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$$M = \left(\begin{array}{cc|cc|cc|cc|cc} -b_1 &0 &b_2 &0 &0 &0 &\ldots &\ldots &0 &0\\ 0 &-a_1 &0 &a_2 &0 &0 &\ldots &\ldots &0 &0\\ \hline 0 &0 &-b_2 &0 &b_3 &0 &\ddots &\ddots &\vdots &\vdots\\ 0 &0 &0 &-a_2 &0 &a_3 &\ddots &\ddots &\vdots &\vdots\\ \hline \vdots &\vdots &\ddots &\ddots &\ddots &\ddots &\ddots &\ddots &0 &0\\ \vdots &\vdots &\ddots &\ddots &\ddots &\ddots &\ddots &\ddots &0 &0\\ \hline 0 &0 &\ldots &\ldots &0 &0 &-b_{N-1} &0 &b_N &0\\ 0 &0 &\ldots &\ldots &0 &0 &0 &-a_{N-1} &0 &a_N\\ \hline b_1 &0 &0 &0 &\ldots &\ldots &0 &0 &-b_N &0\\ 0 &a_1 &0 &0 &\ldots &\ldots &0 &0 &0 &-a_N \end{array}\right) $$ For $N\ge3$, assume that at least one of $a_i$ is strictly negative, and at least one of $b_{i}$ is strictly positive. For the matrix $M$ above, can I say there exists an eigenvalue whose real part is strictly positive?

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The answer is no. Your matrix is permutation-similar to $A\oplus B$, where $$ A=\pmatrix{-a_1&a_2\\ &-a_2&a_3\\ &&\ddots&\ddots\\ &&&-a_{N-1}&b_N\\ a_1&&&&-a_N}, \ B=\pmatrix{-b_1&b_2\\ &-b_2&b_3\\ &&\ddots&\ddots\\ &&&-b_{N-1}&b_N\\ b_1&&&&-b_N}. $$ Now, consider $$ A=B=\pmatrix{1&2&0\\ 0&-2&3\\ -1&0&-3}. $$ It has both positive and negative diagonal entries, so that the requirements on $a_i$s and $b_i$s are satisfied. However, the eigenvalues of $A$ (or $B$, or $M$, neglecting multiplicities) are $0$ and $-2\pm\sqrt{3}$. None of them is positive.

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  • $\begingroup$ exactly you're right. thank you. $\endgroup$ – BARK May 21 '14 at 6:50

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