Exercise 7.D.13

Proof. Suppose $T$ is invertible. Then

where the first equality follows from 7.7 and the second because $T$ is surjective. This shows that $T^{\text{∗}}$ is also invertible. Therefore $0$ is not an eigenvalue of $T^{\text{∗}}T$ and so, by 7.52, it cannot be a singular value of $T$.

Conversely, suppose $0$ is not a singular value of $T$. Then $T^{\text{∗}}\mathit{Tv}\ne 0$ for all non-zero $v\in V$. This implies that $\mathit{Tv}\ne 0$ for all non-zero $v\in V$. Thus $T$ is invertible. □