Derivative of trace and norm

Question

I need to find the gradient with respect to $W$ of this equation

$\text{tr}(A’ C^{-\frac{1}{2}} W C^{-\frac{1}{2}} A) + \frac{a}{2} ||W-D+Z/a||_{F}^2$

Where $A, C, W, D$, and $Z$ are matrices, and $a$ is a constant

Answer 1

For ease of typing, define the matrix variables $$\eqalign{ B &= \left(C^{-1/2}AA^TC^{-1/2}\right)^T \\ Y &= W - D + \frac 1aZ &\quad\implies\quad dY = dW \\ }$$ and let's use a colon to denote the trace/Frobenius product $$\eqalign{ B:A &= {\rm Tr}(B^TA) \\ A:A &= \big\|A\big\|^2_F \\ }$$ Write the objective function in terms of these new variables.
Then calculate its differential and gradient. $$\eqalign{ \phi &= B:W + \tfrac a2(Y:Y) \\ d\phi &= B:dW + \tfrac a2(2Y:dY) \\ &= (B + aY):dW \\ \frac{\partial\phi}{\partial W} &= B+aY \\ &= \left(C^{-1/2}AA^TC^{-1/2}\right)^T + \Big(aW - aD + Z\Big) \\ }$$

Answer 2

It is important to know, which matrix norm is used, let me consider the Frobenius norm here \begin{align} ||W||_{\text{F}}=\sqrt{\sum_{i,j} w_{ij}^2} \end{align} Then one can derive these two formulae: \begin{align} \frac{\partial\, \text{tr}(AWC)}{\partial W}=A^{\text{T}}C^{\text{T}},~\frac{\,\partial ||W||_{\text{F}}}{\partial W}=\frac{W}{||W||_{\text{F}}} \end{align} by which you can finish your case. The derivation is easy, it follows the rule $\left(\frac{\partial s(W)}{\partial W}\right)_{ij}=\frac{\partial s(W)}{\partial w_{ij}}$ for any scalar function $s$.