1
S
S
⊥
.
0
Example:
Example:
S
=
R
n
⇒
S
⊥
= {
0
},
S
= {
0
}
⇒
S
⊥
=
R
n
.
Example:
W
= { [
w
1
w
2
0 ]
T

w
1
w
2
∈
R
.}
V
= { [ 0
0
v
3
]
T

v
3
∈
R
.}
⇒
V
=
W
⊥
:
(1) for all
v
∈
V
and
w
∈
W
,
v
•
w
= 0
⇒
V
⊆
W
⊥
;
(2) since
e
1
,
e
2
∈
W
, all
z
= [
z
1
z
2
z
3
]
T
∈
W
⊥
must have
z
1
=
z
2
= 0 by
z
•
e
1
=
z
•
e
2
= 0
⇒
W
⊥
⊆
V
.
Proposition: For any nonempty subset
S
of
R
n
, (Span
S
)
⊥
=
S
⊥
.
Proof
You show it.
Corollary:
W
: a subspace of
R
n
, and
B
: a basis of
W
.
Then
B
⊥
=
W
⊥
.
Example: For
W
= Span{
u
1
,
u
2
}, where
u
1
= [ 1
1
−
1
4 ]
T
and
u
2
=[ 1
−
1
1
2 ]
T
,
v
∈
W
⊥
if and only if
u
1
•
v
=
u
2
•
v
= 0
i.e.,
v
= [
x
1
x
2
x
3
x
4
]
T
satisfies
Property:
1.
0
∈
S
⊥
for every nonempty subset
S
of
R
n
.
2. If
v
,
w
∈
S
⊥
for a nonempty subset
S
of
R
n
, then
v
+
w
∈
S
⊥
and
c
v
∈
S
⊥
for any scalar
c
∈
R
.
Proof
First part of 2: (
v
+
w
)
•
u
=
v
•
u
+
w
•
u
= 0 + 0 = 0,
∀
u
∈
S
.
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⇔
⇔
is a basis for
W
⊥
.
Proof
v
∈
(Row
A
)
⊥
⇔
w
•
v
= 0 for all
w
∈
Span{rows of
A
}
⇔
A
v
=
0
.
Also, (Col
A
)
⊥
= (Row
A
T
) = Null
A
T
.
For any real matrix
A
, (Row
A
)
⊥
= Null
A
, or (Col
A
)
⊥
= Null
A
T
.
*
For a complex matrix
A
and orthogonal complement defined by
the complex dot product, (Row
A
)
⊥
= Null
A
and (Col
A
)
⊥
= Null
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 Spring '09
 Fong
 Linear Algebra, orthogonal projection, Rn ⇒

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