
#1
Aug810, 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)!!! 



#2
Aug810, 02:57 PM

P: 341

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




#3
Aug810, 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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