Arrangement of singular values in the SVD of A^{-1} I know that the SVD of the inverse of A may be expressed as:
$$
A^{-1} = (U\Sigma V^T)^{-1} = (V^T)^{-1}\Sigma^{-1}U^{-1},
$$
where the inverse of a diagonal matrix $\Sigma$ is just the reciprocal of each diagonal element. But if this is the case, then the diagonal elements of $\Sigma^{-1}$ (the singular values) are in ascending order. I thought they should be arranged in descending order for the SVD. Is this still considered a valid SVD? Or is it only valid with the correct arrangement of singular values?
P.S. For motivation, most textbooks speak of the $\sigma_1$ as the max singular value of the SVD, whereas of course here $\sigma_1$ is the minimum value. I haven't found any other questions on SE relating to the specific arrangement of singular values in the SVD of the inverse of A.
 A: In my view, a singular value decomposition just requires the diagonal entries of $\Sigma$ to be nonnegative, so $A^{-1} = V \Sigma^{-1} U^\top$ does indeed constitute a singular value decomposition. This is also the view Wikipedia takes, though some authors may adopt a different definition. If this is something you might see on a true/false question on an exam, I would certainly consult the reference being used in your particular course to see what their official definition of an SVD is.
The ordering of the singular values in descending order is a useful convention, which is worth enforcing if one does not have a good reason not to. Wanting to order the singular values in ascending order for the inverse of a matrix so the SVD of a matrix and its inverse is certainly a good reason to loosen this convention. In any kind of mathematical exposition, it is helpful to make explicit note of any of these convention issues to avoid possible confusion with the reader.
Certainly by reversing the order of the columns in $U$ and $V$ and the diagonal entries of $\Sigma$ to matrices $\tilde{U}$, $\tilde{V}$, and $\tilde{\Sigma}$, one easily converts the (reverse-ordered) SVD of $A^{-1} = V \Sigma^{-1} U^\top$ to an equivalent SVD $A^{-1} = \tilde{V} \tilde{\Sigma}^{-1} \tilde{U}^\top$ with singular values in descending order.
