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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
  • $\begingroup$ I cannot even do that. $\endgroup$ – Zoltán Csáti Feb 25 '16 at 15:57
  • $\begingroup$ 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.

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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

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