Math 417 homework 8 solutions
Problem 1
The matrix is
cos
α

sin
α
sin
α
cos
α
.
Characteristic polynomial:
(cos
α

z
)
2
+ sin
2
α
Roots: solve
(cos
α

λ
)
2
+ sin
2
α
= 0
.
Get:
(cos
α

λ
)
2
=

sin
2
α,
so
cos
α

λ
=
±
i
sin
α,
so
λ
= cos
α
±
i
sin
α.
Problem 2
Result: eigenvalues

1 (algebraic multiplicity 2) and 2 (algebraic multiplicity
1). For the first eigenvalue, we obtain an eigenvector
0
1
0
.
The dimension of this eigenspace, i.e. the geometric multiplicity of

1, is 1, less
than the algebraic multiplicity. For the other eigenvalue, an eigenvector is
1
0

1
.
Again, geometric multiplicity 1.
In this case the matrix is not diagonalizable because we cannot find a third
eigenvector that is not redundant.
1
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Problem 3
First write the righthand side in polar form:
w
=

16 = 16
·
e
iπ
=
re
iφ
with
r
= 16,
φ
=
π
. Now use the formula from class: the
n
solutions are
z
=
r
1
/n
exp
i
(
φ
+ 2
πk
)
n
where
k
= 0
, . . . , n

1 (with
n
= 8). So:
k
z
0
√
2
e
iπ
8
1
√
2
e
3
iπ
8
2
√
2
e
5
iπ
8
3
√
2
e
7
iπ
8
4
√
2
e
9
iπ
8
5
√
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 Fall '07
 ELLING
 Math, Matrices

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