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# 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 $Cn n$ $∀i,:⟨ b i , b i ⟩=1 i i 0 1 … n 1 b i b i 1$ $⁢i≠j⟨ b i , b j ⟩= b j H⁢ b i =0 i j b i b j b j b i 0$

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.