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How to approximate sqrt(A+B)

  1. Feb 1, 2014 #1
    Assume that [itex]A,B\in\mathbb{R}^{N\times N}[/itex] are some matrices which do not commute. Is there any way to approximate the matrix

    [tex]
    \sqrt{A + B}
    [/tex]

    under the assumption that [itex]B[/itex] is smaller than [itex]A[/itex]? 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

    [tex]
    (A + B)^{-1}
    [/tex]

    We know for complex numbers [itex]|z|<1[/itex] the series

    [tex]
    \frac{1}{1 + z} = 1 - z + \sum_{n=2}^{\infty}(-1)^n z^n
    [/tex]

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

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

    We also have Taylor series for square root:

    [tex]
    \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
    [/tex]

    Could it be, that this series could be used get something for the [itex]\sqrt{A+B}[/itex]? Actually the previous trick doesn't work now. The problem is that [itex]\sqrt{XY}[/itex] is not related to [itex]\sqrt{X}\sqrt{Y}[/itex] or [itex]\sqrt{Y}\sqrt{X}[/itex] in any obvious way. So I was left without further ideas here.
     
    Last edited: Feb 1, 2014
  2. jcsd
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