linear-guest

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8.
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
1)
1)
respectively, so are v 2 = and v −2 =
( x
M 12 v −2 = (−2) 12 v −2 . Now,
y)
= x−y
4
( 5
1)
− x−5y
4
. Hence M 12 v 2 = 2 12 v 2 and
( 1
(this was obtained
1)
by solving the linear system av 2 + bv −2 = for a and b). Thus
( x
M =
y)
x − y Mv 2 − x − 5y Mv −2
4
4
Thus
= 2 12( x − y
4
M 12 =
v 2 − x − 5y ) (
v −2 = 2 12 x
4
y
( )
4096 0
.
0 4096
)
.
If you understand the above explanation, then you have a good understanding ( ) 4 0
of diagonalization. A quicker route is simply to observe that M 2 = .
0 4
Thus
( )
P M (λ) = (−1) 2 a − λ b
det
= (λ − a)(λ − d) − bc .
c d − λ
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).
Now if A = P −1 DP then A 2 = P −1 DP P −1 DP = P −1 D 2 P . Similarly
A k = P −1 D k P . So for any matrix polynomial we have
A n + c 1 A n−1 + · · · c n−1 A + c n I
= P −1 D n P + c 1 P −1 D n−1 P + · · · c n−1 P −1 DP + c n P −1 P
= P −1 (D n + c 1 D n−1 + · · · c n−1 D + c n I)P .
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