# Differentiability of the Schatten $p$-norm on positive definite matrices

Let $$V$$ be the vector space of symmetric matrices in $$\Bbb R^{n\times n}$$. For $$p\in (1,\infty)$$, the Schatten $$p$$-norm of $$M\in V$$ is defined as $$\|M\|_p =(\sum_{i=1}^n \sigma_i(M)^p)^{1/p}$$ where $$\sigma_1(M),\ldots,\sigma_n(M)$$ are the singular values of $$M$$. Now, let $$C\subset V$$ be the cone of positive semi-definite matrices. It follows from this old post that $$\nabla \|M\|_p = \|M\|_p^{1-p}M^{p-1}\qquad \forall M\in C.$$

Where is a reference for the above statement?

• As the deadline approach, I should say that I am even fine with just a citable reference. – Surb Mar 13 at 14:19

Let $$M\in C$$. Then $$M$$ is positive semi-definite, thus $$|M|=M$$ and $$\|M\|_p = \big(\operatorname{tr}M^p\big)^\frac 1p.$$ A scalar-by-matrix derivative using the chain rule gives $$\frac{\partial\;\;}{\partial M}\,\|M\|_p \:=\:\left.\frac{d\;}{dx}x^\frac 1p\right|_{x\,=\,\operatorname{tr}M^p} \cdot\frac{\partial\;\;}{\partial M}\,\operatorname{tr}M^p \\[3ex] \qquad\quad=\:\frac 1p\big(\operatorname{tr}M^p\big)^{\frac 1p-1}\cdot pM^{p-1} \\[3ex] =\:\|M\|_p^{1-p}M^{p-1}$$ This is a straightforward application of matrix calculus, and might be the reason why there is no quick $$W^3$$ hit for this derivation.
For the derivative of $$\,\operatorname{tr}(M^p)\,$$ in the second term you may also look at page 155 here .
• Thank you for your answer! This is pretty much what I am looking for. What is a "quick $W^3$ hit"? – Surb Mar 14 at 15:34
• Sounds good $+\;\ddot\smile\quad W^3 = WWW$ – Hanno Mar 14 at 15:46