Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question
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

1 Answer 1

up vote 3 down vote accepted

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

share|cite|improve this answer
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 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 at 22:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.