= Solution
For a nonzero <eigenvector> $u$ of $S^*S$,
$$
\|Su\|^2=\langle Su,Su\rangle
=\langle u,S^*Su\rangle
=\langle u,\lambda u\rangle
=\overline\lambda\|u\|^2.
$$
Therefore $\overline\lambda=\|Su\|^2/\|u\|^2$ is real and nonnegative, and so \b[every <eigenvalue> $\lambda$ is real and nonnegative].
The preceding adjoint identities give $(S^*S)^*=S^*S$, so this <linear operator> is <self-adjoint>. The stated finite-dimensional <spectral theorem> supplies an <orthonormal basis> $e_j$ with $S^*Se_j=\lambda_j e_j$. If $v=\sum_jc_je_j$, then
$$
\|v\|^2=\sum_j|c_j|^2,\qquad
\|Sv\|^2=\langle v,S^*Sv\rangle=\sum_j\lambda_j|c_j|^2.
$$
Since $\lambda_-\leq\lambda_j\leq\lambda_+$,
$$
\lambda_-\|v\|^2\leq\|Sv\|^2\leq\lambda_+\|v\|^2.
$$
Taking nonnegative <square roots> proves
$$
\boxed{\sqrt{\lambda_-}\|v\|\leq\|Sv\|\leq\sqrt{\lambda_+}\|v\|.}
$$
The 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>.
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