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How to show the square root function of a positive semidefinite matrix is differentiable? In this context PSD means symmetric PSD.

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    $\begingroup$ Certainly not at $0$. $\endgroup$ – abx Jan 7 '15 at 6:45
  • $\begingroup$ Did you try applying the definition? $\endgroup$ – András Bátkai Jan 7 '15 at 8:20
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Let $U$ be the set of all positive definite (not semidefinite) matrices, it is an open subset of the vector space $S$ of symmetric matrices. Let $F:U \to U$ be the square root map and $Q: U \to U$ the squaring map. Then $F$ is differentiable.

Proof: Show that $Q$ is a diffeomorphism and appeal to the inverse function theorem. To see that $Q$ is a diffeo, show that $DQ_A$ is injective, for positive definite $A$. Let $X \in S$ be such that $DQ_A (X)=0$. But $DQ_A (X) = AX+XA$. Let $(v_1, \ldots, v_n)$ be a ONB of eigenvectors of $A$, $Av_i =a_i v_i$. Then $AX v_i = - XAv_i = -a_i XA$, so $Xv_i$ is an eigenvector of $A$ with negative eigenvalue. Therefore $Xv_i=0$, hence $X=0$.

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