Relation between zero singular values and eigenvalues

Question

Given a non-normal diagonalizable square matrix $A\in \mathbb{R}^{N\times N}$, we know its Eigenvalues $\lambda_i$ and its singular values $\sigma_i, i=1,\dots, N$.

Say, I approximate $A$ by $\sigma_i\rightarrow 0\ \forall i\ge k$ for some cutoff index $k$, how can I prove that $\lambda_i=0\ \forall i\ge k$? I did some numerical tests and it very much seems to be the case, but I struggle proving it.

Also, can we say something about the error in the remaining $\lambda_i$ when performing this approximation? Is there some kind of upper limit to the error we make?

Answer 1

Zero is a singular value of $A$ iff there exists $u\neq 0\in\mathbb{R}^N$ such that $ A^TAu = 0$. However, $$ A^TAu=0\implies u^TA^TAu = \|Au\|_2^2=0\implies Au=0, $$ so $u$ is an eigenvector of $A$ corresponding to $\lambda=0$. This trick can be used in the other direction to show that $\lambda=0 \iff \sigma=0$.

As for the sensitivity of $\lambda_i$ to perturbing $A$, I suggest you check out section 7.2 of Matrix Computations by Golub and Van Loan. They show that $\mathcal{O}(\epsilon)$ perturbations in $A$ can induce $\mathcal{O}(\epsilon^{1/p})$ errors in $\lambda$, where $p$ is the multiplicity of $\lambda$, with more detailed estimates in the case where $p=1$.

Golub, Gene; Van Loan, Charles F., Matrix computations., Baltimore, MD: The Johns Hopkins Univ. Press. xxvii, 694 p. (1996). ZBL0865.65009.