Given $A\in\mathbb R^{n\times n}$ and $B_0,B_1\in\mathbb R^{m\times n}$, let us consider the block matrix: $$ M= \begin{pmatrix} A &B_1^T \\ B_0 & 0 \end{pmatrix}, $$ arising in the discretisation of some generalise saddle point problems. I am looking for necessary and sufficient conditions on $A,B_0,B_1$ guaranteeing the invertibility of the matrix $M$. Any ideas?
Moreover I am wondering if there is a flaw in the following reasoning, providing sufficient conditions.
- $\ker B_1^T =\{ 0\}$
- $A|_{\ker B_1}\to \ker B_0$ is injective.
Assume indeed that 1. and 2. hold. Let us consider $\left(u,p\right)\in \mathbb R^{n}\times \mathbb R^m$ such that \begin{equation}\label{eq:matrix_weak_pb_fe_hom} \begin{pmatrix} A & B_1^T \\ B_0 & 0 \end{pmatrix} \begin{pmatrix} u \\ p \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}. \end{equation} First of all, we observe that the second equation implies $u\in \ker B_0$. By projecting, via the orthogonal projection $\pi_{K_1}:\mathbb R^n\to \ker B_1$, the first ''block equation'' of the linear system onto $\ker B_1$ and using $\ker B_1 \cong \left(\operatorname{Im} B_1^T\right)^\perp$, we get $\pi_{K_1}A u =0$. Thanks to 1., we conclude $u=0$. Hence, the first ''block equation'' of the linear system implies $B_1^Tp=0$, from which we deduce $p=0$ by using the injectivity of $B_1^T$.