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
I assume that $s\in\mathbb C^n$. 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.