Exercise 7.D.10

Proof. Suppose $\lambda$ is an eigenvalue of $T$ and $v$ a corresponding eigenvector. Then

where the last equality follows because the eigenvalues of self-adjoint operators are real (see 7.13). Therefore, the eigenvectors of $T$ are also eigenvectors of $T^{\text{∗}}T$ with corresponding eigenvalues squared. Since $T$ has a basis consisting of eigenvectors, so does $T^{\text{∗}}T$ and thus all eigenvalues of $T^{\text{∗}}T$ are squares of the absolute values of eigenvalues of $T$. 7.52 now implies that the singular values of $T$ are the absolute values of the eigenvalues of $T$. □