Page 263 - DMTH502_LINEAR_ALGEBRA

Unit 24: Inner Product and Inner Product Spaces

Notes
= ( m 1 | j ) ( m 1 | j )
= 0.

Therefore { , . . . , } is an orthogonal set consisting of m + 1 non-zero vectors in the subspace
1 m+1
spanned by , . . . , . By theorem 2, it is a basis for this subspace. Thus the vectors , . . . ,
1 m+1 1 n
may be constructed one after the other in accordance with (9). In particular, when n = 4, we have
=
1 1
( | )
= 2 1 … (10)
2 2 2 1
1
( | ) ( | )
= 3 1 3 2
3 3 2 1 2 2
1 2
( | ) ( | ) ( | )
= 4 1 4 2 4 3 . … (11)
4 4 2 1 2 2 2 3
1 2 3
Corollary: Every finite-dimensional inner product space has an orthonormal basis.
Proof: Let V be a finite-dimensional inner product space and { , . . . , } a basis for V. Apply the
1 n
Gram-Schmidt process to construct an orthogonal basis { , . . . , }. Then to obtain an orthonormal
1 n
basis, simply replace each vector by / .
k k k
One of the main advantages which orthonormal bases have over arbitrary bases is that
computations involving coordinates are simpler. To indicate in general terms why this is true,
suppose that V is a finite-dimensional inner product space. Then, as in the last section, we may
use Equation (5) to associate a matrix G with every ordered basis  = { , . . . , } of V. Using this
1 n
matrix
G = ( | )
jk k j
we may compute inner products in terms of coordinates, If  is an orthonormal basis, then G is
the identity matrix, and for any scalars x and y
j k

x y x y .
j j k k j j
j k j
Thus in terms of an orthonormal basis, the inner product in V looks like the standard inner
product in F .
n
Although it is of limited practical use for computations, it is interesting to note that the
Gram-Schmidt process may also be used to test for linear dependence. For suppose , …, are
1 n
linearly dependent vectors in an inner product space V. To exclude a trivial case, assume that
0. Let m be the largest integer for which , …, are independent. Then 1 m < n. Let
1 1 m
, …, be the vectors obtained by applying the orthogonalization process to …, . Then
1 m 1 m
the vector given by (9) is necessarily 0. For is in the subspace spanned by , , …,
m+1 m+1 1 2 n
and orthogonal to each of these vectors; hence it is 0 by (6). Conversely it , …, are different
1 n
from 0 and = 0, then , , …, are linearly dependent.
m+1 1 2 m+1
Example 12: Consider the vectors
= (4, 0, 3)
1

LOVELY PROFESSIONAL UNIVERSITY 257