Page 121 - Linear Algebra Done Right

Chapter 6. Inner-Product Spaces
108
Proof: Let v ∈ V. Because (e 1 ,...,e n ) is a basis of V, there exist
scalars a 1 ,...,a n such that
v = a 1 e 1 + ··· + a n e n .
Take the inner product of both sides of this equation with e j , get-
ting v, e j = a j . Thus 6.18 holds. Clearly 6.19 follows from 6.18
and 6.15.
Now that we understand the usefulness of orthonormal bases, how
do we go about finding them? For example, does P m (F), with inner
product given by integration on [0, 1] (see 6.2), have an orthonormal
basis? As we will see, the next result will lead to answers to these ques-
tions. The algorithm used in the next proof is called the Gram-Schmidt
The Danish procedure. It gives a method for turning a linearly independent list into
mathematician Jorgen an orthonormal list with the same span as the original list.
Gram (1850–1916) and
the German 6.20 Gram-Schmidt: If (v 1 ,...,v m ) is a linearly independent list
mathematician Erhard of vectors in V, then there exists an orthonormal list (e 1 ,...,e m ) of
Schmidt (1876–1959) vectors in V such that
popularized this
algorithm for 6.21 span(v 1 ,...,v j ) = span(e 1 ,...,e j )
constructing
orthonormal lists. for j = 1,...,m.

Proof: Suppose (v 1 ,...,v m ) is a linearly independent list of vec-
tors in V. To construct the e’s, start by setting e 1 = v 1 / v 1 . This
satisfies 6.21 for j = 1. We will choose e 2 ,...,e m inductively, as fol-
lows. Suppose j> 1 and an orthornormal list (e 1 ,...,e j−1 ) has been
chosen so that

6.22 span(v 1 ,...,v j−1 ) = span(e 1 ,...,e j−1 ).

Let
v j −Pv j ,e 1 e 1 −· · ·−Pv j ,e j−1 e j−1
6.23 e j = .
v j −Pv j ,e 1 e 1 −· · ·−Pv j ,e j−1 e j−1
Note that v j ∉ span(v 1 ,...,v j−1 ) (because (v 1 ,...,v m ) is linearly inde-
pendent) and thus v j ∉ span(e 1 ,...,e j−1 ). Hence we are not dividing
by 0 in the equation above, and so e j is well defined. Dividing a vector
by its norm produces a new vector with norm 1; thus e j = 1.