Eigenvalues of $A^\ast A$

Question

How can I show that if $A$ is unitarily diagonalisable, then an eigenvector $v$ of $A$ is also an eigenvector of $A^*A$ and eigenvalue of $A^*A = |\lambda|^2$, where $\lambda$ is the eigenvalue of $A$?

Answer 1

And for the eigenvectors, the eigenvectors of a diagonal matrix are $(1,0,\cdots ,0)^T$, $(0,1,0,\cdots ,0)^T$ , ... $(0,\cdots ,0,1)^T$

And since $A$ and $A^*A$ are similar to a diagonal matrix using the same diagonalizing transformation (as Jan explained), thus they have the same eigenvectors.