Linear Algebra and Analysis

11.4.5 Orthonormal Bases

Not all bases for an inner product space H are created equal. The best bases are orthonormal.

Every orthonormal set of vectors is automatically linearly independent. Indeed, if

∑n akvk = 0, k=1

then taking the inner product with v_j, yields 0 = ∑ _k=1ⁿa_k

= a_j. Thus each a_j = 0. We will use this simple observation whenever convenient.

Proof: Suppose ∑ _i=1^kc_iv_i = 0. Then taking inner products with v_j,

∑ ∑ 0 = (0,vj) = ci(vi,vj) = ciδij = cj. i i

Since j is arbitrary, this shows the set is linearly independent as claimed. ■

It turns out that if X is any subspace of H, then there exists an orthonormal basis for X.

Proof: Let

be a basis for X. Let u₁ ≡ x₁∕. Thus for k = 1, span = span and is an orthonormal set. Now suppose for some k < n, u₁, , u_k have been chosen such that = δ_jl and span = span. Then define

∑ xk+1 − kj=1(xk+1,uj)uj uk+1 ≡ ||------∑k--------------||, (11.12) |xk+1 − j=1(xk+1,uj)uj|

(11.12)

where the denominator is not equal to zero because the x_j form a basis and so

xk+1 ∕∈ span (x1,⋅⋅⋅,xk) = span (u1,⋅⋅⋅,uk)

Thus by induction,

uk+1 ∈ span (u1,⋅⋅⋅,uk,xk+1) = span (x1,⋅⋅⋅,xk,xk+1) .

Also, x_k+1 ∈ span

which is seen easily by solving 11.12 for x_k+1 and it follows

span (x1,⋅⋅⋅,xk,xk+1) = span(u1,⋅⋅⋅,uk,uk+1).

If l ≤ k, then denoting by C the scalar

⁻¹,

The vectors, _j=1ⁿ, generated in this way are therefore an orthonormal basis because each vector has unit length. ■

The process by which these vectors were generated is called the Gram Schmidt process.

The following corollary is obtained from the above process.