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Let $A$ and $B$ be Hermitian positive-definite matrices. I am looking for a perturbative expansion of $$ (A+\epsilon B)^{-1/2} $$ in orders of $\epsilon$. I suspect such an expansion is possible since for real $a,b$, $$ (a+\epsilon b)^{-1/2} = \frac{1}{\sqrt{a}}-\epsilon\frac{b}{2 a^{3/2}} + \epsilon^2\frac{3b^2}{8a^{5/2}} + \mathcal{O}(\epsilon^3) $$ It is not obvious to me how to derive an analogous result for matrices.

ABW
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cf. my answer in Derivative (or differential) of symmetric square root of a matrix

For $A$ symmetric $>0$ and $H$ small symmetric

$(A+H)^{-1/2}=A^{-1/2}-A^{-1/2}\int_0^{+\infty}e^{-t\sqrt{A}}He^{-t\sqrt{A}}dtA^{-1/2}+O(||H||^2)$

When $H=\epsilon B$ where $\epsilon$ is small and $B$ symmetric

$(A+\epsilon B)^{-1/2}=A^{-1/2}-\epsilon A^{-1/2}\int_0^{+\infty}e^{-t\sqrt{A}}Be^{-t\sqrt{A}}dtA^{-1/2}+O(\epsilon ^2)$

  • How does the formula change if $A$ is positive semidefinite? – Doris May 16 '19 at 23:57
  • Then, there is no more formula. –  May 17 '19 at 11:07
  • I want to calculate $\sqrt{A+\epsilon B}$ (not its inverse), but the support of $A$ is smaller than the support of $B$. There is clearly some answer for this question but I am having trouble finding the answer. – Doris May 19 '19 at 00:28