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Thanks guys for the previous answer, Now suppose if I have a matrix e.g

$$M_1 = \begin{pmatrix} \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} & B \\ B' & D \end{pmatrix}$$ and $M_2$ as

$$M_2 = \begin{pmatrix} \begin{pmatrix} a_{11} & -a_{12} \\ -a_{21} & a_{22} \end{pmatrix} & B \\ B' & D \end{pmatrix}$$

How can i prove for this as eig($M_1$) = eig($M_2$), can this be proven?

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2 Answers 2

Thanks for your answer. Can you tell me one more thing, for example, i have

C = [17 12 15 11];
Cr = [2.5 8 10 20];
CC = ([C -Cr].')*([C -Cr]);
eig(CC);
rank(CC);

why there is only one non-zero eigen values, rest is all zero? This is true of any C and Cr i used here. How can i understand this thing?

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This isn't true. Consider the matrix $$M_1 = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 3 & 3 \end{pmatrix}$$ with determinant $9$ and $$M_2 = \begin{pmatrix} 1 & -2 & 3 \\ -2 & 1 & 3 \\ 3 & 3 & 3 \end{pmatrix}$$ with determinant $-63$.

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