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?
Best Answer
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 coincide, each being equal to the eigenvectors. In this case the singular values are just the moduli of the eigenvalues.