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6.6 Orthogonal Complements and Projections 177

Finally, let

X 3 · V 1 X 3 · V 2
V 3 =X 3 − V 1 − V 2
V 1 2 V 2 2
= < 1,0,0,0,−5,0,0 > + < 1,2,0,0,2,0,0 >
63
+ < −8/9,−7/9,0,0,11/9,0,0 >
26
= < −2/13,3/26,0,0,−1/26,0,0 >.
Then V 1 ,V 2 ,V 3 form an orthogonal basis for S.

SECTION 6.5 PROBLEMS

In each of Problems 1 through 8, use the Gram-Schmidt 5. < 0,0,2,2,1 >,< 0,0,1,−1,5 >,< 0,1,−2,1,0 >,
process to find an orthogonal basis spanning the same < 0,1,1,2,0 > in R 5
n
subspace of R as the given set of vectors.
6. < 1,2,0,−1,2,0 >,< 3,1,−3,−4,0,0 >,< 0,−1,
3
1. < 1,4,0 >,< 2,−5,0 > in R . 0,−5,0,0 >,< 1,−6,4,−2,−3,0 > in R 6
2. < 0,−1,2,0 >,< 0,3,−4,0 > in R 4
7. < 0,0,1,1,0,0 >,< 0,0,−3,0,0,0 > in R 6
3. < 0,2,1,−1 >,< 0,−1,1,6 >,< 0,2,2,3 > in R 4
4. <−1,0,3,0,4>,<4,0,−1,0,3>,<0,0,−1,0,5> 8. <0,−2,0,−2,0,−2>,<0,1,0,−1,0,0>,<0,−4,
in R 5 0,0,0,6 > in R 6

6.6 Orthogonal Complements and Projections

The Gram-Schmidt process serves as a springboard to an important concept that has practical
consequences, including the rationale for least squares approximations (see Section 7.8).

n
n
⊥
Let S be a subspace of R . Denote by S the set of all vectors in R that are orthogonal to
n
⊥
every vector in S. S is called the orthogonal complement of S in R .
3
For example, in R , suppose S is the two-dimensional subspace having < 1,0,0 > and
⊥
< 0,1,0 > as basis. We think of S as the x, y - plane. Now S consists of all vectors in 3-space
that are perpendicular to this plane, hence all constant multiples of k.
3
In this example, S is a subspace of R . We claim that this is always true.
⊥
THEOREM 6.6

n
If S is a subspace of R , then S is also a subspace of R . Further, the only vector in both S and
n
⊥
S is the zero vector.
⊥
Proof The zero vector is certainly in S because O is orthogonal to every vector, hence to
⊥
every vector in S.

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October 14, 2010 14:21 THM/NEIL Page-177 27410_06_ch06_p145-186