Page 235 - A Course in Linear Algebra with Applications

7.2: Inner Product Spaces 219

Proof
To show that S 1 is a subspace we need to verify that it con-
tains the zero vector and is closed with respect to addition
and scalar multiplication. The first statement is true since
the zero vector is orthogonal to every vector. As for the re-
1 1
maining ones, take two vectors v and w in S - , let s be an
arbitrary vector in S and let c be a scalar. Then

< cv, s > = c < v, s > = 0,

and
< v + w, s > = < v , s > + < w , s > = 0 .
1
Hence cv and v + w belong to S -.
1
Now suppose that v belongs to the intersection S P\ S -.
Then v is orthogonal to itself, which can only mean that v =
0.
Finally, assume that v i , . . . , v m are generators of S and
that v is orthogonal to each v^. A general vector of S has the
c v
form YlT=i i i f° r s o m e scalars Q . Then
m m
V
< V, ^ °i i > = ^ Q < V, Vj > = 0.
i=l i=l

Hence v is orthogonal to every vector in S and so it belongs
1
to S- . The converse is obvious.
Example 7.2.11
In the inner product space ^ ( R ) with

< f,9> = / f{x)g{x)dx,
Jo

find the orthogonal complement of the subspace S generated
by 1 and x.