 Solution

For a nonzero eigenvector $u$ of $S^{\ast }S$,Therefore $\overline{\lambda }=\Vert Su{\Vert }^2/\Vert u{\Vert }^2$ is real and nonnegative, and so every eigenvalue $\lambda$ is real and nonnegative.

The preceding adjoint identities give $(S^{\ast }S{)}^{\ast }=S^{\ast }S$, so this linear operator is self-adjoint. The stated finite-dimensional spectral theorem supplies an orthonormal basis $e_j$ with $S^{\ast }S{e}_j={\lambda }_j{e}_j$. If $v={\sum }_j{c}_j{e}_j$, thenSince ${\lambda }_{-}\le {\lambda }_j\le {\lambda }_{+}$,Taking nonnegative square roots provesThe square roots of these extreme eigenvalues are the smallest and largest singular values of $S$; the lower bound can be zero if $S$ has a nontrivial kernel of a linear map.