Does $|M|=|\frac{M+M^T}{2}|$ hold for any matrix $M$ or only some special $M$, and if so, what're they?

Question

Today when I read some materials on multivariate Gaussian distribution, I hit upon a seemingly correct and very simple identity on determinants of two highly related matrices by accident:

If $M$ is a square matrix, $|M|=|\frac{M+M^T}{2}|$.

For the moment, I'm only sure that if $M$ is a positive definite matrix, the above proposition is correct (which can be proved in a zigzag way based on the fact that the $pdf$ of multivariate Gaussian distribution is normalized). Anyone know how to prove this directly? And whether is the condition that $M$ is positive definite necessary?

UPDATE:

Since someone has given a simple counterexample showing that not all $M$ is qualified, to make the problem more meaningful, the focus now is that:

(1) What if $M$ is positive definite?

(2) If (1) can make the proposition become correct, is it necessary? If not, can we give a sufficient and necessary condition on $M$?

Answer 1

That's clearly not true. Consider $$M = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},$$ then $\det(M) = 0$, whereas $\det(\frac{M+M^T}{2}) = -1/4$.

As explained in a comment, a similar example shows that "positive definite" isn't enough either, consider $$M = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}.$$ This example can easily be generalized to any dimension. The only "general" condition I can honestly think of is "$M$ is symmetric", in which case it's trivial.

Answer 2

Let's try with $2\times 2$ matrices; if $$ M=\begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ then $$ \frac{1}{2}(M+M^T)=\begin{bmatrix} 2a & b+c \\ b+c & 2d \end{bmatrix} $$ so its determinant is $$ \frac{1}{4}(4ad-b^2-2bc-c^2) $$ and equality of determinants becomes $$ b^2+2bc+c^2=4bc $$ so $b=c$. Hence the statement only holds when $M$ is already symmetric.