Suppose we're given the following (physics-based) system of two matrix equations, involving an invertible block lower triangular matrix: \begin{align} \begin{bmatrix} \mathbf{A}_{1,1} & \mathbf{0}\\ \mathbf{A}_{2,1} & \mathbf{A}_{2,2} \end{bmatrix} \begin{bmatrix} \mathbf{x}_1 \\ \mathbf{x}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{0} \\ \mathbf{b}_2 \end{bmatrix}, \end{align} where $\mathbf{A}_{1,1}$ has size $m \times m$, $\mathbf{A}_{2,2}$ has size $n \times n$. The above equation represents two systems of equations which were discretized via the boundary element method.
Since we know that $\mathbf{A}_{1,1}$ must be invertible (because we're told the entire system matrix is invertible, e.g.), then it means $\mathbf{A}_{1,1}$ should not have a null space, therefore the only solution to this system should be \begin{align} \mathbf{x}_1 &= \mathbf{0},\\ \mathbf{x}_2 &= \mathbf{A}_{2,2}^{-1}\,\mathbf{b}_2. \end{align} Assuming the above makes sense, my issue is that the physics dictates that $\mathbf{x}_1 \neq \mathbf{0}$. Does this mean that I have set up my system wrong, and am missing some terms? Or is it possible somehow to impose a constraint on $\mathbf{x}_1$ so that it isn't $\mathbf{0}$? I'm not very experienced in linear algebra, unfortunately.
Thanks.
A block matrix like yours \begin{bmatrix} A & 0 \\ B & C \end{bmatrix} is invertible if and only if $A$ and $C$ are. Its inverse is \begin{bmatrix} A^{-1} & 0 \\ X & C^{-1} \end{bmatrix} provided $$ \begin{bmatrix} A & 0 \\ B & C \end{bmatrix} \begin{bmatrix} A^{-1} & 0 \\ X & C^{-1} \end{bmatrix} = \begin{bmatrix} I & 0 \\ BA^{-1}+CX & I \end{bmatrix} $$ is the identity matrix, which means $X=-C^{-1}BA^{-1}$.
Thus the unique solution to $$ \begin{bmatrix} A & 0 \\ B & C \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix} $$ is $$ \begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} A^{-1} & 0 \\ -C^{-1}BA^{-1} & C^{-1} \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} A^{-1}a \\ -C^{-1}BA^{-1}a + C^{-1}b \end{bmatrix} $$ If $a=0$, then $x=0$. So if you want $x\ne0$, the upper block of $\left[\begin{smallmatrix} a \\ b \end{smallmatrix}\right]$ has to be nonzero.