Solutions to Homework 12.
Math 110, Fall 2006.
Prob 6.1.1.
(a)
True, directly by definition.
(b)
True, this is also required by the definition.
(c)
False, it
is linear in the first component and conjugate linear in the second component.
(d)
False, there are in fact
infinitely many inner products.
(e)
False, it also holds in infinite dimensional spaces (see the proof, which
does not use finitedimensionality).
(f)
False, every rectangular matrix has a conjugate transpose (defined
the usual way).
(g)
False when
x
is fixed.
(h)
True when
x
runs over the whole space.
Prob 6.1.3.
We have:
f, g
=
1
0
te
t
dt
= (
t

1)
e
t
1
0
= 1
,
f
2
=
1
0
t
2
dt
=
t
3
3
1
0
=
1
3
,
so
f
=
1
√
3
,
g
2
=
1
0
e
2
t
dt
=
e
2
t
2
1
0
=
e
2

1
2
,
so
g
=
e
2

1
2
,
f
+
g
2
=
1
0
(
t
+
e
t
)
2
dt
=
1
0
(
t
2
+ 2
e
t
+
e
2
t
)
dt
=
3
e
2
+ 11
6
,
so
f
+
g
=
3
e
2
+ 11
6
.
Since
e
2
>
7, we can confirm that the CauchySchwarz inequality holds:

f, g

= 1
≤
e
2

1
6
. Likewise,
f
+
g
=
3
e
2
+ 11
6
≤
1
√
3
+
e
2

1
2
=
f
+
g .
The intermediate inequality can be checked numerically or by squaring both sides:
1
3
+
e
2

1
2
+ 2
e
2

1
6
=
3
e
2

1
6
+ 2
e
2

1
6
≥
3
e
2

1
6
+ 2 =
3
e
2
+ 11
6
.
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 Spring '08
 VALDIMARSSON
 Math, Linear Algebra, dt, PROB

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