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Suppose there is a (non-symmetric) real square matrix $A$ with symmetric part $A+A^T$.

What are some conditions on $A$ that are sufficient for $A+A^T$ to be positive definite?

For example, if the eigenvalues of $A$ are strictly positive is $A+A^T$ positive definite? (EDIT: This part of the question is answered in the negative in the comments).

This would then give the result I actually want which is that given two positive definite matrices $C$ and $D$ it follows that the symmetric part of $CD$ is also positive definite. (EDIT: But I think it is still not clear if $CD+DC>0$ - this is (perhaps) a slightly more special case than $A+A^T$ with $A$ having positive eigenvalues.)

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wat can you say about eigen values of $A^T$ –  Praphulla Koushik Jan 11 '14 at 12:42
oh yes.. that does not help! I am helpless!! –  Praphulla Koushik Jan 11 '14 at 13:09
Consider $A = \begin{pmatrix}\frac14 & 1\\0 & \frac14\end{pmatrix}$, the eigenvalues of $A$ are all positive but $A + A^T$ are not positive definite. –  achille hui Jan 11 '14 at 13:43
A matrix is sum of symmetric and anti symmetric parts. $x' A x$ is zero if $A$ is anti symmetric. So a matrix is pd, iff its symmetric part is pd. –  user114628 Jan 11 '14 at 14:12
Also note that eigenvalues of a non-symmetric pd matrix are not necessarily positive. –  user114628 Jan 11 '14 at 14:17

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