- #1
ali987
- 9
- 0
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)!
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)!