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I have a following block matrix

M1 =([a11 a12;a21 a22] [b13;b23];[b13;b23]' d33) + ([c -cr]'*[c -cr] 0;0 I) 

now what i observe is that whether i use [c -cr]'*[c -cr] or [c cr]'*[c cr] (the sign change in cr the eigen value of M1 doesn't change. How can i prove this in general? Or this is just a co-incidence that, with the examples i am running there is no change in eigen values of M1 with the sign changes. Please I am really looking forward for your answers. Thanks in advance.

All the elements of M1 are matrices of appropriate dimensions.

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    The above notations are strange, although I could try to guess what you mean. Can you explain more? Is this some language in Maple, Sage, Mathematica, etc?2012-07-13
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    Thanks for the reply, you can understand it like there is a M1 which is the sum of two matrices e.g., M1 = Ma + Mb. Now for Mb matrix the eigen values doesn't change with changing signs of cr (i can prove this by using a similar matrix technique which P = [I 0 0;0 -I 0;0 0 I] where I is the identity matrix of appropriate dimension... Now the question is that what will be the net effect of no change of eigen value of Mb on M1? I hope it clearfy the question a bit2012-07-13
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    Does $M_1$ look like$$\begin{pmatrix}a_{11}&a_{12}&b_{13}\\a_{21}&a_{22}&b_{23}\\b_{13}&b_{23}&d_{33}\end{pmatrix}$$?2012-07-13
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    its like &M1& \begin{align}\begin{pmatrix}a_{11}&0&b_{13}\\0&a_{22}&b_{23}\\b_{13}^\top &b_{23}^\top&d_{33}\end{pmatrix} + \begin{pmatrix}c^\top c &-cc_r&0\\-c_r c&c_rc_r&0\\0 &0 &I\end{pmatrix} \end{align}2012-07-13
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    Also, what's meant by (') followed by a star. I assume (') is transpose?2012-07-13
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    I am new here, unable to type the problem correctly. The second matrix in the sum is \begin{pmatrix} c^\top c&-c^\top c_r&0\\-c_r^\top c&c_r^\top c_r&0\\0&0&I \end{pmatrix}2012-07-13
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    Oh I understand now that star is matrix multiplication! Thanks!2012-07-13
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    Now the problem is with the signs of 'c^\top c_r' and 'c_r^\top c' as they change their signs the eigen values doesn't change. what will be the effect of this on the whole M1... this is my question. I hope that i am able to put my question correctly. I feel extremely sorry for this mess of comments :(2012-07-13

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