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Help me on finding spectral radius!

by ali987
Tags: radius, spectral
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ali987
#1
Aug8-10, 02:21 PM
P: 9
Hi everyone,

consider two following partitioned matrices:

[tex]\begin{array}{l}
{M_1} = \left[ {\begin{array}{*{20}{c}}
{ - \frac{1}{2}{X_1}} & {{X_2}} \\
{{X_3}} & { - \frac{1}{2}{X_4}} \\
\end{array}} \right] \\
{M_2} = \left[ {\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{{X_1}} & { - I} \\
I & 0 \\
\end{array}} & {\begin{array}{*{20}{c}}
0 & 0 \\
0 & 0 \\
\end{array}} \\
{\begin{array}{*{20}{c}}
0 & 0 \\
0 & 0 \\
\end{array}} & {\begin{array}{*{20}{c}}
{{X_4}} & { - I} \\
I & 0 \\
\end{array}} \\
\end{array}} \right] \\
\end{array}[/tex]

I want to show that spectral radius (maximum absolute value of eigenvalues) of M1 and M2 are equal, but I don't know how!!!!!!!!!

this is general form of my problem the real one is somewhat easier (or maybe more complex)!!!
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trambolin
#2
Aug8-10, 02:57 PM
P: 341
you need more constraints on the problem, one can find counterexamples such that this is not true.
ali987
#3
Aug8-10, 03:25 PM
P: 9
[tex]\begin{array}{l}
{X_1} = A_2^{ - 1}{B_2} \\
{X_2} = \Delta A_2^{ - 1}L \\
{X_3} = \Delta A_1^{ - 1}{L^T} \\
{X_4} = A_1^{ - 1}{B_1} \\
\end{array}[/tex]

In which

[tex]\begin{array}{l}
{A_1} = G + \frac{{{\Delta ^2}}}{4}{L^T}{C^{ - 1}}L \\
{A_2} = C + \frac{{{\Delta ^2}}}{4}L{G^{ - 1}}{L^T} \\
{B_1} = 2\left( {\frac{{{\Delta ^2}}}{4}{L^T}{C^{ - 1}}L - G} \right) \\
{B_2} = 2\left( {\frac{{{\Delta ^2}}}{4}L{G^{ - 1}}{L^T} - C} \right) \\
\end{array}[/tex]

where

[tex]\Delta = [/tex] positive constant coefficient

C and G are symmetric and positive definite.


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