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I wonder whether the following property holds true: For every real symmetric matrix $S$, which is positive in both senses: $$\forall x\in{\mathbb R}^n,\,x^TSx\ge0,\qquad\forall 1\le i,j\le n,\,s_{ij}\ge0,$$ then $\sqrt S$ (the unique square root among positive semi-definite symmetric matrices) is positive in both senses too. In other words, it is entrywise non-negative.

At least, this is true if $n=2$. By continuity of $S\mapsto\sqrt S$, we may assume that $S$ is positive definite. Denoting $$\sqrt S=\begin{pmatrix} a & b \\ b & c \end{pmatrix},$$ we do have $a,c>0$. Because $s_{12}=b(a+c)$ is $\ge0$, we infer $b\ge0$.

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No. If $$A = \begin{pmatrix}10&-1&5\\-1&10&5\\5&5&10\end{pmatrix},$$ then $A$ is positive definite but does not have all entries positive, while $$ A^2 = \begin{pmatrix}126&5&95\\5&126&95\\95&95&150\end{pmatrix} $$ is positive in both senses.

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Robert Bryant already showed via an example that the answer is "no". To come up with lots of counterexamples, recall that (under some mild assumptions) if $A$ has maximal eigenvalue $\lambda_{\text{max}}$ and corresponding unit eigenvector $\mathbf{v}$ then $(A/\lambda_{\text{max}})^k \rightarrow \mathbf{v}\mathbf{v}^*$ as $k \rightarrow \infty$.

So if you pick any matrix that is (a) positive semidefinite with a negative entry, and (b) has a unique maximal eigenvalue with corresponding entrywise positive eigenvector, then repeatedly squaring $A$ will eventually give a counterexample to the original question.

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    $\begingroup$ I like this conceptual analysis; it's what I had in mind when I constructed the example that I did. In order to really complete this as an argument, though, one needs to explain that, while there is no symmetric $n$-by-$n$ matrix $A$ that satisfies both conditions (a) and (b) when $n=2$, there is such a matrix when $n=3$ (and hence, when $n\ge 3$). Of course, once one has produced a counterexample when $n=3$, there's no need for the general argument. $\endgroup$ – Robert Bryant Oct 19 at 11:48

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