9.5. THE JORDAN CANONICAL FORM 249

The Jordan canonical form is very significant when you try to understand powers of amatrix. There exists an n× n matrix S1 such that

A = S−1JS.

Therefore, A2 = S−1JSS−1JS = S−1J2S and continuing this way, it follows

Ak = S−1JkS.

where J is given in the above corollary. Consider Jk. By block multiplication,

Jk =

Jk1 0

. . .

0 Jkr

 .

The matrix Js is an ms ×ms matrix which is of the form

Js = D +N

for D a multiple of the identity and N an upper triangular matrix with zeros down themain diagonal. Thus Nms = 0. Now since D is just a multiple of the identity, it followsthat DN = ND. Therefore, the usual binomial theorem may be applied and this yields thefollowing equations for k ≥ ms.

Jks = (D +N)

k=

k∑j=0

(k

j

)Dk−jN j

=

ms∑j=0

(k

j

)Dk−jN j , (9.9)

the third equation holding because Nms = 0. Thus Jks is of the form

Jks =

αk · · · ∗...

. . ....

0 · · · αk

 .

Lemma 9.5.3 Suppose J is of the form Js, a Jordan block where the constant α, on themain diagonal is less than one in absolute value. Then

limk→∞

(Jk)ij= 0.

Proof: From 9.9, it follows that for large k, and j ≤ ms,(k

j

)≤ k (k − 1) · · · (k −ms + 1)

ms!.

Therefore, letting C be the largest value of∣∣∣(N j

)pq

∣∣∣ for 0 ≤ j ≤ ms,∣∣∣(Jk)pq

∣∣∣ ≤ msC

(k (k − 1) · · · (k −ms + 1)

ms!

)|α|k−ms

1The S here is written as S−1 in the corollary.