TITLE: How to prove that there exists a $B$ such that $A=B^2$
QUESTION [7 upvotes]: 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

REPLY [2 votes]: 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.