Page 195 - Advanced engineering mathematics

6.5 Orthogonalization 175

Draw a parallelogram law diagram justification for 24. Show that any finite set of vectors that includes the
2
this conclusion, for the case that the vectors are in R . zero vector is linearly dependent.
n
n
22. Let V 1 ,··· ,V k be mutually orthogonal vectors in R . 25. Let S be a nontrivial subspace of R . Show that any
n
Show that, for any X in R , spanning set of S must contain a basis for S.
k 26. Let u 1 ,··· ,u k be linearly independent vectors in

2
2
(X · V j ) ≤ X . R , with k < n. Show that there are n − k vectors
n
j=1 v 1 ,··· ,v n−k such that
This is known as Bessel’s inequality for vectors. A
version for Fourier series and eigenfunction expan-
u 1 ,··· ,u k ,v 1 ,··· ,v n−k
sions will be seen in Chapter Fifteen. Hint Let Y =
k 2
X − (X · V j )V j and compute Y .
j=1 n
form a basis for R . This states that any linearly inde-
n
23. Suppose V 1 ,··· ,V n are a basis for R , consisting of pendent set of vectors in R is either a basis, or can
n
mutually orthogonal unit vectors. Show that, if X is be expanded into a basis by adjoining more vectors.
n
any vector in R ,then Hint: Choose v 1 in R but not in the span of u,··· ,u k .
n
n
n If u 1 ,··· ,u k ,v 1 span R , stop. Otherwise, there is
2 2
(X · V j ) = X . some v 2 in R but not in the span of u 1 ,··· ,u k ,v 1 .
n
n
j=1 If u 1 ,··· ,u k ,v 1 ,v 2 span R , stop. Otherwise continue
This is a vector version of Parseval’s equality. this process.
6.5 Orthogonalization
n
Suppose X 1 ,···, X m form a basis for a subspace S of R , with m ≥ 2. We would like to replace
this basis with an orthogonal basis V 1 ,··· ,V m for S.
We will build an orthogonal basis one vector at a time. Begin by setting
V 1 = X 1 .
Now look for a nonzero V 2 that is in S and orthogonal to V 1 .One waytodothisistoattempt V 2
of the form
V 2 = X 2 − cV 1 .
Choose c so that V 2 is orthogonal to V 1 . For this, we need
V 2 · V 1 = X 2 · V 1 − cV 1 · V 1 = 0.

This will be true if
X 2 · V 1
c = .
V 1 2
Therefore set
X 2 · V 1
V 2 = X 2 − V 1 .
V 1 2
Observe that V 2 is X 2 , minus the projection of X 2 onto V 1 .
If m =2 we are done. If m ≥3, produce nonzero V 3 in S orthogonal to V 1 and V 2 as follows.
Try
V 3 = X 3 − dV 1 − hV 2 .
We need

V 3 · V 2 = X 3 · V 2 − dV 1 · V 2 − hV 2 · V 2 = 0,

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October 14, 2010 14:21 THM/NEIL Page-175 27410_06_ch06_p145-186