Page 337 - 35Linear Algebra

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The associated eigenvalues solve the homogeneous systems (in augmented
matrix form)

1 −5 0 1 −5 0 5 −5 0 1 −1 0
∼ and ∼ ,
1 −5 0 0 0 0 1 −1 0 0 0 0

5 1
12
12
respectively, so are v 2 = and v −2 = . Hence M v 2 = 2 v 2 and
1 1

x x−y 5 x−5y 1
12
12
M v −2 = (−2) v −2 . Now, = − (this was obtained
y 4 1 4 1
by solving the linear system av 2 + bv −2 = for a and b). Thus

x x − y x − 5y
M = Mv 2 − Mv −2
y 4 4

x − y x − 5y x
= 2 12 v 2 − v −2 = 2 12 .
4 4 y
Thus

4096 0
12
M = .
0 4096
If you understand the above explanation, then you have a good understanding

4 0
2
of diagonalization. A quicker route is simply to observe that M = .
0 4
8.

a − λ b
2
P M (λ) = (−1) det = (λ − a)(λ − d) − bc .
c d − λ
Thus
P M (M) = (M − aI)(M − dI) − bcI

a b a 0 a b d 0 bc 0
= − − −
c d 0 a c d 0 d 0 bc

0 b a − d b bc 0
= − = 0 .
c d − a c 0 0 bc
Observe that any 2 × 2 matrix is a zero of its own characteristic polynomial
(in fact this holds for square matrices of any size).
2
2
Now if A = P −1 DP then A = P −1 DPP −1 DP = P −1 D P. Similarly
k
k
A = P −1 D P. So for any matrix polynomial we have
n
A + c 1 A n−1 + · · · c n−1 A + c n I
n
= P −1 D P + c 1 P −1 D n−1 P + · · · c n−1 P −1 DP + c n P −1 P
n
= P −1 (D + c 1 D n−1 + · · · c n−1 D + c n I)P .
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