434 CHAPTER 18. LINEAR TRANSFORMATIONS
columns is orthogonal to the rows of Ci since CiB j = 0 if i ̸= j. Therefore,
S−1AS =
C1...
Cr
A(
B1 · · · Br
)=
C1...
Cr
( AB1 · · · ABr
)
=
C1AB1 0 · · · 0
0 C2AB2 · · · 0... 0
. . . 00 · · · 0 CrABr
and Crk ABrk is an rk× rk matrix.
What about the eigenvalues of Crk ABrk ? The only eigenvalue of A restricted to Vλ kis
λ k because if Ax = µx for some x ∈Vλ kand µ ̸= λ k, then
(A−λ kI)rk x = (A−µI +(µ−λ k) I)rk x
=rk
∑j=0
(rk
j
)(µ−λ k)
rk− j (A−µI) j x = (µ−λ k)rk x ̸= 0
contrary to the assumption that x ∈ Vλ k. Suppose then that Crk ABrk x = λx where x ̸= 0.
Why is λ = λ k? Let y = Brk x so y ∈Vλ k. Then
S−1Ay = S−1AS
0...x...0
=
0...
Crk ABrk x...0
= λ
0...x...0
and so
Ay = λS
0...x...0
= λy.
Therefore, λ = λ k because, as noted above, λ k is the only eigenvalue of A restricted to Vλ k.
Now let Pk =Crk ABrk . ■The above theorem contains a result which is of sufficient importance to state as a
corollary.
Corollary 18.4.3 Let A be an n× n matrix and let Dk denote a basis for the generalizedeigenspace for λ k. Then {D1, · · · ,Dr} is a basis for Cn.