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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 $$x=\begin{pmatrix}{x}_{0}\\ {x}_{1}\\ \vdots \\ {x}_{n-1}\\ \end{pmatrix}$$ $$y=\begin{pmatrix}{y}_{0}\\ {y}_{1}\\ \vdots \\ {y}_{n-1}\\ \end{pmatrix}$$ $$x^Ty=\begin{pmatrix}{x}_{0} & {x}_{1} & \dots & {x}_{n-1}\\ \end{pmatrix}\begin{pmatrix}{y}_{0}\\ {y}_{1}\\ \vdots \\ {y}_{n-1}\\ \end{pmatrix}=\sum x_{i}y_{i}=y\cdot x$$ 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: $$B=\begin{pmatrix}\vdots & \vdots & & \vdots \\ {b}_{0} & {b}_{1} & \dots & {b}_{n-1}\\ \vdots & \vdots & & \vdots \\ \end{pmatrix}$$ Now, $$(B)B=\begin{pmatrix}\dots & ({b}_{0}) & \dots \\ \dots & ({b}_{1}) & \dots \\ & \vdots & \\ \dots & ({b}_{n-1}) & \dots \\ \end{pmatrix}\begin{pmatrix}\vdots & \vdots & & \vdots \\ {b}_{0} & {b}_{1} & \dots & {b}_{n-1}\\ \vdots & \vdots & & \vdots \\ \end{pmatrix}=\begin{pmatrix}({b}_{0}){b}_{0} & ({b}_{0}){b}_{1} & \dots & ({b}_{0}){b}_{n-1}\\ ({b}_{1}){b}_{0} & ({b}_{1}){b}_{1} & \dots & ({b}_{1}){b}_{n-1}\\ \vdots \\ ({b}_{n-1}){b}_{0} & ({b}_{n-1}){b}_{1} & \dots & ({b}_{n-1}){b}_{n-1}\\ \end{pmatrix}$$
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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