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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... – J. M. May 13 '12 at 11:52
@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. – Tarek May 13 '12 at 12:21
Well, if $x$ is one eigenvalue of your matrix, then yes, that inequality you have in your post should be satisfied... – J. M. May 13 '12 at 12:23
up vote 3 down vote accepted

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... – J. M. May 13 '12 at 12:38
Is there a citation for this equation? – Alec Jacobson Jul 8 '12 at 16:17
@mangledorf Yes, the expansion of $\sqrt{1-x}$ at $x=0$. – Did Jul 8 '12 at 19:42
You are rather using the expansion of $\sqrt x=\sqrt{1-(1-x)}$ near $x=0$ which is very slowly converging, aren't you? – Tarek Aug 27 '15 at 20:57
@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... – Did Aug 27 '15 at 22:07

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