# Relation between zero singular values and eigenvalues

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?

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 $$p=1$$.