# 15.11 Orthonormal bases in real and complex spaces

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

## Notation

Transpose operator $A^T$ flips the matrix across it's diagonal. $A=\begin{pmatrix}a_{1, 1} & a_{1, 2}\\ a_{2, 1} & a_{2, 2}\\ \end{pmatrix}$ $A^T=\begin{pmatrix}a_{1, 1} & a_{2, 1}\\ a_{1, 2} & a_{2, 2}\\ \end{pmatrix}$ Column $i$ of $A$ is row $i$ of $A^T$

Recall, inner product $x=\begin{pmatrix}{x}_{0}\\ {x}_{1}\\ ⋮\\ {x}_{n-1}\\ \end{pmatrix}$ $y=\begin{pmatrix}{y}_{0}\\ {y}_{1}\\ ⋮\\ {y}_{n-1}\\ \end{pmatrix}$ $x^Ty=\begin{pmatrix}{x}_{0} & {x}_{1} & \dots & {x}_{n-1}\\ \end{pmatrix}\begin{pmatrix}{y}_{0}\\ {y}_{1}\\ ⋮\\ {y}_{n-1}\\ \end{pmatrix}=\sum x_{i}y_{i}=y\dot x$ on $\mathbb{R}^{n}$

Hermitian transpose $(A)$ , transpose and conjugate $(A)=\overline{A^T}$ $y\dot x=(x)y=\sum x_{i}\overline{y_{i}}$ on $\mathbb{C}^{n}$

Now, let $\{{b}_{0}, {b}_{1}, \dots , {b}_{n-1}\}$ be an orthonormal basis for $\mathbb{C}^{n}$ $\forall i, \colon {b}_{i}\dot {b}_{i}=1$ $(i\neq j, {b}_{i}\dot {b}_{j}=({b}_{j}){b}_{i}=0)$

Basis matrix: $B=\begin{pmatrix}⋮ & ⋮ & & ⋮\\ {b}_{0} & {b}_{1} & \dots & {b}_{n-1}\\ ⋮ & ⋮ & & ⋮\\ \end{pmatrix}$ Now, $(B)B=\begin{pmatrix}\dots & ({b}_{0}) & \dots \\ \dots & ({b}_{1}) & \dots \\ & ⋮ & \\ \dots & ({b}_{n-1}) & \dots \\ \end{pmatrix}\begin{pmatrix}⋮ & ⋮ & & ⋮\\ {b}_{0} & {b}_{1} & \dots & {b}_{n-1}\\ ⋮ & ⋮ & & ⋮\\ \end{pmatrix}=\begin{pmatrix}({b}_{0}){b}_{0} & ({b}_{0}){b}_{1} & \dots & ({b}_{0}){b}_{n-1}\\ ({b}_{1}){b}_{0} & ({b}_{1}){b}_{1} & \dots & ({b}_{1}){b}_{n-1}\\ ⋮\\ ({b}_{n-1}){b}_{0} & ({b}_{n-1}){b}_{1} & \dots & ({b}_{n-1}){b}_{n-1}\\ \end{pmatrix}$

For orthonormal basis with basis matrix $B$ $(B)=B^{(-1)}$ ( $B^T=B^{(-1)}$ in $\mathbb{R}^{n}$ ) $(B)$ is easy to calculate while $B^{(-1)}$ is hard to calculate.

So, to find $\{{\alpha }_{0}, {\alpha }_{1}, \dots , {\alpha }_{n-1}\}$ such that $x=\sum {\alpha }_{i}{b}_{i}$ Calculate $(\alpha =B^{(-1)}x)\implies (\alpha =(B)x)$ Using an orthonormal basis we rid ourselves of the inverse operation.

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