If $E$ is an $n$-dimensional euclidean space then for all $u\in\Hom(E)$:
$$
|\det u|=\inf\Big\{\prod_j\norm{u(e_j)}: e_1,\ldots,e_n\mbox{ is an ONB}\Big\}~.
$$
By Hadamards inequality the left hand side is dominated by the right hand side. Thus it suffices to find an orthonormal basis $e_1,\ldots,e_n$ such that
$$
|\det u|=\prod_j\norm{u(e_j)}~.
$$
1. We have $|\det u|^2=\det(u^*u)$. So let $e_1,\ldots,e_n$ be an orthonormal basis of eigen vectors of $u^*u$. Since $u^*u$ is positive (definite) all eigen values $\l_j$ are positive (i.e. non negative) and
$$
\det(u^*u)
=\prod_j\l_j
=\prod_j\la u^*u(e_j),e_j\ra
=\prod_j\Vert u(e_j)\Vert^2~.
$$
2. This is more or less the singular value decomposition of $u$: there are orthonormal bases $e_1,\ldots,e_n$ and $b_1,\ldots,b_n$ and positive numbers $\l_1,\ldots,\l_n$ such that for all $j=1,\ldots,n$: $u(e_j)=\l_jb_j$.