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MATH 223, Linear Algebra
Winter, 2008
Solutions to Assignment 9
1. Let
V
be the vector space of realvalued functions deﬁned and continuous
on [

π
2
,
π
2
]
.
(a) Show that, for any
f
∈
V
,
(
Z
π
2

π
2
f
(
x
)cos
xdx
)
2
≤
2
Z
π
2

π
2
f
(
x
)
2
cos
xdx.
Identify those functions
f
for which equality holds.
Solution: We have seen that
< f,g >
=
R
π
2

π
2
f
(
x
)
g
(
x
)cos
xdx
de
ﬁnes an inner product on the space. So CauchySchwarz applies. In
particular,
< f,
1
>
2
≤
< f,f ><
1
,
1
>
(with equality if and only
if
f
is a constant multiple of 1; that is,
f
is a constant function).
Now
< f,
1
>
=
R
π
2

π
2
f
(
x
)cos
xdx
,
< f,f >
=
R
π
2

π
2
f
(
x
)
2
cos
xdx
and
<
1
,
1
>
=
R
π
2

π
2
cos
xdx
= sin
x
]
π
2

π
2
= 2. The given inequality follows,
and is an equality if and only if
f
is a constant function.
(b) Show that, for any
f
∈
V
,
(
Z
π
2

π
2
f
(
x
)cos
2
xdx
)
2
≤
4
3
Z
π
2

π
2
f
(
x
)
2
cos
xdx.
Identify those function
f
for which equality holds.
Solution: The same basic idea as above, except that now, instead of
taking
g
(
x
) = 1, we take
g
(
x
) = cos
x
. So
< f,
cos
x >
2
≤
< f,f ><
cos
x,
cos
x > . < f,
cos
x >
=
R
π
2

π
2
f
(
x
)cos
2
xdx
,
< f,f >
is still
R
π
2

π
2
f
(
x
)
2
cos
xdx
and
<
cos
x,
cos
x >
=
R
π
2

π
2
cos
3
xdx
=
4
3
. We have equality if and only if
f
(
x
) =
r
cos
x
for some constant
r
.
[Incidentally, to do the integral
R
π
2

π
2
cos
3
xdx
, ﬁrst write
cos
3
x
= cos
x
(1

sin
2
x
) = cos
x

cos
x
sin
2
x
. I presume you can
integrate cos
x
; for the other part, let
u
= sin
x
, so
du
= cos
xdx
and
you’re oﬀ to the races.]
2. (a) Suppose that
V
is any inner product space, and that
~v
and
~w
are
orthogonal vectors in
V
. Show that

~v
+
~w

2
=

~v

2
+

~w

2
.
Solution:

~v
+
~w

2
=
< ~v
+
~w,~v
+
~w >
=
< ~v,~v
+
~w >
+
< ~w,~v
+
~w >
=
< ~v,~v >
+
< ~v, ~w >
+
< ~w,~v >
+
< ~w, ~w > .
As we are assuming
that
~v
and
~w
are orthogonal,
< ~v, ~w >
=
< ~w,~v >
= 0. So

~v
+
~w

2
=
< ~v,~v >
+
< ~w, ~w >
=

~v

2
+

~w

2
,
as advertized.
1
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View Full Document(b) Now suppose that
V
is ﬁnitedimensional and
W
is a subspace of
V
. For any
~u
∈
V
, show that
Proj
W
~u
is the “closest” vector to
~u
that’s in
W
; that is, if
~x
=
Proj
W
~u
and
~
y
6
=
~x
but
~
y
∈
W
, then

~u

~x

<

~u

~
y

.
Solution: We know that
~u

~x
is orthogonal to every vector in
W
.
In particular, it is orthogonal to
~x

~
y
, which is in
W
since both
~x
and
~
y
are. We let
~v
=
~u

~x
and
~w
=
~x

~
y
, so that
~v
and
~w
are
orthogonal, and
~v
+
~w
=
~u

~
y
; note that
~w
6
=
~
0, so that

~w

>
0.
Part (a) applies, so

~v
+
~w

2
=

~v

2
+

~w

2
>

~v

2
; taking square
roots, we see that

~v
+
~w

>

~v

. That is,

~u

~x

<

~u

~
y

.
3. Let
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 Fall '07
 Loveys
 Linear Algebra, Algebra, Vector Space

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