Inverse of a self-adjoint operator

Question

Let $A \colon \mathbb{R}^{n} \to \mathbb{R}^{k}$ and $B \colon \mathbb{R}^{m} \to \mathbb{R}^{k}$ be two bounded linear operators. Denote $$ V \colon \left( x , y , z \right) \mapsto \left( \left( \dfrac{1}{s} - rA^{*}A \right) x + \dfrac{r}{2} A^{*}By , \dfrac{r}{2} B^{*}Ax + \left( \dfrac{1}{t} - rB^{*}B \right) y , \dfrac{1}{r} z \right) . $$

The question is

Under which condition $V$ is invertible and find $V^{-1}$

I can show that $V$ is a self-adjoint operator and when $$ 0 < t < \dfrac{2}{3r \left\lVert A \right\rVert ^{2}} , 0 < s < \dfrac{2}{3r \left\lVert B \right\rVert ^{2}} $$ it is $\rho-$strongly positive therefore $V^{-1}$ exists. The remain part is to find its inverse. I don't know how to find it. I tried to cheat by consider $A$ and $B$ like a matrix instead of operator then we can rewrite $$ V = \begin{pmatrix} \dfrac{1}{s} - rA^{T}A & \dfrac{r}{2} A^{T}B & 0 \\ \dfrac{r}{2} B^{T}A & \dfrac{1}{t} - rB^{T}B & 0 \\ 0 & 0 & \dfrac{1}{r} I \end{pmatrix} $$ but it leads to nowhere and I also not sure that it is an appropriate way...

Answer 1

It's neither cheating nor inappropriate, just using block matrices. First of all, the inverse will be of the form $$ \pmatrix{M^{-1} & 0\cr 0 & r I}$$ where $M$ is the block matrix $$ \pmatrix{1/s - r A^T A & r A^T B/2\cr r B^T A/2 & 1/t - r B^T B\cr} $$

Now use the $2 \times 2$ block matrix inversion formula

Answer 2

One can rewrite $V$ as identity plus 'rank one' as: $$ V = \pmatrix{ s^{-1}I && \\& t^{-1}I & \\&& r^{-1}I&&} - r\pmatrix{A^T\\B^T\\0}\pmatrix{A&B&0} \\ = \pmatrix{ s^{-1}I && \\& t^{-1}I & \\&& r^{-1}I&&} \left( I - r\pmatrix{\sqrt s A^T\\\sqrt t B^T\\0}\pmatrix{\sqrt s A&\sqrt t B&0} \right) $$ Then one can apply the Woodbury matrix identity.