15.11 Orthonormal bases in real and complex spaces

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 $$x=\begin{pmatrix}{x}_0\\ x_1\\ ⋮\\ x_{n-1}\\ \end{pmatrix}$$ $$y=\begin{pmatrix}{y}_0\\ y_1\\ ⋮\\ y_{n-1}\\ \end{pmatrix}$$ $$x^Ty=\begin{pmatrix}{x}_0 & x_1 & \dots & x_{n-1}\\ \end{pmatrix}\begin{pmatrix}{y}_0\\ y_1\\ ⋮\\ y_{n-1}\\ \end{pmatrix}=\sum x_{i}y_i=y\cdot x$$ on $\mathbb{R}^n$

Hermitian transpose $(A)$ , transpose and conjugate $$(A)=\overline{A^T}$$ $$y\cdot x=(x)y=\sum x_i\overline{y_i}$$ on $\mathbb{C}^n$

Now, let $\{b_0, b_1, \dots , b_{n-1}\}$ be an orthonormal basis for $\mathbb{C}^n$ $$\forall i, \colon b_i\cdot b_i=1$$ $$(i\neq j, b_i\cdot b_j=(b_j)b_i=0)$$

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.