What is the Spectral Radius of Partitioned Matrices M1 and M2?

[ali987]

Hi everyone,

consider two following partitioned matrices:

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

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)!

 

[trambolin]

you need more constraints on the problem, one can find counterexamples such that this is not true.

 

[ali987]

$$\begin{array}{l}<br /> {X_1} = A_2^{ - 1}{B_2} \\ <br /> {X_2} = \Delta A_2^{ - 1}L \\ <br /> {X_3} = \Delta A_1^{ - 1}{L^T} \\ <br /> {X_4} = A_1^{ - 1}{B_1} \\ <br /> \end{array}$$

In which

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

where

$$\Delta =$$ positive constant coefficient

C and G are symmetric and positive definite.