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I'm working on a problem and was lead to trying to find an approximation for the square root of a matrix. I came across a way of doing this using holomorphic functional calculus. However, my first attempt at a computation resulted in failure. Did I miss something? --

Wikipedia gives the expansion of the square root matrix as follows: $$\frac{A^{1/2}}{\|A\|^{1/2}} = I - \sum_{n = 1}^{\infty}\left\lvert {1/2 \choose n}\right\rvert \left(I-\frac{A}{\|A\|}\right)^n = I - \frac 1 2\left(I-\frac{A}{\|A\|}\right) - \frac 1 8 \left(I-\frac{A}{\|A\|}\right)^2 \,...$$ In this post, I'm taking the matrix norm to be the max of the absolute values of the column-sums of matrix $A$.

I thought I might be able to get an estimate of square root matrix by taking the first two terms and using the third term as an error bound. But a simple numerical example appears to not work. I'm not sure if I'm using the expansion incorrectly, or there's some hypothesis I'm missing. In particular for $A = \begin{pmatrix} 4 & 0 \\ 0 & 16 \end{pmatrix}, \|A\| = 16$.

I get an error:$$\big\|A^{1/2}/\|A\|^{1/2} - I + (I-A/\|A\|)/2 \big\| = 1/8.$$ Error bound using the third term: $$\big\|(I-A/\|A\|)^2/8 \big\| = 9/128.$$

[Remark: This is cross posted.]

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    $\begingroup$ the Wikipedia formula you cite says that it needs the eigenvalues of $A$ to lie in the interval (0,2), which does not apply to your choice of $A$. $\endgroup$
    – Carlo Beenakker
    Commented Sep 26, 2020 at 14:22
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    $\begingroup$ doesn't it state the requirement that the spectrum of $A$ is contained within a unit disc in the complex plane? $\endgroup$
    – Carlo Beenakker
    Commented Sep 26, 2020 at 15:08
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    $\begingroup$ As another approach, see mathoverflow.net/questions/301272/… $\endgroup$
    – Mahdi - Free Palestine
    Commented Sep 26, 2020 at 19:03
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    $\begingroup$ Why do you expect the first omitted term to bound the error ? This series does not have alternating signs, and as far as I understand this result is not true even for scalar inputs. $\endgroup$
    – Federico Poloni
    Commented Sep 28, 2020 at 14:14
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    $\begingroup$ No, the $n+1$st term does not estimate the error in Taylor series, in general. Case in point: $1/(1-x)$. $\endgroup$
    – Federico Poloni
    Commented Sep 29, 2020 at 10:02

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