We know that when projector onto positive semidefinite cone $K$ is differentiable at $X$ then its derivative is $D \Pi_K(X)[H]= Q f^{(1)}(\lambda) \circ (Q^T H Q) Q^T$,
where $X=Q \lambda Q^T$ is eigenvalue decomposition of $X$, $f(x)=[x]_+$,
$ (f^{(1)}(d))_{ij}= \left\{\begin{array}{ll} \cfrac{[d_i]_+ - [d_j]_+}{d_i - d_j} & \text{if}\; d_i \ne d_j \\ f'(d_i) &\text{if}\; d_i =d_j. \end{array} \right.$
We note that $f'(x) =\left\{\begin{array}{ll} 1 &\text{if} \; x>0\\ 0 & \text{if} \; x<0 \end{array} \right.$.
I try to explain why the formula of the derivative does not depend on the choice of orthogonal matrices $Q$, but have no idea. Would you please help me?
Thanks.
