Math 311 Lecture 21
NOTATION. u7v means u and v are orthogonal. Once
again, we'll write u v instead of (u, v).
DEFINITION. A vector is orthogonal to W iff it is
orthogonal to every vector in W. The orthogonal
complement of W is the set W
7
of all vectors which
are orthogonal to W.
WW
7
THEOREM. For any subspace W of a vector space V,
W
7
is a subspace.
WQW
7
= {0}.
dim(W) + dim(W
7
) = dim(V).
PROOF. Suppose u,vLW
7
& wLW. Want u+v, cuLW
7
.
u,vLW
7
î u, v are 7 to W î u w = v w = 0
î(u+v)
w = u
w + v
w= 0+0 = 0 & (cu)
w= c(u
w)=c
0 = 0
î u+v, cu are 7 to W î u+v, cu L W
7
.
uLWQW
7
î u is perpendicular to itself
î u u = 0 î u = 0. W WQW
7
= {0}. E
CLet V = R3. Let W be the subspace spanned {w1, w2} =
{ }, find a basis for W
7
. Note dim(W
7
) = 3
2 = 1.
1
1
0
,
1
0
1
v = LW
7
iff v7w1 & v7w2 iff v w1 = v w2 = 0 iff
x
y
z
= 0 & = 0 iff .
x
y
z
⋅
1
1
0
x
y
z
⋅
1
0
1
x+y=0
x+z=0
Reducing the augmented matrix gives
é . Thus x = -z, y = z, z arbitrary.
110
101
10 1
01−1
Hence elements of W
7
are of the form .
−z
z
z
=z
−1
1
1
Hence is a basis for W
7
.
−1
1
1
DEFINITION. Suppose W is a subspace with orthonomal
basis {w1, w2, ..., wn}. For any vector v, let
projWv = (v w1)w1+(v w2)w2+...+(v wn)wn.
projWv is the orthogonal projection of v on W.
This definition only applies when the basis is
orthonormal. If you are given a basis which isn’t (as in
homework problems 12 and 14) you must first convert
the basis to an orthonormal basis before using the
definition of projWv.
CLet V = R3. Let W be the subspace spanned {w1, w2} =
. Let v = . Find projWv.
1
1
0
,
1
0
1
1
2
3
Before we can use the definition above, we must get an
orthonormal basis {u1, u2}.
u1 = .
1
2
1
1
0
v2 = w1 (w1 u1)u1 =
1
0
1
−
1
0
1
⋅1
2
1
1
0
1
2
1
1
0
= simplify é
1
2
1
−1
2
1
−1
2
u2 = v2/||v2|| = .
1
6
1
−1
2
Now back to finding projWv, projWv = (v u1)u1 + (v u2)u2
=
1
2
3
⋅1
2
1
1
0
1
2
1
1
0
+
1
2
3
⋅1
6
1
−1
2
1
6
1
−1
2
= = = .
3/2
3/2
0
+
5/6
−5/6
5/3
7/3
2/3
5/3
1
3
7
2
5
THEOREM. For v and W as in the definition, the
component of v orthogonal to W is u = v projWv. u is
inW
7
and
the distance between v and W is ||u|| = ||vprojWv||.
v
proj v
W
W
W
u=v-proj v
CFor W and v as in the previous example, find the
distance between v and W.
The distance = ||v projWv|| =
.
1
2
3
−
7/3
2/3
5/3
=
−4/3
4/3
4/3
=16
9+16
9+16
9=4
33
Hw 19 Answers
3DJH
. Straight forward verification.
. D 7 E 0
. D 13/6 E 3
. D -1/2 E 1
. E O3¯
. D neither E orthonormal