11.4.5 Orthonormal Bases
Not all bases for an inner product space H are created equal. The best bases are orthonormal.
Definition 11.4.12 Suppose
is a set of vectors in an inner product space H. It is an
orthonormal set if
{
(vi,vj) = δij = 1 if i = j
0 if i ⁄= j
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Every orthonormal set of vectors is automatically linearly independent. Indeed, if
then taking the inner product with vj, yields 0 = ∑
k=1nak
=
aj. Thus each
aj = 0. We will use
this simple observation whenever convenient.
Proposition 11.4.13 Suppose
is an orthonormal set of vectors. Then it is linearly
independent.
Proof: Suppose ∑
i=1kcivi = 0. Then taking inner products with vj,
∑ ∑
0 = (0,vj) = ci(vi,vj) = ciδij = cj.
i i
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Since j is arbitrary, this shows the set is linearly independent as claimed. ■
It turns out that if X is any subspace of H, then there exists an orthonormal basis for
X.
Lemma 11.4.14 Let X be a subspace of dimension n whose basis is
. Then there
exists an orthonormal basis for X, which has the property that for each k ≤ n,
span =
span.
Proof: Let
be a basis for
X. Let
u1 ≡ x1∕. Thus for
k = 1
, span =
span
and
is an orthonormal set. Now suppose for some
k < n, u1,
,
uk have been chosen such that
=
δjl and
span =
span. Then define
∑
xk+1 − kj=1(xk+1,uj)uj
uk+1 ≡ ||------∑k--------------||, (11.12)
|xk+1 − j=1(xk+1,uj)uj|
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where the denominator is not equal to zero because the xj form a basis and so
xk+1 ∕∈ span (x1,⋅⋅⋅,xk) = span (u1,⋅⋅⋅,uk)
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Thus by induction,
uk+1 ∈ span (u1,⋅⋅⋅,uk,xk+1) = span (x1,⋅⋅⋅,xk,xk+1) .
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Also, xk+1 ∈ span
which is seen easily by solving
11.12 for
xk+1 and it follows
span (x1,⋅⋅⋅,xk,xk+1) = span(u1,⋅⋅⋅,uk,uk+1).
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If l ≤ k, then denoting by C the scalar
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||x − ∑k (x ,u )u ||
k+1 j=1 k+1 j j
−1,
( )
∑k
(uk+1,ul) = C ((xk+1,ul)− (xk+1,uj)(uj,ul))
j=1
( k )
= C ((xk+1,ul)− ∑ (xk+1,uj)δlj)
j=1
= C ((xk+1,ul)− (xk+1,ul)) = 0.
The vectors,
j=1n, generated in this way are therefore an orthonormal basis because each vector has
unit length.
■
The process by which these vectors were generated is called the Gram Schmidt process.
The following corollary is obtained from the above process.
Corollary 11.4.15 Let X be a finite dimensional inner product space of dimension n whose basis
is
{u ,⋅⋅⋅,u ,x ,⋅⋅⋅,x }
1 k k+1 n
. Then if is orthonormal, then the Gram Schmidt process
applied to the given list of vectors in order leaves unchanged.