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Chapter 4 of a linear algebra course text associated with Matthews, covering the Jordan canonical form. It defines the subspaces N_{h,p} = Im p^{h-1}(T) ∩ Ker p(T), proves they are nested, and introduces the Matthews dot diagram and conjugate partitions. It then proves the secondary decomposition of Ker p^b(T) into cyclic subspaces, with an inductive independence lemma, and concludes with the Jordan basis and the form P^-1AP = J.
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4 The Jordan Canonical Form
The following subspaces are central for our treatment of the Jordan and
rational canonical forms of a linear transformation T:V!V.
DEFINITION 4.1
WithmT=pb1
1:::pbt
tas before and p=pi,b=bifor brevity, we de ne
Nh;p= Imph 1(T)\Kerp(T):
REMARK. In numerical examples, we will need to nd a spanning family
forNh;p. This is provided by Problem Sheet 1, Question 11(a): we saw
that ifT:U!VandS:V!Ware linear transformations, then If
Kerph(T) =hu1;:::;uni, then
Nh;p=hph 1u1;:::;ph 1uni;
where we have taken U=V=Wand replaced SandTbyp(T) and
ph 1(T) respectively, so that ST=ph(T). Also
dim( ImT\KerS) =(ST) (T):
Hence
h;p= dimNh;p
= dim( Im ph 1(T)\Kerp(T))
=(ph(T)) (ph 1(T)):
THEOREM 4.1
N1;pN2;pNb;p6=f0g=Nb+1;p=:
PROOF. Successive containment follows from
ImLh 1ImLh
withL=p(T).
The fact that Nb;p6=f0gand thatNb+1;p=f0gfollows directly from
the formula
dimNh;p=(ph(T)) (ph 1(T)):
For simplicity, assume that pis linear, that is that p=x c. The general
story (when deg p>1) is similar, but more complicated; it is delayed until
the next section.
Telescopic cancellation then gives
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THEOREM 4.2
1;p+2;p++b;p=(pb(T)) =a;
wherepais the exact power of pdividingchT.
Consequently we have the decreasing sequence
1;p2;pb;p1:
EXAMPLE 4.1
SupposeT:V7!Vis a LT such that p4jjmT; p=x cand
(p(T)) = 3; (p2(T)) = 6;
(p3(T)) = 8; (p4(T)) = 10:
So
Kerp(T)Kerp2(T)Kerp3(T)Kerp4(T) = Kerp5(T) =:
Then
1;p= 3; 2;p= 6 3 = 3;
3;p= 8 6 = 2; 4;p= 10 8 = 2
so
N1;p=N2;pN3;p=N4;p6=f0g:
4.1 The Matthews' dot diagram
We would represent the previous example as follows:
4;p
3;p
2;p
1;pDots represent dimension:
3 + 3 + 2 + 2 = 10
)10 = 4 + 4 + 2
The conjugate partition of 10 is 4+4+2 (sum of column heights of diagram),
and this will soon tell us that there is a corresponding contribution to the
Jordan canonical form of this transformation, namely
J4(c)J4(c)J2(c):
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In general,
heightb8
>>>>>>>><
>>>>>>>>:b;p
1;p
|{z}
columns
and we label the conjugate partition by
e1e2e
:
Finally, note that the total number of dots in the dot diagram is (pb(T)),
by Theorem 4.2.
THEOREM 4.3
9v1;:::;v
2Vsuch that
pe1 1v1;pe2 1v2;:::;pe
1v
form a basis for Kerp(T).
PROOF. Special case, but the construction is quite general.
choose a basis p3v1;p3v2forN4;p
extend to a basis p3v1;p3v2;pv3forN2;p
Thenp3v1; p3v2; pv3is a basis for N1;p= Kerp(T).
THEOREM 4.4 (Secondary decomposition)
(i)
mT;vi=pei
(ii)
Kerpb(T) =CT;v1CT;v
PROOF.
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(i) We have pei 1vi2Kerp(T), sopeivi= 0 and hence mT;vijpei. Hence
mT;vi=pf, where 0fei.
Butpei 1vi6= 0, as it is part of a basis. Hence feiandf=eias
required.
(ii) (a)
CT;viKerpb(T):
Forpeivi= 0 and so pei(fvi) = 08f2F[x]. Hence as eib,
we have
pb(fvi) =pb ei(peifvi) =pb ei0 = 0
andfvi2Kerpb(T). Consequently CT;viKerpb(T) and hence
CT;v1++CT;v
Kerpb(T):
(b) We presently show that the subspaces CT;vj,j= 1;:::;
are in-
dependent, so
dim(CT;v1++CT;v
) =
X
j=1dimCT;vj
=
X
j=1degmT;vj=
X
j=1ej
=(pb(T))
= dim Ker pb(T):
Hence
Kerpb(T) =CT;v1++CT;v
=CT;v1CT;v
:
The independence of the CT;viis stated as a lemma:
Lemma : Letv1;:::;v
2V; e 1e
1;
mT;vj=pej1j
;p=x c;
Alsope1 1v1;:::;pe
1v
are LI. Then
f1v1++f
v
= 0 ;f1;:::;f
2F[x]
)pejjfj1j
:
Proof : (induction on e1)
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Firstly, consider e1= 1. Then
e1=e2==e
= 1:
NowmT;vj=pejand
pe1 1v1;:::;pe
1v
are LI, sov1;:::;v
are LI. So assume
f1v1++f
v
= 0f1;:::;f
2F[x]: (7)
and by the remainder theorem
fj= (x c)qj+fj(c): (8)
Thus
fjvj=qj(x c)vj+fj(c)vj
=fj(c)vj:
So (7) implies
f1(c)v1++f
(c)v
(c) = 0
)fj(c) = 08j= 1;:::;
and (8) implies
(x c)jfj8j
which is the result.
Now lete1>1 and assume the lemma is true for e1 1. If
mT;vj=pej;
pe1 1v1;:::;pe
1v
are LI,
andf1v1++f
v
= 0 (9)
as before, we have
f1(pv1) ++f
(pv
) = 0 (10)
wheremT;pvj=pej 1.
Now letbe the greatest positive integer such that e>1; i.e.
e+1= 1, bute>1. Applying the induction hypothesis to (10),
in the form
f1(pv1) ++f(pv) = 0
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we obtain
pej 1jfj8j= 1;:::;;
so we may write
fj=pej 1gj;
(where ifgj=fjifj > ). Now substituting in (9),
g1pe1 1v1++g
pe
1v
= 0: (11)
But
mT;pej 1vj=p
so (11) and the case e1= 1 give
pjgj8j;
as required.
A summary :
IfmT= (x c1)b1:::(x ct)bt=pb1
1:::pbt
t;then there exist vectors vij
and positive integers eij(1it;1j
i), where
i=(T ciIV);
satisfying
bi=ei1ei
i; mT;vij=peij
i
and
V=tM
i=1
iM
j=1CT;vij:
We choose the elementary Jordan bases
ij:vij;(T ciIV)(vij);:::; (T ciIV)eij 1(vij)
forCT;vij. Then if
=t[
i=1
i[
j=1 ij;
is a basis for Vand we have
[T]
=tM
i=1
iM
j=1Jeij(ci) =J:
A direct sum of elementary Jordan matrices such as Jis called a Jordan
canonical form of T.
IfT=TAandP= [v11j. . . . . .jvt
t], then
P 1AP=J
andJis called a Jordan canonical form of A.
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