043 uniq
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Section 4.3 of a linear algebra text, apparently from Matthews' notes, in a folder of linear algebra materials. It orders the Jordan blocks by eigenvalue and decreasing size, defines the Segre and Weyr characteristics and elementary divisors, and uses nullities of powers of (T - c_k) to prove the block sizes are determined. It includes a lemma on nilpotent Jordan blocks, a similarity remark, an example showing equal characteristic and minimum polynomials do not imply similarity, and an exercise on 2x2 and 3x3 matrices.
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4.3 Uniqueness of the Jordan form
Let be a basis for Vfor which [T]
is in Jordan canonical form
J=Je1(1)Jes(s):
If we change the order of the basis vectors in , we produce a corresponding
change in the order of the elementary Jordan matrices. It is customary to
assume our Jordan forms arranged so as to group together into a block those
elementary Jordan matrices having the same eigenvalue ci:
J=J1Jt;
where
Ji=
iM
j=1Jeij(ci):
Moreover within this i{th blockJi, we assume the sizes ei1;:::;ei
iof the
elementary Jordan matrices decrease monotonically:
ei1:::ei
i:
We prove that with this convention, the above sequence is uniquely deter-
mined byTand the eigenvalue ci.
We next observe that
chT=chJ=tY
i=1chJi=tY
i=1
iY
j=1(x ci)eij=tY
i=1(x ci)ei1++ei
i:
Hencec1;:::;ctare determined as the distinct eigenvalues of T.
DEFINITION 4.2
The numbers ei1;:::;ei
i;1it, are called the Segre characteristic
ofT, while the numbers 1;x ci;:::;bi;x ci;1itare called the Weyr
characteristic of T.
The polynomials (x ci)eijare called the elementary divisors ofT.
LEMMA 4.1
Let
A=Je(0) =2
666666640 0 0
1 0
0 1
.........
0 00 0
0 0 1 03
77777775:
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Then
(Ah) =hif1he 1;
eifeh:
Proof.Ahhas 1 on the h{th sub{diagonal, 0 elsewhere, if 1 he 1,
whereasAh= 0 ifhe.
Consequently
(Ah) (Ah 1) =1 if 1he;
0 ife<h:
We now can prove that the sequence ei1:::ei
iis determined uniquely
byTand the eigenvalue ci.
Letpk=x ckand
A= [T]
=tM
i=1
iM
j=1Jeij(ci):
Then
(ph
k(T)) =(ph
k(A))
=(tM
i=1
iM
j=1ph
k(Jeij(ci)))
=(tM
i=1
iM
j=1Jh
eij(ci ck))
=tX
i=1
iX
j=1(Jh
eij(ci ck));
where we have used the fact that
pk(Jeij(ci)) =Jeij(ci) ckIn=Jeij(ci ck):
HoweverJeij(ci ck) is a non{singular matrix if i6=k, so
(Jh
eij(ci ck)) = 0
ifi6=k. Hence
(ph
k(T)) =
kX
j=1(Jh
ekj(0)):
76
Hence
h;x ck=(ph
k(T)) (ph 1
k(T)) =
kX
j=1
(Jh
ekj(0)) (Jh 1
ekj(0))
=
kX
j= 1
hekj1:
Consequently h;x ck h+1;x ckis the number of ekjwhich are equal to
h. Hence by taking h= 1;:::; we see that the sequence ek1;:::;ek
kis
determined by Tandckand is in fact the contribution of the eigenvalue ck
to the Segre characteristic of T.
REMARK. If AandBare similar matrices over F, thenB=P 1APsay.
AlsoAandBhave the same characteristic polynomials. Then if ckis an
eigenvalue of AandBandpk=x ck, we have
ph
k(TB) =ph
k(B) =P 1ph
k(A)P=P 1ph
k(TA)P
and hence
(ph
k(TB)) =(ph
k(TA))
for allh1.
Consequently the Weyr characteristics of TAandTBwill be identical.
Hence the corresponding dot diagrams and so the Segre characteristics will
also be identical. Hence TAandTBhave the same Jordan form.
EXAMPLE 4.2
LetA=J2(0)J2(0)andB=J2(0)J1(0)J1(0). Then
chA=chB=x4andmA=mB=x2:
HoweverAis not similar to B. For both matrices are in Jordan form and
the Segre characteristics for TAandTBare2;2and2;1;1, respectively.
EXERCISE List all possible Jordan canonical forms of 2 2 and 33
matrices and deduce that if AandBhave the same characteristic and same
minimum polynomials, then AandBare similar if AandBare 22 or
33.
REMARK. Of course if AandBhave the same Jordan canonical form, then
AandBare similar.
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