Can a matrix not admitting an LU decomposition be shifted so that it does admit an LU decomposition?

Question

Almost all square matrices have an (unpivoted) LU decomposition, but some don't. The question I have is whether - assuming $M$ is a square matrix over $\mathbb R$ or $\mathbb C$ which does not admit an LU decomposition - there is some scalar $\lambda$ such that $M + \lambda I$ does have an LU decomposition. I suspect that the answer is yes, but I'm not confident, and I don't know how to show it.

Answer 1

I would use Doolittle's algorithm (which is equivalent to Gaussian elimination without the pivoting) to prove that the answer is yes over $\mathbb R$ or $\mathbb C$, but not in a way that's valid for an arbitrary field. One can pick $\lambda$ very large, so that after performing the shift $M \leftarrow M + \lambda I$, each row-addition in the algorithm $R_i \leftarrow R_i + x_j R_j$ is done with very small $x_j$. This results in the diagonal entries of the matrix not changing much.

So the answer is yes over $\mathbb R$ and $\mathbb C$.