Linear Algebra, Theory And Applications, 2012a

12.2. THE GRAM SCHMIDT PROCESS 289
Proposition 12.1.8 Foraninnerproductspace,|x| ≡(x, x) 1/2 does specify a norm.
Proof: All the axioms are obvious except the triangle inequality. To verify this,
|x + y| 2 ≡ (x + y, x + y) ≡|x| 2 + |y| 2 +2Re(x, y)
≤ |x| 2 + |y| 2 +2|(x, y)|
≤ |x| 2 + |y| 2 +2|x||y| =(|x| + |y|) 2 .
The best norms of all are those which come from an inner product because of the following
identity which is known as the parallelogram identity.
Proposition 12.1.9 If (V,(·, ·)) is an inner product space then for |x| ≡ (x, x) 1/2 , the
following identity holds.
|x + y| 2 + |x − y| 2 =2|x| 2 +2|y| 2 .
It turns out that the validity of this identity is equivalent to the existence of an inner
product which determines the norm as described above. These sorts of considerations are
topics for more advanced courses on functional analysis.
Definition 12.1.10 A basis for an inner product space, {u 1 , ··· ,u n } is an orthonormal
basis if
{ 1 if k = j
(u k ,u j )=δ kj ≡
.
0 if k ≠ j
Note that if a list of vectors satisfies the above condition for being an orthonormal set,
then the list of vectors is automatically linearly independent. To see this, suppose
n∑
c j u j =0
Then taking the inner product of both sides with u k ,
n∑
n∑
0= c j (u j ,u k )= c j δ jk = c k .
j=1
j=1
j=1
12.2 The Gram Schmidt Process
Lemma 12.2.1 Let X be a finite dimensional inner product space of dimension n whose
basis is {x 1 , ··· ,x n } . Then there exists an orthonormal basis for X, {u 1 , ··· ,u n } which has
the property that for each k ≤ n, span(x 1 , ··· ,x k ) = span (u 1 , ··· ,u k ) .
Proof: Let {x 1 , ··· ,x n } be a basis for X. Let u 1 ≡ x 1 / |x 1 | . Thus for k =1, span (u 1 )=
span (x 1 )and{u 1 } is an orthonormal set. Now suppose for some k