Page 257 - 35Linear Algebra

14.2 Orthogonal and Orthonormal Bases 257

where h·, ·i is the inner product, you might ask whether this can be related
to a dot product? The answer to this question is yes and rather easy to
understand:
0
Given an orthonormal basis, the information of two vectors v and v in V
can be encoded in column vectors
   
hv, u 1 i hv, u 1 i
. .

 . .   . .  ,
 = 
v = hv, u 1 iu 1 + · · · + hv, u n iu n = (u 1 , . . . , u n ) 
hv, u n i hv, u n i
O
 0   0 
hv , u 1 i hv , u 1 i
0
0
v 0 = hv , u 1 iu 1 + · · · + hv , u n iu n = (u 1 , . . . , u n ) . . .   . . .  .


 = 
0
0
hv , u n i hv , u n i
O
The dot product of these two column vectors is
   
0
hv, u 1 i hv , u 1 i
. .
0
0
 .   .  = hv, u 1 ihv , u 1 i + · · · + hv, u n ihv, u n i .

 .  ·  .
0
hv, u n i hv , u n i
0
This agrees exactly with the inner product of v and v because
0
0
0
hv, v i =
hv, u 1 iu 1 + · · · + hv, u n iu n , hv , u 1 iu 1 + · · · + hv , u n iu n
0
0
= hv, u 1 ihv , u 1 ihu 1 , u 1 i + hv, u 2 ihv , u 1 ihu 2 , u 1 i + · · ·
0
0
· · · + hv, u n−1 ihv , u n ihu n−1 , u n i + hv, u n ihv , u n ihu n , u n i
0
0
= hv, u 1 ihv , u 1 i + · · · + hv, u n ihv , u n i .
The above computation looks a little daunting, but only the linearity prop-
erty of inner products and the fact that hu i , u j i can equal either zero or
one was used. Because inner products become dot products once one uses
an orthonormal basis, we will quite often use the dot product notation in
situations where one really should write an inner product. Conversely, dot
product computations can always be rewritten in terms of an inner product,
if needed.
Example 133 Consider the space of polynomials given by V = span{1, x} with inner
0
0
product hp, p i = R 1 p(x)p (x)dx. An obvious basis to use is B = (1, x) but it is not
0
hard to check that this is not orthonormal, instead we take
√

O = 1, 2 3 x − 1 .
2
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