Eigenvalues of A when AA'A=A

Question

In some presentation the speaker said that pushing AA'A-A to 0 makes encourages singular values of A to be either 0 or 1, can anyone tell me where this follows from?

Answer 1

Your statement implies that $(AA')^2 = AA'$, so $AA'$ is a projection. Projections only have eigenvalues of 1 or 0. Since the singular values of $A$ are the eigenvalues of $AA'$ the original statement follows.