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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

math110
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  • It says a lot about the relation between being positive definite and having the symmetric part positive definite here: http://mathworld.wolfram.com/PositiveDefiniteMatrix.html – String Dec 17 '13 at 15:25
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    Only no proof are given there. Actually the statements there solve your question. It states that a matrix $A$ is positive definite iff $\frac{1}{2}(A+A^T)$ is positive definite. Also it states that any positive definite matrix has exactly one positive definite square root. These two combined solves the problem. – String Dec 17 '13 at 15:27
  • Also note that you cannot necessarily diagonalize your matrix. Being positive definite does NOT imply being diagonalizable... – String Dec 17 '13 at 15:31

1 Answers1

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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.

Ewan Delanoy
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