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$\textbf{Question: }$ If all the eigenvalues $\lambda_i$ of an $n\times n$ matrix $A$, have a strictly negative real part then prove that all the coefficients $a_j$ of the characteristic polynomial $f_A(\lambda)= \lambda^n + a_{n-1}\lambda^{n-1}+\cdots+a_0$ of that matrix are strictly positive. Does the converse hold true?

Edit:(Ignore my comments and still looking for some hints or ways how to approach this problem)

$\textbf{Comments: }$ I need some help with this problem. I tried to solve it but I don't have any idea. I even think the question is wrong(probably I am wrong) but consider $f_A(\lambda)= (\lambda - \lambda_1) \cdots (\lambda - \lambda_1)$, this implies that $a_0=(-1)^n\lambda_1 \cdots \lambda_n$ which I think can't be positive all the time!

As for the converse I think it doesn't hold. We can just take as an example a diagonal matrix with strictly positive values in the diagonal. Am I right?

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  • $\begingroup$ Regarding your first comment; why not? Suppose the eigenvalues are real. If $n$ is odd, then $a_0 = -\lambda_1\cdots\lambda_n > 0$. If $n$ is even, then $a_0 = \lambda_1\cdots\lambda_n > 0$ as well. $\endgroup$ – Lord Soth Nov 5 '13 at 19:33
  • $\begingroup$ For your second comment, i.e. for the converse, you cannot start with a diagonal matrix with positive values as its characteristic polynomial will have a negative coefficient (why?). Note that to prove the converse, you assume that you have a characteristic polynomial with strictly positive coefficients, and then prove that the roots of that polynomial will have strictly negative real parts. $\endgroup$ – Lord Soth Nov 5 '13 at 19:41
  • $\begingroup$ ok thnx, I understood my mistakes but I still need help cause I can't come up with a solution $\endgroup$ – Albanian_EAGLE Nov 11 '13 at 0:30

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