Exercise 6.3.8

Proof. Suppose $v$ is an eigenvector of $A^{\text{∗}}A$ corresponding to a non-zero eigenvalue $\lambda$, i.e., $A^{\text{∗}}Av=\lambda v$. Then $A{A}^{\text{∗}}Av=A(\lambda v)=\lambda (Av)$, i.e., $\lambda$ is an eigenvalue of $A{A}^{\text{∗}}$ with corresponding eigenvector $Av$. Similarly, we can show that the non-zero eigenvalues of $A{A}^{\text{∗}}$ are also eigenvalues of $A^{\text{∗}}A$. Thus non-zero eigenvalues of the matrices $A^{\text{∗}}A$ and $A{A}^{\text{∗}}$ coincide. □