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I have a linear* problem which arises from an integro-differential system, and writes: $$ (\mathbf{I}+\lambda \mathbf{A})x = b $$ where $\mathbf{A}$ is a real full matrix, size $n\times n$, but is not symmetric and has some zero eigenvalues. I know what these eigenvalues correspond to.

When $\lambda$ increases (but is still much smaller than 1) the condition number of $\mathbf{I}+\lambda \mathbf{A}$ (let's call it $\mathbf{K}$) deteriorates extremely rapidly (to $10^{10}$ and more). Even with a direct solver (luckily $n=100$ is enough for me) the results become meaningless.

I have no certainty, but I believe this is linked with the zero eigenvalues of $\mathbf{A}$. Is there a way I could exploit my knowledge of the corresponding eigenvectors to regularize the numerical resolution?

(* note that this is actually the fixed-point linear problem that I am solving, the original problem has $\lambda=\lambda(x)$)

EDIT: $\mathbf{A}$ itself has indeed a very large condition number (numerically evaluated by numpy with L2-norm), $10^{19}$, with $\sigma_{\min} = 1.8e-07$ and $\sigma_{\max} = 3.2e11$ ($\sigma_{\min}$ should analytically be 0 of course, since as said above, $\mathbf{A}$ has a 0 eigenvalue). Its largest eigenvalue has modulus about 600, here is the complete spectrum for $n=102$

enter image description here

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    $\begingroup$ If matrix $A$ has large real negative eigenvalues, then for some small $\lambda$, 0<$\lambda \ll$1 the eigenvalues of $I+\lambda A$ can become zero , so the matrix $I+\lambda A$ becomes singular. Is that what is going on? $\endgroup$
    – Maxim Umansky
    Oct 19, 2021 at 14:30
  • $\begingroup$ Interesting. Does the problem come from some imaging, tomography etc? What is the physical meaning of A, $\lambda$? $\endgroup$
    – Maxim Umansky
    Oct 19, 2021 at 14:33
  • $\begingroup$ No, it's a mechanical problem in 2D half-space that we've brought to a 1D segment using kernels. $A$ includes in some involved way a 3rd order differentiation, which is partly borne on the kernels, and $x$ contain degrees of freedom of a $P_3-C^1$ (Hermite) finite element representation of the velocity. $\endgroup$
    – Joce NoToPutinsWarInUkraine
    Oct 19, 2021 at 14:36
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    $\begingroup$ If there are large singular values in $K$, then $A$ must have large norm itself. What is the range of $\sigma_{\min}(A), \sigma_{\max}(A)$? What are the maximum and minimum norm of its eigenvalues? $\endgroup$
    – Federico Poloni
    Oct 19, 2021 at 14:36
  • $\begingroup$ My suspicion is that the very large s.v. of $K$ "correspond" to the 0 eigen values of $A$, which themselves correspond to the fact that constant or linear velocities are associated with 0 eigen value. $\endgroup$
    – Joce NoToPutinsWarInUkraine
    Oct 19, 2021 at 14:37

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