Math 33a/2, Quiz 6bd, December 6, 2007
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1. Let
A
be the matrix
±
1
3
2

4
²
. Find a general expression for
A
t
±
14
7
²
, where
t
is an
arbitrary integer.
Solution.
Since
A
is a 2 by 2 matrix, we can quickly write down its characteristic
polynomial:
f
A
(
λ
) =
λ
2

(Trace
A
)
λ
+ (det
A
) =
λ
2
+ 3
λ

10 = 0. This factors as (
λ
+
5)(
λ

2) = 0, so the eigenvalues are

5
,
2.
To ﬁnd an eigenvector for
λ
=

5, we compute ker(
A

(

5)
I
) = ker
±
6 3
2 1
²
. Using
the usual methods, we obtain an eigenvector
±

1
2
1
²
.
Since any nonzero multiple is also
acceptable, we will instead take the eigenvector
v
1
=
±

1
2
²
.
To ﬁnd an eigenvector for
λ
= 2, we compute ker(
A

2
I
) = ker
±

1
3
2

6
²
. We obtain
an eigenvector
v
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 Fall '08
 lee
 Math, Linear Algebra, eigenvector, matrix arbitrary integer

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