# Lysario – by Panagiotis Chatzichrisafis

"ούτω γάρ ειδέναι το σύνθετον υπολαμβάνομεν, όταν ειδώμεν εκ τίνων και πόσων εστίν …"

## Archive for August, 2010

In [1, p. 35 exercise 2.18] we are asked to prove that the inverse of a complex matrix $\mathbf{A}=\mathbf{A}_{R}+j\mathbf{A}_{I}$ may be found by first inverting
 $\mathbf{V}=\left( \begin{array}{cc} \mathbf{A}_{R} & -\mathbf{A}_{I} \\ \mathbf{A}_{I} & \mathbf{A}_{R} \\ \end{array} \right)$ (1)

to yield
 $\mathbf{V}^{-1}=\left( \begin{array}{cc} \mathbf{B}_{R} & -\mathbf{B}_{I} \\ \mathbf{B}_{I} & \mathbf{B}_{R} \\ \end{array} \right)$ (2)

and then letting $\mathbf{A}^{-1}=\mathbf{B}_{R}+j\mathbf{B}_{I}$. read the conclusion >
In [1, p. 35 exercise 2.17] we are asked to verify the alternative expression [1, p. 33 (2.77)] for a hermitian function. read the conclusion >
In [1, p. 35 exercise 2.16] we are asked to verify the formulas given for the complex gradient of a hermitian and a linear form [1, p. 31 (2.70)]. To do so, we are instructed to decompose the matrices and vectors into their real and imaginary parts as $\mathbf{x}=\mathbf{x}_{R}+j\mathbf{x}_{I}, \mathbf{A^{\prime}}= \mathbf{A}_{R}+j\mathbf{A}_{I}, \mathbf{b^{\prime}}=\mathbf{b}_{R}+j\mathbf{b}_{I}$ read the conclusion >