EE263s Summer 200910
Laurent Lessard
EE263s homework 4
1.
Orthogonal matrices.
(a) Show that if
U
and
V
are orthogonal, then so is
UV
.
(b) Show that if
U
is orthogonal, then so is
U

1
.
(c) Suppose that
U
∈
R
2
×
2
is orthogonal. Show that
U
is either a rotation or a reflection. Make
clear how you decide whether a given orthogonal
U
is a rotation or reflection.
Solution.
(a) To prove that
UV
is orthogonal we have to show that (
UV
)
T
(
UV
) =
I
given
U
T
U
=
I
and
V
T
V
=
I
. We have
(
UV
)
T
(
UV
)
=
V
T
U
T
UV
=
V
T
V
(since
U
T
U
=
I
)
=
I
(since
V
T
V
=
I
)
and we are done.
(b) Since
U
is square and orthogonal we have
U

1
=
U
T
and therefore by taking inverses of both
sides
U
= (
U
T
)

1
or equivalently
U
= (
U

1
)
T
(the inverse and transpose operations commute.)
But
U
T
U
=
I
and by substitution
U

1
(
U

1
)
T
=
I
. Since
U

1
is square this also implies that
(
U

1
)
T
U

1
=
I
so
U

1
is orthogonal.
(c) Suppose that
U
=
bracketleftbigg
a
b
c
d
bracketrightbigg
∈
R
2
×
2
is orthogonal. This is true if and only if
•
columns of
U
are of unit length,
i.e.
,
a
2
+
c
2
= 1 and
b
2
+
d
2
= 1,
•
columns of
U
are orthogonal,
i.e.
,
ab
+
cd
= 0.
Since
a
2
+
c
2
= 1 we can take
a
and
c
as the cosine and sine of an angle
α
respectively,
i.e.
,
a
= cos
α
and
c
= sin
α
.
For a similar reason, we can take
b
= sin
β
and
d
= cos
β
.
Now
ab
+
cd
= 0 becomes
cos
α
sin
β
+ sin
α
cos
β
= 0
or
sin(
α
+
β
) = 0
.
The sine of an angle is zero if and only if the angle is an integer multiple of
π
. So
α
+
β
=
kπ
or
β
=
kπ
−
α
with
k
∈
Z
. Therefore
U
=
bracketleftbigg
cos
α
sin(
kπ
−
α
)
sin
α
cos(
kπ
−
α
)
bracketrightbigg
.
Now two things can happen:
•
k
is even so sin(
kπ
−
α
) =
−
sin
α
and cos(
kπ
−
α
) = cos
α
, and therefore
U
=
bracketleftbigg
cos
α
−
sin
α
sin
α
cos
α
bracketrightbigg
.
Clearly, from the lecture notes, this represents a rotation.
Note that in this case det
U
=
cos
2
α
+ sin
2
α
= 1.
1
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•
k
is odd so sin(
kπ
−
α
) = sin
α
and cos(
kπ
−
α
) =
−
cos
α
, and therefore
U
=
bracketleftbigg
cos
α
sin
α
sin
α
−
cos
α
bracketrightbigg
.
From the lecture notes, this represents a reflection. The determinant in this case is det
U
=
−
cos
2
α
−
sin
2
α
=
−
1.
Therefore we have shown that any orthogonal matrix in
R
2
×
2
is either a rotation or reflection
whether its determinant is +1 or
−
1 respectively.
2.
Projection matrices.
A matrix
P
∈
R
n
×
n
is called a
projection matrix
if
P
=
P
T
and
P
2
=
P
.
(a) Show that if
P
is a projection matrix then so is
I
−
P
.
(b) Suppose that the columns of
U
∈
R
n
×
k
are orthonormal. Show that
UU
T
is a projection matrix.
(Later we will show that the converse is true: every projection matrix can be expressed as
UU
T
for some
U
with orthonormal columns.)
(c) Suppose
A
∈
R
n
×
k
is full rank, with
k
≤
n
. Show that
A
(
A
T
A
)

1
A
T
is a projection matrix.
(d) If
S
⊆
R
n
and
x
∈
R
n
, the point
y
in
S
closest to
x
is called the
projection of
x
on
S
. Show
that if
P
is a projection matrix, then
y
=
Px
is the projection of
x
on
R
(
P
). (Which is why such
matrices are called projection matrices . . . )
Solution.
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 Summer '08
 BOYD,S
 Linear Algebra, Eigenvalue, eigenvector and eigenspace, Orthogonal matrix, projection matrix

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