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Orthonormal Bases in Real and Complex Spaces

By Justin Romberg

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 xTy=( x 0 x 1 x n - 1 )( y 0 y 1 y n - 1 )= i 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= i xiyi¯ 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,: b i , b i =1 i i 0 1 n 1 b i b i 1 ij 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, BHB=( 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-1x)(α=BHx) α B x α B x Using an orthonormal basis we rid ourselves of the inverse operation.