If matrix $A$ is such that $A+A^T$ is a positive definite matrix, show that there exists a $B$ such that $A=B^2$, where $B+B^T$ is a positive definite matrix.
My try: since $A+A^T$ is positive matrix, then exists $Q$ such $$Q^{-1}(A+A^T)Q=diag(a_{1},a_{2},\cdots,a_{n})$$ where $a_{i}>0$, $i=1,2,\cdots,n$ then I can't，Thank you
As noted in String’s comment, it suffices to show the property claimed in the Wolfram link, i.e. a real matrix $A$ is positive definite if and only if the symmetric part $S=\frac{A+A^T}{2}$ is positive definite.
We have for any vector $X$,
$$ (SX|X)=\frac{1}{2}.\bigg( (AX|X)+(A^TX|X)\bigg)= \frac{1}{2}.\bigg( (AX|X)+(X|AX)\bigg)=(AX|X) $$
so that the equivalence is clear.
