2
$\begingroup$

Does the product of singular values of a rectangular matrix have a simple expression?

For a square matrix with an eigenvalue decomposition, the product of the singular values is the product of the eigenvalues is the determinant, which has a simple expression that is a weighted sum of the products of the elements of the matrix, with those weights being given by the Levi-Civita symbol.

Is there similarly a simple expression for the product of singular values? I have no idea how to find this in general, but feeding small rectangular matrices into Mathematica suggests that at least for real matrices there is a similar expression, raised to some power.

My motivation is that I have a rectangular matrix whose elements are functions of a variable. I want to know for what values of that variable there exist zero modes corresponding to zero singular values. For square matrices, the most convenient way to check this is to see when the determinant is zero (implying a zero eigenvalue). I'm hoping there is some equivalent convenient algebraic expression for rectangular matrices.

$\endgroup$
0
$\begingroup$

For square matrices, the product of the singular values is technically the absolute value of the determinant, not necessarily equal to the determinant itself.

I don't know if there is a simple way to express this algebraically in terms of only $A$ itself, but there is at least a simple conceptual way to think about it: we have that the product of the singular values is:

$$\prod_{i=1}^r \sigma_i = \prod_{i=1}\sqrt{\sigma_i^2} = \sqrt{\prod_{i=1}^r \sigma_i^2 }= \sqrt{\det A A^T} = \sqrt{\det A^T A} \,. $$

Of course both $AA^T$ and $A^TA$ are both square even if $A$ is not. Then if one of the two is full rank at least then we can identify the product of the singular values of $A$ with the square root of the "volume" corresponding to the determinant of which ever of $AA^T$ or $A^TA$ is full rank (if either is full rank, which need not be true).

If neither $AA^T$ nor $A^TA$ is full rank, there are probably still convoluted ways to use the "thin SVD" to relate the product of the singular values to the determinant of the determinant of some operator on $\mathbb{R}^r$ (where $r$ is the rank of $A$), although I do not know if any would be particularly elegant or "standard".

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.