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Orthonormal Bases in Real and Complex Spaces

Module by: Justin Romberg. E-mail the author

Summary: This module defines the terms transpose, inner product, and Hermitian transpose and their use in finding an orthonormal basis.

Notation

Transpose operator AT A flips the matrix across it's diagonal. A=( a1,1a1,2 a2,1a2,2 ) A a 1 1 a 1 2 a 2 1 a 2 2 AT=( a1,1a2,1 a1,2a2,2 ) A a 1 1 a 2 1 a 1 2 a 2 2 Column i i of A A is row i i of AT A

Recall, inner product x=( x 0 x 1 ⋮ x n - 1 ) x x 0 x 1 ⋮ x n - 1 y=( y 0 y 1 ⋮ y n - 1 ) y y 0 y 1 ⋮ y n - 1 xT⁢y=( x 0 x 1 … x n - 1 )⁢( y 0 y 1 ⋮ y n - 1 )=∑ i xi⁢yi=〈y,x〉 x y x 0 x 1 … x n - 1 y 0 y 1 ⋮ y n - 1 i x i y i y x on Rn n

Hermitian transpose AH A , transpose and conjugate AH=AT* A A 〈y,x〉=xH⁢y=∑ i xi⁢yi* y x x y i x i y i on Cn n

Now, let b 0 b 1 … b n - 1 b 0 b 1 … b n - 1 be an orthonormal basis for Cn n i=01…n−1〈 b i , b i 〉=1  ,     i i 0 1 … n 1 b i b i 1 ⁢i≠j〈 b i , b j 〉= b j H⁢ b i =0 i j b i b j b j b i 0

Basis matrix: B=( ⋮⋮⋮ b 0 b 1 … b n - 1 ⋮⋮⋮ ) B ⋮ ⋮ ⋮ b 0 b 1 … b n - 1 ⋮ ⋮ ⋮ Now, BH⁢B=( … b 0 H… … b 1 H… ⋮ … b n - 1 H… )⁢( ⋮⋮⋮ b 0 b 1 … b n - 1 ⋮⋮⋮ )=( b 0 H⁢ b 0 b 0 H⁢ b 1 … b 0 H⁢ b n - 1 b 1 H⁢ b 0 b 1 H⁢ b 1 … b 1 H⁢ b n - 1 ⋮ b n - 1 H⁢ b 0 b n - 1 H⁢ b 1 … b n - 1 H⁢ b n - 1 ) B B … b 0 … … b 1 … ⋮ … b n - 1 … ⋮ ⋮ ⋮ b 0 b 1 … b n - 1 ⋮ ⋮ ⋮ b 0 b 0 b 0 b 1 … b 0 b n - 1 b 1 b 0 b 1 b 1 … b 1 b n - 1 ⋮ b n - 1 b 0 b n - 1 b 1 … b n - 1 b n - 1

For orthonormal basis with basis matrix B B BH=B-1 B B ( BT=B-1 B B in Rn n ) BH B is easy to calculate while B-1 B is hard to calculate.

So, to find α 0 α 1 … α n - 1 α 0 α 1 … α n - 1 such that x=∑ i α i ⁢ b i x i α i b i Calculate (α=B-1⁢x)⇒(α=BH⁢x) α B x α B x Using an orthonormal basis we rid ourselves of the inverse operation.

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