# Necessary (and sufficient) conditions for the following matrix product to be symmetric positive definite?

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.

• If $C$ is positive and commutes with $A$ then $AC$ will be positive. – Chris Ramsey Aug 15 at 21:24
• 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? – Yemon Choi Aug 15 at 21:25
• @ChrisRamsey thanks for your answer. Any idea if something similar could be a necessary condition? – adenali Aug 15 at 22:09
• @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. – adenali Aug 15 at 22:11
• The obvious necessary and sufficient condition on $C$ is that $C = A^{-1} B$ for some symmetric positive definite $B$. – Robert Israel Aug 16 at 18:20

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.