Math 417 homework 9 solutions
Problem 1
Characteristic polynomial:
χ
A
(
z
) = det(
A

zI
) = det

3

z

2
0
2

3

z
0
0
0
4

z
=
(
(

3

z
)
2

2(

2)
)
(4

z
) = ((3 +
z
)
2
+ 4)(4

z
)
.
One root:
z
= 4. The other two are obtained by solving
0 = (3 +
z
)
2
+ 4
which yields
z
=

3
±
2
i.
The matrix is simple: there are three different eigenvalues, each has algebraic
multiplicity 1.
Geometric multiplicity is always
≤
algebraic, so it must be 1
as well.
Hence each eigenspace has dimension 1; it is sufficient to find one
eigenvector.
Eigenvector for first eigenvalue:
A

4
I
=

3

2
0
2

3
0
0
0
0
.
The kernel is spanned by
~v
1
=
0
0
1
.
Eigenvector for second eigenvalue:
A

(

3 + 2
i
)
I
=

2
i

2
0
2

2
i
0
0
0
7

2
i
.
One kernel vector:
~v
2
=
1

i
0
.
1
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Instead of calculating the third kernel vector, we use from class:
Fact: if
A
is real and
A~v
=
λ~v
, then
A
~v
=
λ
~v
. The third eigenvalue is the
conjugate of the second, so the third eigenvector is a conjugate of the second
(or any nonzero multiple of that).
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 Fall '07
 ELLING
 Math, Linear Algebra, Algebra, Matrices, Eigenvalue, eigenvector and eigenspace, kernel vector

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