Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 34 exercise 2.2

Author: Panagiotis

21 Sep

In exercise [1, p. 34 exercise 2.2] we are asked to prove that the rows and columns of a unitary matrix are orthonormal as per [1, p. 21,(2.21)].
Solution: A complex square matrix is unitary if $\mathbf{A}^{-1}=\mathbf{A}^H$ . Let $\mathbf{a}_i$ be the $i^{th}$ column vector of $\mathbf{A}$ then

$\mathbf{A}$

$=$

$\left[ \begin{array}{*{20}c} \mathbf{a}_{1} & \mathbf{a}_{2} & \cdots & \mathbf{a}_{N} \\ \end{array} \right]$

(1)

Computing the matrix product of the hermitian transpose and the matrix results in:

$\mathbf{A}^H \mathbf{A}$	$=$	$\left[ \begin{array}{{20}c} \mathbf{a}_1^H \\ \mathbf{a}_2^H \\ \vdots \\ \mathbf{a}_N^H \\ \end{array} \right] \left[ \begin{array}{{20}c} \mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{a}_N \\ \end{array} \right]$	(2)
	$=$	$\left[ \begin{array}{*{20}c} \mathbf{a}_1^H \mathbf{a}_1 & \mathbf{a}_1^H \mathbf{a}_2 & \cdots & \mathbf{a}_1^H \mathbf{a}_N \\ \mathbf{a}_2^H \mathbf{a}_1 & \mathbf{a}_2^H \mathbf{a}_2 & \cdots & \mathbf{a}_2^H \mathbf{a}_N \\ \vdots & \vdots & \vdots & \vdots \\ \mathbf{a}_N^H \mathbf{a}_1 & \mathbf{a}_N^H \mathbf{a}_2 & \cdots & \mathbf{a}_N^H \mathbf{a}_N \\ \end{array} \right]$	(3)
	$=$	${\mathbf I}$	(4)