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?
power series expansion of the square root of a Hermitian matrix
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linear-algebra
matrices
power-series
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1Well, if $x$ is one eigenvalue of $y$our matri$x$, then yes, that inequality you have in your post should be satis$f$ied... – 2012-05-13
1 Answers
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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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0The 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