Linear Algebra, Theory And Applications, 2012a

290 INNER PRODUCT SPACES
Thus by induction,
u k+1 ∈ span (u 1 , ··· ,u k ,x k+1 ) = span (x 1 , ··· ,x k ,x k+1 ) .
Also, x k+1 ∈ span (u 1 , ··· ,u k ,u k+1 ) which is seen easily by solving (12.1) for x k+1 and it
follows
span (x 1 , ··· ,x k ,x k+1 ) = span (u 1 , ··· ,u k ,u k+1 ) .
If l ≤ k,
⎛
⎞
k∑
(u k+1 ,u l ) = C ⎝(x k+1 ,u l ) − (x k+1 ,u j )(u j ,u l ) ⎠
⎛
= C ⎝(x k+1 ,u l ) −
j=1
⎞
k∑
(x k+1 ,u j ) δ lj
⎠
j=1
= C ((x k+1 ,u l ) − (x k+1 ,u l )) = 0.
The vectors, {u j } n j=1
, 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 12.2.2 Let X be a finite dimensional inner product space of dimension n whose
basis is {u 1 , ··· ,u k ,x k+1 , ··· ,x n } . Then if {u 1 , ··· ,u k } is orthonormal, then the Gram
Schmidt process applied to the given list of vectors in order leaves {u 1 , ··· ,u k } unchanged.
Lemma 12.2.3 Suppose {u j } n j=1
is an orthonormal basis for an inner product space X.
Then for all x ∈ X,
n∑
x = (x, u j ) u j .
j=1
Proof: By assumption that this is an orthonormal basis,
n∑ { }} {
(x, u j ) (u j ,u l )=(x, u l ) .
j=1
Letting y = ∑ n
k=1 (x, u k) u k , it follows
δ jl
(x − y, u j ) = (x, u j ) −
n∑
(x, u k )(u k ,u j )
k=1
= (x, u j ) − (x, u j )=0
for all j. Hence, for any choice of scalars c 1 , ···, c n ,
⎛
⎞
n∑
⎝x − y, c j u j
⎠ =0
j=1
and so (x − y, z) = 0 for all z ∈ X. Thus this holds in particular for z = x − y. Therefore, x
= y.
The following theorem is of fundamental importance. First note that a subspace of an
inner product space is also an inner product space because you can use the same inner
product.