Linear Algebra, Theory And Applications, 2012a

260 LINEAR TRANSFORMATIONS CANONICAL FORMS
⎛
B ≡
⎜
⎝
J ′ (λ 1 ) − λ k I 0
. ..
J ′ (λ k ) − λ k I
. ..
0 J ′ (λ r ) − λ k I
and consequently, rank ( A k) = rank ( B k) for all k ∈ N. Also, both J (λ j ) − λ k I and
J ′ (λ j ) − λ k I are one to one for every λ j ≠ λ k . Since all the blocks in both of these matrices
are one to one except the blocks J ′ (λ k ) − λ k I, J (λ k ) − λ k I, it follows that this requires the
two sequences of numbers {rank ((J (λ k ) − λ k I) m )} ∞ m=1 and { rank ( (J ′ (λ k ) − λ k I) m)} ∞
m=1
must be the same.
Then
⎛
J (λ k ) − λ k I ≡ ⎜
⎝
and a similar formula holds for J ′ (λ k )
⎛
J ′ (λ k ) − λ k I ≡ ⎜
⎝
J k1 (0) 0
J k2 (0)
. ..
0 J kr (0)
J l1 (0) 0
J l2 (0)
. ..
0 J lp (0)
and it is required to verify that p = r and that the same blocks occur in both. Without
loss of generality, let the blocks be arranged according to size with the largest on upper left
corner falling to smallest in lower right. Now the desired conclusion follows from Corollary
10.4.5.
Note that if any of the generalized eigenspaces ker (A − λ k I) m k
has a basis of eigenvectors,
then it would be possible to use this basis and obtain a diagonal matrix in the
block corresponding to λ k . By uniqueness, this is the block corresponding to the eigenvalue
λ k . Thus when this happens, the block in the Jordan canonical form corresponding to λ k
is just the diagonal matrix having λ k down the diagonal and there are no generalized
eigenvectors.
The Jordan canonical form is very significant when you try to understand powers of a
matrix. There exists an n × n matrix S 1 such that
A = S −1 JS.
Therefore, A 2 = S −1 JSS −1 JS = S −1 J 2 S and continuing this way, it follows
A k = S −1 J k S.
where J is given in the above corollary. Consider J k . By block multiplication,
⎛
⎞
J1 k 0
J k ⎜
= ⎝
. ..
⎟
⎠ .
0 Jr
k
1 The S here is written as S −1 in the corollary.
⎞
⎟
⎠
⎞
⎟
⎠
⎞
⎟
⎠