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I have a following block matrix

\begin{align*} M_1 &= M_a + M_b\\\ M_a &= \begin{pmatrix} a_{11} & 0 &b_{13}\\\ 0 & a_{22}& b_{23}\\\ b_{13}^T & b_{23}^T & d_{33} \end{pmatrix}\\\ Mb &= \begin{pmatrix} c^Tc & -c^Tcr & 0\\\ -cr^Tc& cr^Tcr &0\\\ 0 &0& \mathrm{id}\end{pmatrix} \end{align*}

now what i observe is that whether i use $$ \begin{pmatrix} c^Tc & -c^Tcr & 0\\\ -cr^Tc& cr^Tcr &0\\\ 0 &0& \mathrm{id}\end{pmatrix} \quad\text{or}\quad \begin{pmatrix} c^Tc & c^Tcr & 0\\\ cr^Tc& cr^Tcr &0\\\ 0 &0& \mathrm{id}\end{pmatrix}$$

(the sign change, the eigenvalues of $M_b$ don't change. How can i generalize this of the whole $M_1$? That the eigen values of $M_1$ will not change under the change of signs in Mb. Please I am really looking forward for your answers. Thanks in advance.

All the elements of $M_1$ are matrices of appropriate dimensions. As i am new here, I am sorry for the last questions which i made a mess because i couldn't write the question properly :(

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Probably I am just rephrasing your question, but if you can find an invertible matrix $S$ such that $SM_a = M_aS$ and such that $SM_b S^{-1}$ is just $M_b$ with signs of some elements changed then eigenvalues of $M_1$ (with transformed $M_b$) will remain same. –  user10001 Jul 13 '12 at 10:21
I am unable to find such S. I can find an S such that SM_bS^-1 is just M_b with the change of signs as i am mentioning. Another important thing i want to mention is for M_b the rank of \begin{pmatrix}c^\top c&-c^\top c\\-cr^\top c&cr^\top c\end{pmatrix} is always 1. –  Mohsin Jul 13 '12 at 11:01

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