15.12: Orthonormal Bases in Real and Complex Spaces

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Notation

Transpose operator \(A^T\) flips the matrix across it's diagonal.

\[\begin{array}{l}
A=\left(\begin{array}{ll}
a_{1,1} & a_{1,2} \\
a_{2,1} & a_{2,2}
\end{array}\right) \\
A^{T}=\left(\begin{array}{ll}
a_{1,1} & a_{2,1} \\
a_{1,2} & a_{2,2}
\end{array}\right)
\end{array} \nonumber \]

Column \(i\) of \(A\) is row \(i\) of \(A^T\)

Recall, inner product

\[\boldsymbol{x}=\left(\begin{array}{c}
x_{0} \\
x_{1} \\
\vdots \\
x_{n-1}
\end{array}\right) \nonumber \]

\[\boldsymbol{y}=\left(\begin{array}{c}
y_{0} \\
y_{1} \\
\vdots \\
y_{n-1}
\end{array}\right) \nonumber \]

\[\boldsymbol{x}^{T} \boldsymbol{y}=\left(\begin{array}{cccc}
x_{0} & x_{1} & \ldots & x_{n-1}
\end{array}\right)\left(\begin{array}{c}
y_{0} \\
y_{1} \\
\vdots \\
y_{n-1}
\end{array}\right)=\sum_{i} \boldsymbol{x}_{i} \boldsymbol{y}_{i}=\langle\boldsymbol{y}, \boldsymbol{x}\rangle \nonumber \]

on \(\mathbb{R}^n\).

Hermitian transpose \(A^H\), transpose and conjugate

\[\begin{array}{c}
A^{\mathrm{H}}=\overline{A^{T}} \\
\langle \boldsymbol{y}, \boldsymbol{x}\rangle=\boldsymbol{x}^{\mathrm{H}} \boldsymbol{y}=\sum_{i} \boldsymbol{x}_{i} \bar{\boldsymbol{y}}_{i}
\end{array} \nonumber \]

on \(\mathbb{C}^n\).

Now, let \(\left\{b_{0}, b_{1}, \dots, b_{n-1}\right\}\) be an orthonormal basis for \(\mathbb{C}^n\)

\[\begin{array}{c}
i=\{0,1, \ldots, n-1\}\left\langle b_{i}, b_{i}\right\rangle=1 \\
\left(i \neq j,\left\langle b_{i}, b_{i}\right\rangle=b_{j}^{H} b_{i}=0\right)
\end{array} \nonumber \]

Basis matrix:

\[B=\left(\begin{array}{cccc}
\vdots & \vdots & & \vdots \\
b_{0} & b_{1} & \dots & b_{n-1} \\
\vdots & \vdots & & \vdots
\end{array}\right) \nonumber \]

Now,

\[B^{\mathrm{H}} B=\left(\begin{array}{ccc}
\cdots & b_{0}^{\mathrm{H}} & \cdots \\
\cdots & b_{1}^{\mathrm{H}} & \cdots \\
& \vdots \\
\cdots & b_{n-1}^{\mathrm{H}} & \ldots
\end{array}\right)\left(\begin{array}{cccc}
\vdots & \vdots & & \vdots \\
b_{0} & b_{1} & \cdots & b_{n-1} \\
\vdots & \vdots & & \vdots
\end{array}\right)=\left(\begin{array}{cccc}
b_{0}^{\mathrm{H}} b_{0} & b_{0}^{\mathrm{H}} b_{1} & \ldots & b_{0}^{\mathrm{H}} b_{n-1} \\
b_{1}^{\mathrm{H}} b_{0} & b_{1}^{\mathrm{H}} b_{1} & \ldots & b_{1}^{\mathrm{H}} b_{n-1} \\
\vdots & & \\
b_{n-1}^{\mathrm{H}} b_{0} & b_{n-1}^{\mathrm{H}} b_{1} & \ldots & b_{n-1}^{\mathrm{H}} b_{n-1}
\end{array}\right) \nonumber \]

For orthonormal basis with basis matrix \(B\)

\[B^H=B^{-1} \nonumber \]

(\(B^{T}=B^{-1} \text {in } \mathbb{R}^{n}\) in \(\mathbb{R}^n\)) \(B^H\) is easy to calculate while \(B^{-1}\) is hard to calculate.

So, to find \(\left\{\alpha_{0}, \alpha_{1}, \ldots, \alpha_{n-1}\right\}\) such that

\[\boldsymbol{x}=\sum_{i} \alpha_{i} b_{i} \nonumber \]

Calculate

\[\left(\boldsymbol{\alpha}=B^{-1} \boldsymbol{x}\right) \Rightarrow\left(\boldsymbol{\alpha}=B^{H} \boldsymbol{x}\right) \nonumber \]

Using an orthonormal basis we rid ourselves of the inverse operation.