Division of Positive Definite Matrix - Inner Product Space

Question

I am trying to derive a simplified expression of something requiring some linear algebra.  I have gotten the stage below:
$$G = frac{mathbf{1}^H mathbf{A}^{-1} mathbf{B} mathbf{A}^{-1} mathbf{1}}{mathbf{1}^H mathbf{A}^{-1}mathbf{1}}$$
for completeness to provide more details:
$mathbf{A} = mathbf{I} + mathbf{s}mathbf{s}^H$ and
$mathbf{B} = mathbf{s}mathbf{s}^H$.
I have having trouble finding a simplification, if one exists.  I have looked at Sherman-Morrison formula.  Nothing is appearing to me and any assistance is appreciated.  My expectation is that after a simplification, a closed form solution to a problem I am working will pop out and will be greatly helpful.
Thanks in advance

user1551 · Answer

I assume that $sinmathbb C^n$. Let $mathbf1^Hs=z$. Since
begin{aligned}
(I+ss^H)^{-1} &= I-frac{ss^H}{1+s^Hs},\
(I+ss^H)^{-2} &= left(I-frac{ss^H}{1+s^Hs}right)^2
=I-frac{2ss^H}{1+s^Hs}+frac{(s^Hs)ss^H}{(1+s^Hs)^2},\
mathbf1^H(I+ss^H)^{-1}mathbf1 &= n-frac{|z|^2}{1+s^Hs},\
mathbf1^H(I+ss^H)^{-2}mathbf1 &= n-frac{2|z|^2}{1+s^Hs}+frac{(s^Hs)|z|^2}{(1+s^Hs)^2},\
end{aligned}
we have
begin{aligned}
G&=frac{mathbf{1}^H mathbf{A}^{-1} mathbf{B} mathbf{A}^{-1} mathbf{1}}{mathbf{1}^H mathbf{A}^{-1}mathbf{1}}
=frac{mathbf{1}^H mathbf{A}^{-1} (mathbf{A}-I) mathbf{A}^{-1} mathbf{1}}{mathbf{1}^H mathbf{A}^{-1}mathbf{1}}
=1-frac{mathbf{1}^H mathbf{A}^{-2} mathbf{1}}{mathbf{1}^H mathbf{A}^{-1}mathbf{1}}\
&=1-frac{n-frac{2|z|^2}{1+s^Hs}+frac{(s^Hs)|z|^2}{(1+s^Hs)^2}}{n-frac{|z|^2}{1+s^Hs}}.
end{aligned}
You may simplify last expression if you want.

Division of Positive Definite Matrix - Inner Product Space

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