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Is there a power series expansion of the square root of a Hermitian matrix, as a procedure to calculate the square root without taking the inverse or diagonalizing the matrix? I find for scalar number $x$, $$\sqrt{x}=\sum_{k=0}^\infty \frac{(-1)^k \left((-1+x)^k \left(-\frac12\right)_k\right)}{k!}\qquad\text{for }|-1+x|<1$$, under what condition can I use the same expansion for a matrix?

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    The binomial series only works for matrices whose eigenvalues are within the disk of convergence of the usual scalar series...2012-05-13
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    @J.M. You mean the eigenvalues of x or |-1+x| ? In any case, I think the condition on eigenvalues can be met by scaling the appropriate matrix.2012-05-13
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    Well, if $x$ is one eigenvalue of your matrix, then yes, that inequality you have in your post should be satisfied...2012-05-13

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If $H$ is semi-definite positive, choose $c$ positive and large enough so that $H\le2cI$ and use $$ \sqrt{H}=\sqrt{c}\sqrt{I-(I-c^{-1}H)}=\sqrt{c}I-\sqrt{c}\sum_{k=1}^{+\infty}\frac1{2k-1}{2k\choose k}\frac1{4^k}(I-c^{-1}H)^k. $$

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    One might consider letting $c$ be the trace of $\mathbf H$, for instance...2012-05-13
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    Is there a citation for this equation?2012-07-08
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    @mangledorf Yes, the expansion of $\sqrt{1-x}$ at $x=0$.2012-07-08
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    You are rather using the expansion of $\sqrt x=\sqrt{1-(1-x)}$ near $x=0$ which is very slowly converging, aren't you?2015-08-27
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    @Tarek No, as already explained, this uses the expansion of $\sqrt{1-x}$ at $x=0$ (for $x=I-c^{-1}H$), not the expansion of $\sqrt{x}$ at $x=0$ (which would be what, anyway?). After more than two years, this comment of yours is rather disquieting, I must say...2015-08-27
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    How is the condition $2I-c^{-1}H\ge 0$ analogous to the condition $x=0$ for $x=I-c^{-1}H$?2015-08-28
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    The condition $H<2cI$ ensures that every eigenvalue $\lambda$ of $I-c^{-1}H$ is such that $|\lambda|<1$-hence the convergent expansion. There is no condition $x=0$, the expansion of a function $f:x\mapsto f(x)$ at $x=x_0$ refers to a series expansion $f(x)=\sum\limits_{k=0}^\infty c_k(x-x_0)^k$, here $x_0=0$.2015-08-28