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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.

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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 $$ \frac{\partial \|X^TX\|_*}{\partial X} = \frac{\partial \mathrm{trace}(X^TX)}{\partial X} = 2X. $$

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    $\begingroup$ 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

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