Solutions To Linear Algebra Practice Problems Page 11
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18The eigenvector for λ =
i is the complex conjugate. Thus, an
5
eigenvector corresponding to λ =
i is
(and any multi-
3 + i
ple is also an eigenvector).
5
Thus, the eigenvalues and eigenvectors are λ = i,
and
3
i
5
λ =
i,
3 + i
(b)
4
λ
4
2
det(A
λI) =
2
2
λ
2
2
0
0
1
λ
4
λ
4
= (1
λ)
2
2
λ
= (1
λ) ((4
λ)( 2
λ) + 8)
2
= (1
λ)( 2λ + λ
) = λ(1
λ)( 2 + λ)
Thus, the eigenvalues are λ = 0, 1, 2.
Now, we find the eigenvalue corresponding to λ = 0:
4
4
2
1
1
0
A
0I =
2
2
2
0
0
1
0
0
1
0
0
0
1
Thus, an eigenvector corresponding to λ = 0 is
1
. Any multiple
0
of this vector is also an eigenvector.
Now, we find the eigenvalue corresponding to λ = 1:
3
4
2
1
0
2
A
I =
2
3
2
=
0
1
2
0
0
0
0
0
0
11
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