Consider a block matrix $A = \left[ \begin{array}{cc} A_1 & \alpha B_1 \\ \alpha B_2 & A_2\end{array}\right]$, where submatrices $A_1 \in \mathbb{R}^{n \times n}$ and $A_2 \in \mathbb{R}^{m \times m}$ are positive definite. (Notice that $B_1$ may differ from $B_2$ and hence $A$ itself is not spd in general)
This means that there exists constants $\lambda_m,\lambda_M > 0$ such that $\lambda_m \leq ||A_i|| \leq \lambda_M$, defined by the lesser and higher eigenvalues of $A_i$, $i=1,2$.
I want to know if is there a result which I can apply to $A$ to ensure the norm of its inverse is bounded from above, provided that $\alpha$ is small enough. I take it it should be possible, because if $\alpha$ is small, then $A$ is basically the diagonal of the positive definite matrices $A_i$, and the inverse of each exists and is also positive definite. But I don't know how I should accomplish that. I want a result that dictates how small $\alpha$ needs to be in order for it to work, like maybe in terms of the norms of $B_i$ as well, I dunno. Is there such a result?
I was trying to do by definition limiting $||A||$ from below (which would then limit $A^{-1}$ from above), like:
$\sup_{||z||=1} || A_1z_1 + A_2z_2|| = \sup_{||z||=1} || A_1z_1 + A_2z_2 + \alpha(B_1z_2 + B_2z_1) - \alpha(B_1z_2 + B_2z_1)||$
$\leq \sup_{||z||=1} || A_1z_1 + A_2z_2 + \alpha(B_1z_2 + B_2z_1)|| + \alpha \sup_{||z=||1}||B_1z_2 + B_2z_1||$
which implies
$||A|| \geq \sup_{||z=1||}|| A_1z_1 + A_2z_2|| + \alpha \inf_{||z||=1}||B_1z_2 + B_2z_1||$. The first term can be bounded below by $||A_1||$ and hence $\lambda_m$ but I don't know how to limit the second...
