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Let $A\in\mathbb{R}^{n\times n}$ be positive definite in $\mathbb{R}^n$. Prove that the real part $\lambda_{1}=Re\lambda$ of any eigenvalue $\lambda\in\mathbb{C}$ of $A$ is positive.

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    By definition, a Pos. Def. matrix is symmetric and thus all its eigenvalues are real...and non-negative, of course. Am I missing something here? – DonAntonio Mar 31 '14 at 20:26

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The OP does not give the definition of a PD matrix he has chosen, that is:

for every $x\in\mathbb{R}^n\setminus \{0\}$, $x^TAx> 0$ ; in particular $A$ is not assumed to be symmetric.

Then the asked result is true. cf.

Does non-symmetric positive definite matrix have positive eigenvalues?

In these circumstances there is no reason to remove 1 point to ellya; so I add it 1.

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    Didnt see that bit, the answer is no if non symmetric since symmetric and positive definite $\longleftrightarrow$ $n$ real positive eigenvalues. Thank you for the +1 :) – Ellya Apr 02 '14 at 19:43
  • @ Joe Berg , you are not new in this forum. When somebody gives an answer to one of your question, the least you can do is to say thank. –  Sep 17 '14 at 08:03
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If $A\in\mathbb{R^{n\times n}}$ is positive definite symmetric then there exists an orthonormal matrix $S\in\mathbb{R^{n\times n}}$ which is composed of the eigenvectors of $A$, and a diagonal matrix $D\in\mathbb{R^{n\times n}}$ which is composed of the eigenvalues of $A$, then:

$S^{-1}AS=D$ here $S^{-1}=S^T$, Thus $\lambda_i=D_{ii}=[S]_i^TA[S]_i\gt 0$ as $A$ is positive definite, this holds for all eigenvalues. (note that $[S]_i$ is the i-th column of $S$)

Ellya
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