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 '16 at 15:46
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$\begingroup$ I cannot even do that. $\endgroup$ – Zoltán Csáti Feb 25 '16 at 15:57
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$\begingroup$ I removed your signature at the end of the post "Thanks, Zoltán" $\endgroup$ – nicoguaro♦ Feb 25 '16 at 16:04
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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.
answered Feb 25 '16 at 16:20
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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. $\endgroup$ – Zoltán Csáti Feb 25 '16 at 16:31
