If A,B,C,X are matrices, find: $$\frac{\partial \text{tr}[AXBXC^T]}{\partial X}.$$
Here's my initial approach:
$$\partial \text{d} \text{tr}[AXBXC^T] = \text{tr}[\text{d}(AXB)XC^T] + \text{tr}[AXB \text{d} (XC^T)] = \text{tr}[(XC^T(B^T \otimes A) + AXB(B^T \otimes I))\text{d}X]$$
However, this approach doesn't yield a clean and clear derivative. Could someone provide a more straightforward method or correct my approach?
Thank you.
You might want to employ the index notation and make use of the definition of the trace and partial derivative.
Using Einstein summation convention - implied summation over repeated indices - you may rewrite your initial trace as: $$ \text{trace}\left(AXBX{C}^{T}\right)=a_{km}x_{ml}b_{lp}x_{pq}c^{T}_{qk} $$ and then use the matrix derivative definition: $$ \begin{array}{ll} \frac{\partial}{\partial x_{ij}}\left(a_{km}x_{ml}b_{lp}x_{pq}c^{T}_{qk}\right) &=& \left(a_{km}\frac{\partial x_{ml}}{\partial x_{ij}}b_{lp}x_{pq}c^{T}_{qk}\right) + \left(a_{km}x_{ml}b_{lp}\frac{\partial x_{pq}}{\partial x_{ij}}c^{T}_{qk}\right) \\ &=& \left(a_{km}\delta^{im}_{jl}b_{lp}x_{pq}c^{T}_{qk}\right) + \left(a_{km}x_{ml}b_{lp}\delta^{ip}_{jq}c^{T}_{qk}\right) \\ &=& \left(a_{ki}b_{jp}x_{pq}c^{T}_{qk}\right) + \left(a_{km}x_{ml}b_{li}c^{T}_{jk}\right) \\ &=& \left[\text{rearranging terms in products}\right] \\ &=& \left(b_{jp}x_{pq}c^{T}_{qk}a_{ki}\right) + \left(c^{T}_{jk}a_{km}x_{ml}b_{li}\right) \end{array} $$
Notice here, that the resulting elements in both terms read $(j,i)$-th elements of the corresponding matrices, whereas we are seeking for $(i,j)$, as we take the derivative w.r.t. $x_{ij}$. To make the indices conform, you just need to transpose both terms.
By invariance of the trace under rotation of the factors and cancelling of any basis change matrices inside, in Einstein notation in any basis
$$d (A_{ij} X_{jk}B_{kl}X_{lm}C_{im})= A_{ij} dX_{jk}B_{kl}X_{lm}C_{im}+ A_{ij} X_{jk}B_{kl}dX_{lm}C_{im}= Tr\left( B \ X \ C^T \ A \ dX\right)+Tr\left(C^T \ A \ X \ B \ dX\right)$$
\begin{equation} \frac {\partial AXB}{\partial X}=B^T⊗A\quad \text{hence for this problem}\rightarrow \frac{\partial,\left(AXBXC^T\right)}{\partial,XBX} = C ⊗ A \end{equation}
and regarding the second part $\frac{\partial,\left(XBX\right)}{\partial,X}$ in the matrix cookbook in (79)-(80) something similar can be found
– Dennis Marx Jun 13 '24 at 11:00