# How to approximate sqrt(A+B)

1. Feb 1, 2014

### jostpuur

Assume that $A,B\in\mathbb{R}^{N\times N}$ are some matrices which do not commute. Is there any way to approximate the matrix

$$\sqrt{A + B}$$

under the assumption that $B$ is smaller than $A$? Could it be possible to write the square root precisely with some infinite series under some conditions?

Here's a related problem, which I know already. Suppose we were interested in the matrix

$$(A + B)^{-1}$$

We know for complex numbers $|z|<1$ the series

$$\frac{1}{1 + z} = 1 - z + \sum_{n=2}^{\infty}(-1)^n z^n$$

The same series will work for bounded operators with norm less than one, so by using the formula $(XY)^{-1}=Y^{-1}X^{-1}$ we get

$$(A + B)^{-1} = ((\textrm{id} + BA^{-1})A)^{-1} = A^{-1}(\textrm{id} + BA^{-1})^{-1}$$
$$= A^{-1}\Big(\textrm{id} - BA^{-1}+ BA^{-1}BA^{-1} + \sum_{n=3}^{\infty}(-1)^n(BA^{-1})^n\Big)$$
$$= A^{-1} - A^{-1}BA^{-1}+ A^{-1}BA^{-1}BA^{-1} - \cdots$$

We also have Taylor series for square root:

$$\sqrt{1 + z} = 1+ \frac{1}{2}z- \frac{1}{4}z^2 + \frac{3}{8}z^3 + \sum_{n=4}^{\infty}\frac{(-1)^{n+1}3\cdot 5\cdot\ldots\cdot (2n-3)}{2^n}z^n$$

Could it be, that this series could be used get something for the $\sqrt{A+B}$? Actually the previous trick doesn't work now. The problem is that $\sqrt{XY}$ is not related to $\sqrt{X}\sqrt{Y}$ or $\sqrt{Y}\sqrt{X}$ in any obvious way. So I was left without further ideas here.

Last edited: Feb 1, 2014