If we have a discrete saddle point problem with the coefficient matrix
$$ \mathcal{A} =
\begin{bmatrix}
A & B^T \\
B & 0
\end{bmatrix},
$$
then $\mathcal{A}$ is invertible, supposing $B$ has full rank and $A$ is positive semidefinite (which hold in my case), if $\mathrm{ker}(A)\cap\mathrm{ker}(B) = \{0\}$. The basis for the kernel or null space of a matrix can be calculated by null(A)
in MATLAB. But how should I interpret mathematically and calculate the intersection of the two kernels?
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$\begingroup$ Please don't sign your posts. $\endgroup$– nicoguaro ♦Feb 25, 2016 at 15:46
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$\begingroup$ I cannot even do that. $\endgroup$– Zoltán CsátiFeb 25, 2016 at 15:57
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$\begingroup$ I removed your signature at the end of the post "Thanks, Zoltán" $\endgroup$– nicoguaro ♦Feb 25, 2016 at 16:04
1 Answer
In MATLAB, null([A; B]) will find an orthogonal basis for the intersection of the null spaces of A and B.
It seems unlikely that you really want to find this basis, but it's not clear from your question what you're actually trying to do.
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$\begingroup$ I really wanted to find that, because if this is empty, then it means that $\mathcal{A}$ is invertible. Unfortunately, in my code, this is not empty. You are clear, the question was subtle. I am doing domain decomposition using the TFETI method. After prescribing all the constraints, $\mathcal{A}$ should be regular. So I do not want to directly use the inverse, I just wanted to calculate it for debugging purposes. $\endgroup$ Feb 25, 2016 at 16:31