Page 196 - Advanced engineering mathematics

176 CHAPTER 6 Vectors and Vector Spaces

so
X 3 · V 2
h = .
V 2 2
And we need
V 3 · V 1 = X 3 · V 1 − dV 1 · V 1 = 0,
so
V 3 · V 1
d =
V 1 · V 1
V 3 · V 1
= V 1 .
V 1 2
Therefore, choose
X 3 · V 1 X 3 · V 2
V 3 = X 3 − V 1 − V 2 .
V 1 2 V 2 2
This is X 3 , minus the projections of X 3 onto V 1 and V 2 .
This pattern suggests a general procedure. Set V 1 = X 1 and, for j = 2,··· ,m, V j equal to
X j minus the projections of X j onto V 1 ,··· ,V j−1 . This gives us
X j · V 1
V j =X j − V 1
V 1 2
X j · V 2 X j · V j−1
− V 2 − ··· − V j−1 ,
V 2 2 V j−1 2
for j = 2,··· ,m.

This way of forming mutually orthogonal vectors from X 1 ,··· ,X m is called the Gram-
Schmidt orthogonalization process. When we use it, we say that we have orthogonalized
the given basis for S (in the sense of replacing that basis with an orthogonal basis).

The vectors V 1 ,··· ,V m are linearly independent because they are orthogonal. Further, they
n
span the same subspace S of R that X 1 ,··· ,X m span, because each V j is a linear combination
of the X j vectors, which span S. The vectors V j therefore form an orthogonal basis for S.Ifwe
want an orthonormal basis, then divide each V j by its length.

EXAMPLE 6.19

7
Let S be the subspace of R having basis
X 1 =< 1,2,0,0,2,0,0 >,X 2 =< 0,1,0,0,3,0,0 >,X 3 =< 1,0,0,0,−5,0,0 >.
We will produce an orthogonal basis for S. First let
V 1 = X 1 =< 1,2,0,0,2,0,0 >.
Next let
X 2 · V 1
V 2 = X 2 − V 1
V 1 2
8
=< 0,1,0,0,3,0,0 > − < 1,2,0,0,2,0,0 >
9
=< −8/9,−7/9,0,0,11/9,0,0 >.

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October 14, 2010 14:21 THM/NEIL Page-176 27410_06_ch06_p145-186