# Derivative of AXB with respect to X

Assume that $$A\in\mathbb{R}^{m\times m}$$ and $$B\in\mathbb{R}^{n\times n}$$ are two constant matrices. How can I find the partial derivative of $$AXB$$ with respect to $$X$$ in which $$X\in\mathbb{R}^{m\times n}$$? In fact, how to compute $$\frac{\partial(AXB)}{\partial X}.$$

I think that it is easy to see that $$\mathrm{vec}(AXB)=(B^T\otimes A)\mathrm{vec}(X),$$ where the notation $$\otimes$$ denotes the Kronecker product, and $$\mathrm{vec}(X)$$ is the vectorization of the matrix $$X$$. Therefore, we have $$\frac{\partial ((B^T\otimes A)\mathrm{vec}(X))}{\partial \mathrm{vec}(X)}=B^T\otimes A.$$ Here, I can not understand what is the relationship between $$\frac{\partial(AXB)}{\partial X}$$ and $$\frac{\partial ((B^T\otimes A)\mathrm{vec}(X))}{\partial \mathrm{vec}(X)}$$?

Thank you very much for the help.

• $X\mapsto AXB$ is a linear map from $\Bbb R^{m\times n}$ to itself. – Angina Seng Jun 4 '19 at 4:48

## 1 Answer

When vectorizing the $$m\times n$$ matrix $$X$$, you obtain a vector $$v =\mathrm{vec}(X)$$ whose $$i$$th element is given by $$v_i = X_{i\%m, i//m+1},$$ where $$i//m$$ is the integer division and $$i\%m$$ is the remainder of the integer division.

Now consider what you mean by $$\frac{\partial AXB}{\partial X},$$ You are taking the derivative of an object with two indices with respect to an object with two indices, so you are looking at all terms of the form $$\frac{\partial [AXB]_{i,j}} {\partial X_{k,l}},$$ the matrix derivative vectorizes this indexing along two indices, the row is the position along $$i,j$$, the column is the position along $$k,l$$, so $$\frac{\partial [AXB]_{i,j}} {\partial X_{k,l}}= (B\otimes A)_{m(j-1)+i, m(l-1)+k},$$ where the indexing reverses the vectorization operation.

• Thank you so much for your help. – Math-Data Jun 4 '19 at 8:22
• I hope it's clear :) it's a way to make sense of four different integers in a way that we can use what we know about nirmal vectors and matrices – Riccardo Sven Risuleo Jun 4 '19 at 9:38
• Yes, it is completely right. Thanks a lot. – Math-Data Jun 4 '19 at 15:35
• Here's a way to state the result without the Kronecker product or modular arithmetic. $$\frac{\partial(AXB)_{ij}}{\partial X_{kl}} = A_{ik} B_{lj}$$ – greg Jun 4 '19 at 18:34