 Solution

The singular value decomposition of $B$ is $B=U\Sigma V^{†}$, where $U,V$ are unitary and $\Sigma =\mathrm{diag}({\sigma }_1,\dots ,{\sigma }_n)$, ${\sigma }_1\ge \cdots \ge {\sigma }_n>0$ since $B$ is invertible. Its right singular vector $v_j$ and left singular vector $u_j$ are the corresponding unit columns of $V$ and $U$, withEquivalently $B^{†}B{v}_j={\sigma }_j^2{v}_j$ and $B{B}^{†}{u}_j={\sigma }_j^2{u}_j$. The positive numbers ${\sigma }_j$ are the singular values. Degenerate singular subspaces allow any orthonormal choice of paired vectors.