# Regularisation of ill-conditioned matrix-vector problem

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 $$n=102$$

• 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? Oct 19, 2021 at 14:30
• Interesting. Does the problem come from some imaging, tomography etc? What is the physical meaning of A, $\lambda$? Oct 19, 2021 at 14:33
• 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. Oct 19, 2021 at 14:36
• 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? Oct 19, 2021 at 14:36
• 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. Oct 19, 2021 at 14:37