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Le $\mathbf{A}$ and $\mathbf{B}$ two psd matrices of size $n$. Additionally we assume that the entries are real and non-negative. Does the following hold: $$\forall (i,j) \in [n], \mathbf{A}_{ij} \leq \mathbf{B}_{ij} \implies \mathbf{A}^2 \preccurlyeq \mathbf{B}^2$$ where $\preccurlyeq$ is the Loewner partial order on the cone of psd matrices.

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No. E.g. $A=\pmatrix{1&0\\ 0&0}\le \pmatrix{1&1\\ 1&1}=B$ entrywise, but $B^2-A^2=\pmatrix{1&2\\ 2&2}$ is indefinite.

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