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 >

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