A First Course in Linear Algebra, 2017a

4.11. Orthogonality and the Gram Schmidt Process 233
a 1 (⃗w i •⃗w 1 )+a 2 (⃗w i •⃗w 2 )+···+ a k (⃗w i •⃗w k ) = 0
Now since the set is orthogonal, ⃗w i •⃗w m = 0forallm ≠ i,sowehave:
a 1 (0)+···+ a i (⃗w i •⃗w i )+···+ a k (0)=0
a i ‖⃗w i ‖ 2 = 0
Since the set is orthogonal, we know that ‖⃗w i ‖ 2 ≠ 0. It follows that a i = 0. Since the a i was chosen
arbitrarily, the set {⃗w 1 ,⃗w 2 ,···,⃗w k } is linearly independent.
Finally since W = span{⃗w 1 ,⃗w 2 ,···,⃗w k }, the set of vectors also spans W and therefore forms a basis of
W.
♠
If an orthogonal set is a basis for a subspace, we call this an orthogonal basis. Similarly, if an orthonormal
set is a basis, we call this an orthonormal basis.
We conclude this section with a discussion of Fourier expansions. Given any orthogonal basis B of R n
and an arbitrary vector⃗x ∈ R n , how do we express⃗x as a linear combination of vectors in B? The solution
is Fourier expansion.
Theorem 4.127: Fourier Expansion
Let V be a subspace of R n and suppose {⃗u 1 ,⃗u 2 ,...,⃗u m } is an orthogonal basis of V. Then for any
⃗x ∈ V ,
( ) ( ) ( )
⃗x •⃗u1 ⃗x •⃗u2
⃗x •⃗um
⃗x =
‖⃗u 1 ‖ 2 ⃗u 1 +
‖⃗u 2 ‖ 2 ⃗u 2 + ···+
‖⃗u m ‖ 2 ⃗u m .
This expression is called the Fourier expansion of ⃗x, and
j = 1,2,...,m are the Fourier coefficients.
⃗x •⃗u j
‖⃗u j ‖ 2 ,
Consider the following example.
Example 4.128: Fourier Expansion
⎡ ⎤ ⎡ ⎤
1 0
Let ⃗u 1 = ⎣ −1 ⎦,⃗u 2 = ⎣ 2 ⎦, and⃗u 3 =
2 1
⎡
⎣
5
1
−2
⎤
⎡
⎦, andlet⃗x = ⎣
Then B = {⃗u 1 ,⃗u 2 ,⃗u 3 } is an orthogonal basis of R 3 .
Compute the Fourier expansion of⃗x, thus writing⃗x as a linear combination of the vectors of B.
1
1
1
⎤
⎦.
Solution. Since B is a basis (verify!) there is a unique way to express ⃗x as a linear combination of the
vectors of B. Moreover since B is an orthogonal basis (verify!), then this can be done by computing the
Fourier expansion of⃗x.