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Fix some $n\times n$ symmetric positive definite matrix $A$. Consider the following matrix product,

$$B = AC$$

where $C$ is an arbitrary $n\times n$ matrix. Given $A$, I would like to know if there are known necessary and sufficient conditions on all square matrices $C$ such that the resulting matrix $B$ is also symmetric positive definite? I am more interested in knowing (if possible) necessary conditions.

Edit:

I am only concerned with real matrices.

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  • $\begingroup$ If $C$ is positive and commutes with $A$ then $AC$ will be positive. $\endgroup$ – Chris Ramsey Aug 15 at 21:24
  • $\begingroup$ Could you clarify two things: (1) are you only working with real matrices? (2) when you say positive definite, do you mean strictly positive definite, or positive semi-definite? $\endgroup$ – Yemon Choi Aug 15 at 21:25
  • $\begingroup$ @ChrisRamsey thanks for your answer. Any idea if something similar could be a necessary condition? $\endgroup$ – adenali Aug 15 at 22:09
  • $\begingroup$ @YemonChoi I edited my question to clarify that 1) yes, I am working with real matrices. For 2) I mean strictly positive definite not positive semi-definite. $\endgroup$ – adenali Aug 15 at 22:11
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    $\begingroup$ The obvious necessary and sufficient condition on $C$ is that $C = A^{-1} B$ for some symmetric positive definite $B$. $\endgroup$ – Robert Israel Aug 16 at 18:20
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If $C$ is a positive definite real matrix that commutes with $A$ then $AC = C^{1/2}AC^{1/2}$ which is positive definite. So this is certainly a sufficient condition.

However, it is far from necessary. Consider that $$ \left[\begin{matrix}2 & 1 \\ 1 & 2\end{matrix}\right]\left[\begin{matrix}2 & 0 \\ 1 & 4\end{matrix}\right] = \left[\begin{matrix}5 & 4 \\ 4 & 8\end{matrix}\right]. $$

I am not convinced there is going to be a nice condition that completely describes such $C$.

One necessary condition is that $$ AC = (AC)^T = C^TA \ \ \ \ \textrm{or} \ \ \ ACA^{-1} = C^T $$ If in addition $C$ is symmetric then it commutes with $A$ and then $A^{1/2}CA^{1/2} = AC > 0$ which implies that $C$ is positive definite since $A^{-1}$ is positive as well.

Hardly a complete answer, but that's all I have for now.

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  • $\begingroup$ Thank you for your answer. I guess it's not easy to say anything beyond this, but it's still helpful :) $\endgroup$ – adenali Aug 16 at 21:08

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