3
6
=
1
3
1
3

2
3
0
.
As

w
3

=
√
6
3
we have
v
3
=
3
√
6
1
3
1
3

2
3
0
=
1
√
6
1
√
6

2
√
6
0
.
w
4
=
u
4

u
4
, v
1
v
1

u
4
, v
2
v
2

u
4
, v
3
v
3
=
1
0
0
0

1
2
1
2
1
2
1
2
1
2

1
2
√
3
1
2
√
3
1
2
√
3
1
2
√
3

3
2
√
3

1
√
6
1
√
6
1
√
6

2
√
6
0
=
1
0
0
0

1
4
1
4
1
4
1
4

1
12
1
12
1
12

3
12

1
6
1
6

2
6
0
=
1
2

1
2
0
0
.
1
As

w
4

=
1
√
2
we have
v
4
=
√
2
1
2

1
2
0
0
=
√
2
2

√
2
2
0
0
.
3. Let
u
1
= 2 + 3
x
2
,
u
2
= 5
x
2

1 and
u
3
= 10

7
x
+ 2
x
2
.
Now
v
1
=
1

u
1

u
1
=
1
√
13
(2 + 3
x
2
).
w
2
=
u
2

u
2
, v
1
v
1
=
5
x
2

1

(2 + 3
x
2
) = 2
x
2

3
.
As

w
2

=
√
13 we have
v
2
=
1
√
13
(2
x
2

3).
w
3
=
u
3

u
3
, v
1
v
1

u
3
, v
2
v
2
=
10

7
x
+ 2
x
2

2(2 + 3
x
2
) + 2(2
x
2

3) =

7
x.
As

w
3

= 7 we have
v
3
=

x
.
4. We start with the basis of
P
2
given by
{
u
1
= 1
,
u
2
=
x,
u
3
=
x
2
}
and use the GramSchmidt
process to construct an orthonormal basis
{
v
1
,
v
2
,
v
3
}
.
Set
v
1
=
u
1

u
1

. As

1

2
=
1

1
1
dx
= [
x
]
1

1
= 2
,
we have
v
1
=
1
√
2
.
Set
w
2
=
u
2

u
2
,
v
1
v
1
. As
u
2
,
v
1
=
1

1
x
√
2
dx
=
x
2
2
√
2
1

1
= 0
,
we have
w
2
=
u
2
=
x.
Now we set
v
2
=
w
2

w
2

. As

w
2

2
=
1

1
x
2
dx
=
x
3
3
1

1
=
2
3
we have
v
2
=
√
3
√
2
x.
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 Fall '13
 Linear Algebra, Algebra, −1, Orthonormal basis