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Let $B$ be a symmetric $n \times n$ matrix, and let $B=DA$, where $D$ is an $n \times n$ diagonal matrix with positive real values, and $A$ is an $n \times n$ matrix, not necessarily symmetric, but with all ($n$) real eigenvalues. Note that $BD^{-1}=A$ so $D^{1/2}[D^{-1/2}BD^{-1/2}]D^{-1/2}=A$, so if $B$ is positive definite, then $A$ will have $n$ positive real eigenvalues using matrix similarity of $[D^{-1/2}BD^{-1/2}]$ and $A$. My question is, if $A$ has all positive real eigenvalues, is that sufficient for $B$ to be positive definite? What I conjecture is: $B$ positive definite $\Longleftrightarrow$ $A$ has positive real eigenvalues. However, I am stuck on the $\Longleftarrow$ part, and beginning to doubt if it is true.

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  • $\begingroup$ I guess you mean that $A$ has $n$ positive eigenvalues. Then $D^{-1/2}BD^{-1/2}$ has these eigenvalues and is therefore positive definite (symmetric matrices having $n$ positive eigenvalues are positive definite). Let $y\in\mathbb R^n$, $y = D^{-1/2}x$. Then$$(By,y) = (D^{-1/2}BD^{-1/2}x,x)\ge\alpha\|x\|^2 > 0$$for all $y$. $\endgroup$ – Friedrich Philipp Apr 3 '17 at 4:25
  • $\begingroup$ Correct. Edited to make that clearer. $\endgroup$ – agilestats Apr 3 '17 at 4:44
  • $\begingroup$ Instead, you could also write that $A$ has no purely complex eigenvalues. $\endgroup$ – Friedrich Philipp Apr 3 '17 at 4:48

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