CALIFORNIA INSTITUTE OF TECHNOLOGY
Department of Mathematics
Math 1b; Solutions to Homework Set 7
Due: March 3, 2008
1. Let
A
=
a
b
c
d
¶
be a real 2 by 2 matrix. Then
A
is orthogonal iff
AA
t
=
I
, or equivalently
A
t
=
A

1
.
Further as
A
is real, Theorem 5.19 says det(
A
) =
²
=
±
1. Now from Problem 4 on HW1,
det(
A
) =
ad

bc
and
A

1
=
²
·
d

b

c
a
¶
.
Thus
A
is orthogonal iff
a
c
b
d
¶
=
A
t
=
A

1
=
²
·
d

b

c
a
¶
,
which holds iff
a
=
²d
and
b
=

²c
. That is
A
is orthogonal iff
A
=
a

²c
c
²a
¶
and
a
2
+
c
2
= 1 since
²
= det(
A
) =
²
(
a
2
+
c
2
). As
a
and
c
are real, this last equation is
equivalent to
a
= cos(
θ
) and
c
= sin(
θ
) for some angle
θ
. Therefore (a) is established,
since
²
= 1 if
A
is proper.
Next
A
is improper iff
²
=

1 and hence
A
is improper iff
A
=
cos(
θ
)
sin(
θ
)
sin(
θ
)

cos(
θ
)
¶
.
Thus we have our description of improper matrices.
Further the two matrices in (b)
correspond to the cases
θ
= 0 and 180 degrees, respectively, so they are improper.
2.
(a) First char
A
(
x
) =
x
2

tr
(
A
) + det(
A
) =
x
2
+
a
2
= (
x

ia
)(
x
+
ia
), so the
eigenvalues of
A
are
±
ia
, and they are distinct as
a
6
= 0. Thus by Theorem 4.10,
A
is
similar to the diagonal matrix
D
=
ia
0
0

ia
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 Winter '07
 Aschbacher
 Math, Linear Algebra, Algebra, Eigenvalue, eigenvector and eigenspace, Orthogonal matrix, xk, )mk ,

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