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Orthonormal Bases in Real and Complex Spaces

Module by: Justin Romberg

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=a11a12a21a22 A a 1 1 a 1 2 a 2 1 a 2 2 AT=a11a21a12a22 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 xTy= x 0 x 1 … x n - 1 y 0 y 1 ⋮ y n - 1 =∑xiyi=<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>=xHy=∑xiyi¯ y x x y i x i y i on ℂn n

Now, let b0b1…b n - 1 b 0 b 1 … b n - 1 be an orthonormal basis for ℂn n ∀i,:i=01…n-1<bi,bi>=1 i i 0 1 … n 1 b i b i 1 i≠j <bi,bj>=bjHbi=0 i j b i b j b j b i 0

Basis matrix: B=⋮⋮⋮b0b1…b n - 1 ⋮⋮⋮ B ⋮ ⋮ ⋮ b 0 b 1 … b n - 1 ⋮ ⋮ ⋮ Now, BHB=…b0H……b1H…⋮…b n - 1 H…⋮⋮⋮b0b1…b n - 1 ⋮⋮⋮=b0Hb0b0Hb1…b0Hb n - 1 b1Hb0b1Hb1…b1Hb n - 1 ⋮b n - 1 Hb0b n - 1 Hb1…b n - 1 Hb 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=∑ α i b i x i α i b i Calculate α=B-1x⇒α=BHx α B x α B x Using an orthonormal basis we rid ourselves of the inverse operation.

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