# What is the relationship between singular values and eigenvalues of a matrix?

Suppose I have a general $n\times n$ real matrix $A$. And suppose that $A$ has an SVD of the form $A=U^T S V$ with S of the form $I_m \oplus D$ where $I_m$ is the identity $m\times m$ matrix and $D$ is a matrix of size $n-m \times n-m$.

This means that $A$ has $m$ singular values equal to 1. Would this suffice to conclude that $A$ has $m$ eigenvalues of modulus 1? Why? Why not?

In general the eigenvalues have no direct relation to the singular values. The only thing you can really be sure of is that the eigenvalues, in magnitude, lie in the interval $$[\sigma_n,\sigma_1]$$. Also each singular value of zero is in fact an eigenvalue (with the corresponding right singular vector as an eigenvector).
The exception is when $$A$$ is unitarily diagonalizable, which is equivalent to being normal. Then the left singular vectors and the right singular vectors basically coincide (differing by a complex sign at most), and are eigenvectors. In this case the singular values are just the moduli of the eigenvalues.
• @DylanDijk I don't have a reference, but I can give a quick proof. On the high end, $\sigma_1$ is (either by definition or essentially by definition) $\sup_{\| x \|=1} \| Ax \|$. Any eigenvalue's magnitude is at most equal to this, since if $x$ is a corresponding unit eigenvector then $|\lambda|=\| Ax \|$. On the low end, either $\sigma_n=0$ in which case the lower bound is trivial, or it isn't. If it isn't, then $A$ (assumed square to begin with) is invertible.
• @DylanDijk (Cont.) So if $\tau_i$ are the singular values of $A^{-1}$ and $\eta_i$ are its eigenvalues, then the previous argument tells you $\tau_1 \geq |\eta_i|$. Now just express $\tau_1$ and $\eta_i$ in terms of $\sigma_i$ and $\lambda_i$ to get the lower bound.