I am interested in characterizing the subdifferential of $f=|X^TX|_*$, i.e., the nuclear norm of $X^T X$.

Two ways I am looking at it right now. Certainly, the singular values of $X^T X$ are the square of singular values of $X$. So the subdifferential of $f$ has to be somewhere along the line.

I suspect there gotta be something like a chain rule since $f$ is just a composition of the nuclear norm function and the function $X^T X$, which is a non-linear operator. There is such a formula but then it gets involved with differentiability of non-linear operator which I am not an expert. Any help? Thanks a lot.


Recall that the nuclear norm is defined by $\|X\|_*=\mathrm{trace}(\sqrt{(X^TX)})$ so you have $$ \|X^TX\|_* = \mathrm{trace}(X^TX) $$ which is differentiable everywhere. Therefore the sub-differential is the same as the derivative which is simply $$ \frac{\partial \|X^TX\|_*}{\partial X} = \frac{\partial \mathrm{trace}(X^TX)}{\partial X} = 2X. $$

| cite | improve this answer | |
  • 1
    $\begingroup$ I would add that the nuclear norm boils down to the trace because $X^\top X$ is symmetric and positive semidefinite. $\endgroup$ – Rodrigo de Azevedo Jun 30 '19 at 19:08

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.