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.


I am only concerned with real matrices.

  • $\begingroup$ If $C$ is positive and commutes with $A$ then $AC$ will be positive. $\endgroup$ Aug 15, 2020 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, 2020 at 21:25
  • $\begingroup$ @ChrisRamsey thanks for your answer. Any idea if something similar could be a necessary condition? $\endgroup$
    – adenali
    Aug 15, 2020 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, 2020 at 22:11
  • 1
    $\begingroup$ The obvious necessary and sufficient condition on $C$ is that $C = A^{-1} B$ for some symmetric positive definite $B$. $\endgroup$ Aug 16, 2020 at 18:20

1 Answer 1


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.

  • $\begingroup$ 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, 2020 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.