I have a $y \times u$ matrix $A$ and a $u \times y$ sized matrix $B$, where $y \leq u$.

Given $A$, I want to choose $B$ such that $\det(AB)$ is maximised, subject to the constraint that the columns of $B$ have magnitude 1.

Is there a closed form solution to this?


1 Answer 1


Presumably $A$ has full row rank (or else $\det(AB)$ is always zero). Therefore $A$ admits a polar decomposition $PU^T$, where $P$ is positive definite and $U^T$ has orthonormal rows (i.e., $U$ has orthonormal columns). Then $\det(AB)=\det(P)\det(U^TB)$. Since $U^T$ has orthonormal rows and the columns of $B$ are unit vectors, each column of $U^TB$ has norm $\le1$. It follows from Hadamard’s determinant theorem that $\det(U^TB)\le1$. Consequently, $\det(AB)\le\det(P)$. The optimal solutions are the matrices $B$ that make equality hold. In particular, $B=U=A^T(AA^T)^{-1/2}$ is an optimal solution.

  • $\begingroup$ Can the optimal solution also be derived via an SVD decomposition of A? $\endgroup$
    – Jabby
    Commented Jun 24 at 22:37
  • $\begingroup$ @Jabby Yes. If $A=USV^T$ is a SVD, you may also take $B=VU^T$ or $B=V\operatorname{diag}(1,\ldots,1,\det(U))$. The former choice is actually the one in my answer. In the latter choice, the purpose of the diagonal matrix is to make sure that $\det(AB)$ is positive. $\endgroup$
    – user1551
    Commented Jun 24 at 22:47
  • $\begingroup$ I'm slightly confused. $V$ here would be $u \times u$ and $U$ would be $y \times y$, so I don't understand how $B = V U^T$ is something you can compute. $B$ needs to be $u \times y$. $\endgroup$
    – Jabby
    Commented Jun 24 at 22:57
  • $\begingroup$ @Jabby Oh, I mean a compact/economic SVD, not a full SVD. That is, if $A=U\pmatrix{S&0}W^T$ is a full SVD and $V^T$ is the top $y$ rows of $W^T$ (or $V$ is the leftmost $y$ columns of $W$), then $USV^T$ is the compact SVD of $A$, with $V^T$ having the same size as $A$. $\endgroup$
    – user1551
    Commented Jun 24 at 23:02
  • $\begingroup$ Ah ok. For my sake, are you able to spell out the proof that $𝐡=π‘‰π‘ˆ^𝑇$ in this SVD decomposition case? Can you also comment on the pros and cons of computing B via $𝐡=π‘‰π‘ˆ^𝑇$ (SVD decomposition) vs. $𝐡=A^T(AA^T)^{-1/2}$ (polar decomposition) (e.g. in terms of computational efficiency)? $\endgroup$
    – Jabby
    Commented Jun 24 at 23:16

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