Linear Algebra Notes Chapter 9 MULTIPLE EIGENVALUES AND ...

3
as claimed in Proposition 2.
□
so
Now if you want to compute A n , you could first note that
A n = (B
[ ] n [ ]
λ 1 λ
n
nλ
=
n−1
0 λ 0 λ n
[ ]
[ ]
λ 1
B −1 ) n λ
n
nλ
= B
n−1
0 λ
0 λ n B −1 .
(9b)
(9c)
Actually, there is a much better formula for A n , but first let’s have an example to
illustrate what we’ve seen so far.
Example:
A =
[ ]
0 4
.
−1 4
We find P A (x) = (x − 2) 2 so λ = 2 is a multiple eigenvalue of A. Our formula for
eigenvectors gives (b, λ − a) = (4, 2). Scaling, we can take the eigenvector to be
u = (2, 1). Now we choose any another vector v which is not proportional to u.
Let us take v = e 1 , so we have
B 1 =
[ ]
2 1
.
1 0
We then compute
[ ]
B1 −1 0 1
= ,
1 −2
and
[
B1 −1 0 1
AB 1 =
1 −2
] [
0 4
−1 4
] [ ]
2 1
=
1 0
[ ]
2 −1
.
0 2
So g = −1 and we take
[ ]
−1 0
B = B 1 =
0 1
[ ]
−2 1
.
−1 0
To check our calculations we compute
[
−2
]
1
−1 0
B −1 AB =
[ ]
2 1
.
0 2
So B = does the job, as predicted by Proposition 2.
[ ] n
2 1
Continuing further using equation (9c), we have =
0 2
A n = B
[ ] [ ]
2
n
n2 n−1
0 2 n B −1 2
=
n − n2 n n2 n+1
−n2 n−1 2 n + n2 n .
[ ]
2
n
n2 n−1
0 2 n , so