Help me on finding spectral radius!

  1. 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. jcsd
  3. you need more constraints on the problem, one can find counterexamples such that this is not true.
     
  4. [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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