Math 415 — Midterm 2 practice solutions
Problem 1:
The given definition does not satisfy the positivity condition
that
h
f, f
i
>
0 if
f
6
= 0. In fact consider the continuous function defined as
f
(
x
) =
(
x
for

1
< x <
0
0 for 0
< x <
1
Calculating
h
f, f
i
for this function we get
h
f, f
i
=
Z
1

1
f
2
(
x
)
x
d
x
=
Z
0

1
x
2
x
d
x
=

1
/
4
<
0
Problem 2:
First notice that the limits of integration in the definition of
the inner product in this problem are different from the ones in the previous
problem. This one is a valid inner product. Using this inner product we start
by calculating the norm of
w
1
, as
p
h
w
1
,
w
1
i
= 1
/
√
2. Therefore
u
1
=
w
1

w
1

=
√
2
Then we define
v
2
=
w
2
 h
w
2
,
u
1
i
u
1
and divide it by its norm to get
u
2
=
v
2

v
2

= 6(
x

2
3
)
Finally
v
3
=
w
3
 h
w
3
,
u
1
i
u
1
 h
w
3
,
u
2
i
u
2
and
u
3
=
v
3

v
3

= 10
√
6
x
2

6
5
x
+
3
10
Problem 3:
a) The matrix is not symmetric, so it can’t be positive definite
(We consider positive definite matrices to be symmetric).
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 Spring '08
 BERTRANDGUILLOU
 Math, Linear Algebra, Algebra, Vector Motors, inner product, orthogonal complement, positive definite

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