Given matrices $A, B \in \Bbb R^{2 \times 2}$, compute the derivative of $\mbox{tr}(AXB)$ with respect to $X \in \Bbb R^{2 \times 2}$.
I know that $\frac{\partial tr(AXB}{\partial X}$ is same like $tr(\frac{\partial,AXB}{\partial X})$
i am knot sure about the my understand how the derivative works wrt matrix but this is what i got at first and wanted to know if it is correct?
$$ \Large\frac{\partial}{\partial {\bf X}} \mbox{tr} \left( {\bf f} ({\bf X}) \right) = \mbox{tr} \left( \frac{\partial {\bf f} ({\bf X})}{\partial {\bf X}} \right) $$
My idea was to use this formula on $f(X) := AXB$. Then, I wanted to compute its derivative. However, I got a tensor, i.e., a $4$-dimensional matrix. If compute the trace of this tensor, it seems like something different?
More generally, given matrices ${\bf A}, {\bf B} \in \Bbb R^{n \times n}$, let the linear scalar field $f : \Bbb R^{n \times n} \to \Bbb R$ be defined by
$$ f ( {\bf X} ) := \mbox{tr} \left( {\bf A} {\bf X} {\bf B} \right) = \mbox{tr} \left( {\bf B} {\bf A} {\bf X} \right) = \left\langle \color{blue}{{\bf A}^\top {\bf B}^\top}, {\bf X} \right\rangle$$
where the cyclic property of the trace and the Frobenius inner product were used. Hence, the gradient of $f$ with respect to ${\bf X}$ is
$$ \boxed{ \nabla_{{\bf X}} f ({\bf X}) = \color{blue}{{\bf A}^\top {\bf B}^\top} }$$
Addendum
Let $\partial_{ij} := \partial_{x_{ij}}$. Hence,
$$ \partial_{ij} f ({\bf X}) = \partial_{ij} \mbox{tr} \left( {\bf A} {\bf X} {\bf B} \right) = \mbox{tr} \left( {\bf A} \left( \partial_{ij} {\bf X} \right) {\bf B} \right) = \mbox{tr} \left( {\bf A} \, {\bf e}_i {\bf e}_j^\top {\bf B} \right) = \cdots = \left( {\bf A}^\top {\bf B}^\top \right)_{ij} $$
where $\left( {\bf M} \right)_{ij}$ denotes the $(i,j)$-th entry of matrix ${\bf M}$. Thus, the gradient of $f$ is $\nabla_{{\bf X}} f ({\bf X}) = {\bf A}^\top {\bf B}^\top$.