0
$\begingroup$

I've got the following system to solve

\begin{align} \begin{bmatrix} A & B \\ B^* & C \end{bmatrix} \begin{bmatrix} u \\ p \end{bmatrix} = \begin{bmatrix} 0 \\ g \end{bmatrix}, \end{align} with the following matrix/ vector dimensions: \begin{align} A&: (a,a) \ (\text{rows, columns}) \\ B&: (a,b) \\ C&: (b,b) \\ u&: (a,1) \\ p&: (b,1) \\ 0&: (a,1) \\ g&: (b,1) \end{align} After rearranging, row 1 is \begin{align} u = -A^{-1} B p \end{align} and row 2 is \begin{align} B^* u + C p = g. \end{align} Inserting $u$ into row 2 yields \begin{align} &-B^* A^{-1} B p + C p = g \\ \\ \leftrightarrow& (B^* A^{-1} B - C) p = -g \\ \end{align}

I have tried solving row 2 for $p$ and inserted the resulting $p$ into row 1, which worked. But I intend to avoid using the inverse of $A$ due to computational reasons. Row 1 can be solved for $u$ by the system \begin{align} Au = -Bp \end{align} with $p$ from row 2. But I don't know how to compute $p$ in row 2 without using the inverse of $A$.

Does anyone have any ideas?

Thanks in advance!

$\endgroup$
6
  • 2
    $\begingroup$ Maybe relevant: en.wikipedia.org/wiki/Invertible_matrix#Blockwise_inversion $\endgroup$ Commented Nov 18 at 10:46
  • 1
    $\begingroup$ If "computational reasons" are your main concern, you could also try to address the problem without making use of the block structure. The block structure does not seem to offer many advantages, here, except if either $A^{-1}$ or $C^{-1}$ is very easy to compute. You could make use of other properties. Do we know if $A$ and $C$ are Hermitian? In this case, the whole matrix would be Hermitian, and the effort to solve the linear system of equations can be reduced. Is it known if the whole matrix is positive definite? You could use the conjugate gradient method, in this case. $\endgroup$ Commented Nov 18 at 17:03
  • $\begingroup$ Thanks for your responses. Unfortunately, $C^{-1}$ is not computed easily. I will have a look at the link later. $A$ and $C$ are not Hermitian and the whole matrix is not positive definite. $\endgroup$ Commented Nov 19 at 9:18
  • $\begingroup$ In this case, I think you do not gain anything from a block-based approach. With all those matrix multiplications and inversions, it will be numerically unstable and will even require approximately as many operations as solving the system as a whole. Also note that the whole system can be solvable even if $A$ and $C$ are not invertible. In this case, any block-based approach is doomed to failure, anyway - think of $\begin{bmatrix}A & B \\ B^{\star} & C\end{bmatrix} = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$. $A$ and $C$ can also be ill-conditioned while the whole system is not. $\endgroup$ Commented Nov 19 at 13:26
  • $\begingroup$ Are these intended to be columns or rows? Why do you not type the entire matrix? $\endgroup$ Commented Nov 23 at 17:39

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.