# Tag Info

## Hot answers tagged nullspace

4

There is no way because there is no solution to the given system with $x_2 \neq 0$. This is because the second block of the equation system reads $Ax_2 = 0$, which has no non-trivial solutions because $A$ is SPD (Otherwise $x_2$ would be an eigenvector of $A$ corresponding to a zero eigenvalue).

4

Sagemath looks fast enough. Here is an example created by taking a 999x1000 random rational matrix and appending its column sums at the bottom, so that [1, 1, ... , 1, -1] is the kernel. sage: A = random_matrix(QQ, 999, 1000) sage: B = A.stack(sum(A)) sage: %time v = B.kernel() CPU times: user 5.21 s, sys: 36.4 ms, total: 5.24 s Wall time: 5.24 s Vector ...

3

Quick answer to summarize my comments. Keep in mind that a delicate point is the choice of the truncation threshold in the SVD (what is "numerically zero" and what is not). If you do not see a clear drop in the singular values, then it means your precision is insufficient to identify zeros. Since $\|Ax\|_\infty / \|A\|_\infty \|x\|_\infty$ is of ...

3

I have used the method you described for a similar problem. I found that a combination of your Newton iteration, along with Brent's method on minimizing the smallest singular value gave pretty good results. I believe I used Newton's method first for a few iterations, then switched over to Brent's. This paper of mine contains the description for the analogous ...

2

The problem of finding singular points of a parameter-dependent matrix is called the nonlinear eigenvalue problem in the field of numerical linear algebra. I think you could benefit from using developed methods. It is a quite active area of research. You can consider solving it with the Julia software package NEP-PACK (I am a lead developer of NEP-PACK). NEP-...

1

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.

1

If your sparse matrix isn't too large and you can store it in memory, you could use a sparse (rank-revealing) QR (or LU, or SVD) factorization to determine the kernel of your matrix $A$. Alternately, if you know something about the spectrum of your operator, you could apply a shift to it -- solve for eigenvectors associated with an eigenvalue of $\lambda$ ...

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