Math 304 Solutions 7
1
Section 6.1
4. Let
A
be a nonsingular matrix and let
λ
be an eigenvalue of
A
. Show
that 1
/λ
is an eigenvalue of
A

1
.
Answer:
Let
v
be the eigenvector for
A
corresponding to the eigen
value
λ
. Geometrically, the result of multiplying a multiple of
v
by
A
is to multiply its length by
λ
. Thus, multiplying by
A

1
will multiply
the length of the vector by 1
/λ
.
Here is a more algebraic proof:
Consider
A

1
(
A
v
):
A

1
(
A
v
) =
v
Since
A
v
=
λ
v
,
A

1
(
A
v
) =
A

1
(
λ
v
)
Thus, we have that
A

1
(
λ
v
) =
v
Thus,
λ
v
is an eigenvector for
A
with eigenvalue 1
/λ
.
10. Show that the matrix
A
=
cos
θ

sin
θ
sin
θ
cos
θ
will have complex eigenvalues if
θ
is not a multiple of
π
. Give a geo
metric interpretation of this result.
Answer:
det(
λI

A
) = (
λ

cos
θ
)
2
+ sin
2
θ
1
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If det(
λI

A
) = 0, then
(
λ

cos
θ
)
2
=

sin
2
θ
If
λ
is real, the left side of this equation is always greater than or equal
to 0, and the right side of this equation is always less than or equal to
zero. The only way for this to be valid is if both sides equal 0. Thus,
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 Spring '08
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 Linear Algebra, Determinant, Characteristic polynomial, Complex number, θ

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