Next,V= ImT1++ ImTt. For
v2V)v=IV(v) =T1(v) ++Tt(v)2ImT1++ ImTt:
Next assume v1++vt= 0; vi2ImTi;1it:Thenvi=Ti(ui) and
T1(u1) ++Tt(ut) = 0
Ti(T1(u1) ++Tt(ut)) =T(0) = 0
TiTi(ui) = 0
vi=Ti(ui) = 0:
We now show that
ImTi= Kerpbi
i(T):
\" Letv2ImTi. Then
v=fiqiw
)pbi
iv=pbi
ifiqiw
=fi(pbi
iqi)w
= 0:
\" Supposepbi
iv= 0.
Now ifj6=i, we havepbi
ijfjqj, so
Tj(v) =fjqjv= 0:
So
v=IV(v) =T1(v) ++Tt(v) =Ti(v)
2ImTi;
as required.
Finally, let Vi= Kerpbi
i(T) andLi=TVi. Then because V1;:::;Vtare
T{invariant subspaces of V, we have
chT=chL1chLt:
Nowpbi
i(T)(v) = 0 ifv2Vi, sopbi
i(Li) = 0Vi:HencemLihas the form
mLi=pei
i. HencechLihas the form chLi=pdi
i. Hence
chT=pa1
1pat
t=pd1
1pdt
t
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and consequently di=ai.
Finally,
dimVi= degchLi= degpai
i=aidegpi:
(Incidentally, we mention that mT= lcm (mL1;:::;mLt). Hence
mT=pb1
1pbt
t=pe1
1pet
t
and consequently ei=bi. HencemLi=pbi
i.)
THEOREM 3.6 (Commuting diagonable linear transformations)
IfT1;:::;Tn:V!Vare commuting diagonable linear transformations,
then there exists a basis forVsuch that each of [T1]
;:::[Tm]
are each
diagonal.
(Matrix version) If A1;:::;Amare commuting diagonable matrices of the
same size, then there exists a non-singular matrix Psuch that the matrices
P 1A1P;:::P 1AmP
are simultaneously diagonal.
PROOF. (From Samelson page 158). We prove the result when m= 2, the
general case follows by an easy iteration. Suppose T1andT2are commuting
diagonable linear transformations on V. BecausemT1splits as a product
of distinct linear factors, the primary decomposition theorem gives a direct
sum decomposition as a sum of the T1{eigenspaces:
V=U1Ut:
It turns out that not only are the subspaces UiT1{invariant, they are T2{
invariant. For if Ui= Ker (T1 cIV), then
v2Ui)T1(v) =cv
)T2(T1(v)) =cT2(v)
)T1(T2(v)) =cT2(v)
)T2(v)2Ui:
Now because T2is diagonable, Vhas a basis consisting of T2{eigenvectors
and it is an easy exercise to show that in a direct sum of T2{invariant
subspaces, each non-zero "component" of a T2{eigenvector is itself a T2{
eigenvector; moreover each non{zero component is a T1{ eigenvector. Hence
Vis spanned by a family of vectors which are simultaneously T1{eigenvectors
andT2{eigenvectors. If is a subfamily which forms a basis for V, then [T1]
and [T2]
are diagonal.
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THEOREM 3.7 (Fitting's lemma)
SupposeT:V!Vis a linear transformation over Tand
KerKerT2 KerTn= KerTn+1=
ThenV= ImTnKerTn.
COROLLARY 3.1
IfT:V!Vis an indecomposable linear transformation (that is the
onlyT{invariant subspaces of Varef0gandV), thenTis either nilpotent
(that isTn= 0Vfor somen1) orTis an isomorphism.
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