My lecturer said " when the matrix M is square and it has complex-conjugate eigenvalues(I mean conjugate pairs of eigenvalues for any non-real eigenvalue), the singular values of M are the modulus of the eigenvalues"
Since it was an incomplete phrase I don't know it he meant " the modulus of the eigenvalues of $M$" or " the modulus of the eigenvalues of $MM'$. I think he meant the former, but anyway I think is wrong as a general statement. I verified this with a couple of examples in matlab, and found it to be true for M=[1 -1; 1 1] , but wrong for M=[ 4/5 -3/5 0; 3/5 4/5 0 ; 1 2 2] So was the lecturer wrong?, What is a sufficient condition for the singular values of a square matrix to be the modulus of its eigenvalues ? I know the singular values of M are the square root of the eigenvalues of $MM'$ and that M and M' have the same eigenvalues
Let $USV^\ast$ and $QRQ^\ast$ be a SVD and a Schur upper triangularisation of $A$ respectively. Then $\|S\|_F^2=\|R\|_F^2$. Since the eigenvalues of $A$ are the diagonal entries of $R$, if the singular values and eigenvalues of $A$ can be ordered in a manner such that for each $i$ the $i$-th singular value has the same modulus as the $i$-th eigenvalue, all off-diagonal entries of $R$ must be zero, meaning that $A$ is a normal matrix. The converse is obviously also true.