Graphics Reference
In-Depth Information
Then
1
1
11
(
)
u
=
w
=
131
,, .
2
w
One can easily check that
u
1
and
u
2
are what we want.
Definition.
Let
X
be a subspace of an inner product space
V
. The
orthogonal com-
plement
of
X
in
V
, denoted by
X
^
, is defined by
^
{
}
Xv
=Œ
Vv
∑=
w
0 for all
wX
Œ
.
Every vector in
X
^
is called a
normal vector
for
X
.
1.4.4. Theorem.
If
X
is a subspace of an inner product space
V
, then the orthogo-
nal complement
X
^
of
X
is a subspace of
V
and
=≈
^
.
VX X
Conversely, if
VX Y
=≈,
where
Y
is a subspace with the property that every vector in
Y
is normal to
X
, then
Y
=
X
^
.
Proof.
It is an easy exercise, left to the reader, to show that
X
^
is a subspace. We
prove that
V
is a direct sum of
X
and
X
^
. Let
u
1
,
u
2
,...,
u
k
be an orthonormal basis
for
X
. Define a linear transformation T :
V
Æ
V
by
()
=∑
(
)
(
)
(
)
(
)
T
vvuuvuu
+∑
++∑
...
vuu
or
0
if k
=
0
.
11
2 2
kk
It is easy to check that ker(T) =
X
^
and that
v
- T(
v
) belongs to ker(T). We also have
that
()
+-
()
(
)
v
=
T
v
v
T
v
.
These facts imply the first part of the theorem. We leave the reader to fill in the details
and to prove the converse part (Exercise 1.4.1).
Definition.
An inner product space
V
is said to be the
orthogonal direct sum
of
two subspaces
X
and
Y
if it is a direct sum of
X
and
Y
and if every vector of
X
is
orthogonal to every vector of
Y
.
By Theorem 1.4.4, if
V
is an orthogonal direct sum of
X
and
Y
, then
Y
=
X
^
. Another consequence of Theorem 1.4.4 is that subspaces can be defined
implicitly.
