If A is a nxn singular matrix, then it has a singular value = 0

Question

This is a question on a testexam.

But am I correct in assuming that a singular matrix has det = 0, which gives it an eigenvalue of 0 and that gives it a singular value of 0?

Answer 1

A $n\times n$ matrix $A$ is singular if and only if there is a non-zero vector $v$ such that $A.v=0$. But then $v$ is an eigenvector with eigenvalue $0$.

Answer 2

If $A$ is a square matrix, then $ \lambda $ is a singular value of $A$ , then there is an eigenvalue $\mu$ of $A^*A$ such that $\lambda= \mu^{1/2}$.

If $A$ is singular, then $A^*A$ is singular.... Conclusion ?