I am trying to compute the derivative

$\frac{\partial}{\partial W} \text{Tr}(W^\top A (I\otimes W)B),$

where $W\in\mathbb{R}^{D\times d}, I\in\mathbb{R}^{T\times T}$ is an identity matrix, $A\in\mathbb{R}^{D\times DT}$, and $B\in\mathbb{R}^{dT\times d}$.

I have found a similar post: Derivative involving the trace of a Kronecker product

but it seems that the method is not applicable to my problem.

Thank you!


The technique from the linked post can be applied to the current problem.

Write the function in terms of the trace/Frobenius product, and find its differential $$\eqalign{ \phi &= W:A(I\otimes W)B = A^TWB^T:(I\otimes W) \cr d\phi &= A(I\otimes W)B:dW + A^TWB^T:(I\otimes dW) }$$ At this point, we need use the Pitsianis decomposition on that last term. $$\eqalign{ A^TWB^T &= \sum_k Y_k\otimes Z_k \cr }$$ The matrices $(Y_k,Z_k)$ are shaped like $(I,W)$ respectively.
Finish calculating the differential, then on to the gradient. $$\eqalign{ d\phi &= A(I\otimes W)B:dW + \sum_kY_k\otimes Z_k:(I\otimes dW) \cr &= \Big(A(I\otimes W)B + \sum_k(I:Y_k)Z_k\Big):dW \cr \frac{\partial\phi}{\partial W} &= A(I\otimes W)B + \sum_k {\rm tr}(Y_k)\,Z_k \cr\cr }$$

Another technique uses the SVD of $$B=\sum_k\sigma_ku_kv_k^T$$ to handle the second term of $d\phi$ as follows. $$\eqalign{ A^TW:(I\otimes dW)B &= \sum_k\,A^TW:(I\otimes dW)\sigma_ku_kv_k^T \cr &= \sum_k\,(A^TW\sigma_kv_k):(I\otimes dW)u_k \cr &= \sum_k\,q_k:{\rm vec}(dW\,U_k) \cr &= \sum_k\,Q_k:dW\,U_k \cr &= \sum_k\,Q_kU_k^T:dW \cr }$$ where $$\eqalign{ {\rm vec}(Q_k) &= q_k = A^TW\sigma_kv_k \cr {\rm vec}(U_k) &= u_k \cr }$$ Yielding the gradient as $$\eqalign{ \frac{\partial\phi}{\partial W} &= A(I\otimes W)B + \sum_k Q_kU_k^T \cr }$$

  • $\begingroup$ Thanks a lot for the reply! But one question is: $A^\top W B^\top$ is also a function of $W$, so we cannot do the decomposition, am I right? $\endgroup$ – user3138073 Feb 12 '18 at 21:30
  • $\begingroup$ The dependence of the function on the $A^TWB^T$ term was accounted for by the first term in the differential expression. Perform the Pitsianis decomposition using the current value of $W$. $\endgroup$ – greg Feb 13 '18 at 1:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.