What is the (sub-)gradient of $\|W^TW-I\|_*$ w.r.t. the matrix variable $W$? For $\|W^TW-I\|_*$ , $W\in R^{m\times n}$ with $n\leq m$ and $I\in R^{n\times n}$ is an identity matrix, $\|\|_*$ is nuclear norm, also named as trace norm. 
Q1: $\|W^TW-I\|_*$ is usually used in machine learning algorithms as a regularization term. Minimizing such term can be done by calculating its (sub-)gradient with regard to the matrix variable $W$. Then how to do? I only know the (sub-)gradient of $\|W\|_*$ as answered in Derivative of the nuclear norm with respect to its argument.
Q2: Maybe Q2 will be much complicated. The (sub-)gradient of $\|W\|_*$ involves Singular Value Decomposition, and Is there a nuclear norm approximation for stochastic gradient descent optimization? provides some way to make it efficient. I am afraid the solution of Q1 will also be computationally expensive. Then any method to make it efficient?
 A: For typing convenience, define the symmetric matrix
$$\eqalign{
 X &= W^TW-I \\
}$$
Write the nuclear norm in terms of this new variable.
Then find its differential and (sub)gradient wrt $W$.
$$\eqalign{
\lambda &= \|X\|_* \\
 &= {\rm Tr}\big((X^TX)^{1/2}\big) \\
 &= \pm{\rm Tr}(X)\quad\Big({\rm choose\,sign\,such\,that\;}\lambda>0\Big) \\
d\lambda
 &= \pm{\rm Tr}(dX) \\
 &= \pm{\rm Tr}(W^TdW+dW^TW) \\
 &= \pm{\rm Tr}\Big((2W)^TdW\Big) \\
\frac{\partial \lambda}{\partial W}
 &= \pm 2W =
\begin{cases}
+2W &{\rm if\;Tr}(W^TW-I)>0 \\
-2W &{\rm otherwise} 
\end{cases} \\
}$$
If you're concerned about computational expense, this result is as inexpensive as one could reasonably hope to achieve.
Update
The matrix sign function is defined such that
$$\eqalign{
S &= \operatorname{sign}(X) = X(X^2)^{-1/2} \\
I &= S^2 \quad\implies S^{-1} = S \\
XS &= SX \\
}$$
For a symmetric matrix 
$${
(X^TX)^{1/2} = (X^2)^{1/2} = XS^{-1} = SX 
}$$
Therefore $S$ can be used to write the nuclear norm and its gradient as 
$$\eqalign{
\lambda &= \operatorname{Tr}(SX) \\
\frac{\partial \lambda}{\partial W} &= 2WS \\
}$$
The previous result is only valid when $S=\pm I,\,$ which is true for 
some matrices.
