Page 237 - A Course in Linear Algebra with Applications

7.2: Inner Product Spaces 221

S to a basis of V, say v i , . . . , v TO, v m + i , . . . , v n : this possible
by 5.1.4. If v is an arbitrary vector of V, we can write

n

1
By 7.2.3 the vector v belongs to S - if and only if it is orthog-
onal to each of the vectors v i , . . . , v m ; the conditions for this
are
n
2
< v^ v > = 2_. < Vj, Vj > Cj = 0, for i = 1, ,..., m.

Now the above equations constitute a linear system of m equa-
tions in the n unknowns ci, C2,..., c n . Therefore the dimen-
1
sion of S - equals the dimension of the solution space of the
linear system, which we know from 5.1.7 to be n — r where r
is the rank of the m x n coefficient matrix A = [< Vj, Vj >].
Obviously r < m; we shall show that in fact r = m. If this is
false, then the m rows of A must be linearly dependent and
there exist scalars dfa, • • •, d m, not all of them zero, such that

m m
0 = ^d» < Vi, Vj > = < J^djVi, Vj- >
J

for j = 1,... ,n. But a vector which is orthogonal to every
v
vector in a basis of V must be zero. Hence Yl'iLi ^i i = 0'
which can only mean that d\ = d 2 = • • • = d m = 0 since
v v
i,---) m are linearly independent. By this contradiction
r = m.
±
We conclude that dim(5 ) = n — m = n — dim(S'), which
,
L
implies that dim(5 )+dim(5- ) = n — dim(V). It follows from
5.3.2 that
±
1
dim(5 + S- -) = dim(5) + dim(5' ) = dim(V).