= Solution
The <singular value decomposition> of $B$ is $B=U\Sigma V^\dagger$, where $U,V$ are unitary and $\Sigma=\operatorname{diag}(\sigma_1,\ldots,\sigma_n)$, $\sigma_1\ge\cdots\ge\sigma_n>0$ since $B$ is invertible. Its <right singular vector> $v_j$ and <left singular vector> $u_j$ are the corresponding unit columns of $V$ and $U$, with
$$
\boxed{Bv_j=\sigma_ju_j,\qquad B^\dagger u_j=\sigma_jv_j.}
$$
Equivalently $B^\dagger Bv_j=\sigma_j^2v_j$ and $BB^\dagger u_j=\sigma_j^2u_j$. The positive numbers $\sigma_j$ are the <singular values>. Degenerate singular subspaces allow any orthonormal choice of paired vectors.
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