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Partial Solution Set, Leon
§
5.5
5.5.2
(a) Straight forward computations
(b) We have
u
1
=
±
1
3
√
2
,
1
3
√
2

4
3
√
2
²
T
,
u
2
=
(
2
3
,
2
3
,
1
3
)
T
, and
u
3
=
±
1
√
2
,
1
√
2
,
0
²
T
. Let
x
=
(1
,
1
,
1)
T
. Write
x
as a linear combination of
u
1
,
u
2
, and
u
3
, and use Parseval’s formula
to compute

x

.
Solution
: We know from part (a) that [
u
1
,
u
2
,
u
3
] is an orthonormal basis for
R
3
. By
Theorem 5.5.2, we know that
x
= (
x
T
u
1
)
u
1
+ (
x
T
u
2
)
u
2
+ (
x
T
u
3
)
u
3
=

2
3
√
2
u
1
+
5
3
u
2
+ 0
u
3
=

2
3
√
2
u
1
+
5
3
u
2
By Parseval’s formula,

x

=
(
4
18
+
25
9
)
1
/
2
=
√
3.
5.5.3
We are given
S
, the subspace spanned by
u
2
and
u
3
of the preceding exercise, and
x
= (1
,
2
,
2)
T
. We are to ﬁnd the projection
p
of
x
onto
S
, and to verify that
p

x
∈
S
⊥
.
Solution
: The projection is
p
= (
x
T
u
2
)
u
2
+ (
x
T
u
3
)
u
3
=
8
3
u
2

1
√
2
u
3
=
³
23
18
,
41
18
,
8
9
´
T
So
p

x
= (
5
18
,
5
18
,

10
9
)
T
. It is easy to show that
p
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This note was uploaded on 07/17/2011 for the course MATH 311 taught by Professor Anshelvich during the Spring '08 term at Texas A&M.
 Spring '08
 Anshelvich
 Linear Algebra, Algebra

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