Scalar by matrix derivative $\frac{d {\rm tr} (e^{\bf X} {\bf A})}{d {\bf X}}$

Question

I'm trying to figure out what is the scalar by matrix derivative $\frac{d {\rm tr} (e^{\bf X} {\bf A})}{d {\bf X}}$ equals to?

I know that $\frac{d {\rm tr} (e^{\bf{X}})}{d \bf{X}} = e^\bf{X}$ and $\frac{d {\rm tr} (\bf{X} \bf{A})}{d \bf{X}} = \bf{A}$. I would say $\frac{d {\rm tr} (e^{\bf X} {\bf A})}{d {\bf X}} = {\bf A} e^\bf{X}$ but I don't know if it it's correct or not. Thanks

Answer 1

In fact, your considered gradients are $\dfrac{\partial tr(e^X)}{\partial X} = e^{X^T}$, $\dfrac{\partial tr(XA)}{\partial X} =A^T$.

There is no simple closed form for the required result because (for instance) $(tr(X^2A))'=tr(X'XA+XX'A)=tr((XA+AX)X')$ and $\nabla(tr(X^2A))=(XA+AX)^T$ and not $2(XA)^T$. Yet, using the following integral form of the derivative of $exp(.)$, we can obtain a closed form: $(e^X)'=\int_0^1 e^{uX}X'e^{(1-u)X}du$.

Let $f(X)=tr(e^{X}A)$. One has $f'(X)=tr((e^X)'A)=tr(\int_0^1 e^{uX}X'e^{(1-u)X}du.A)=\int_0^1 tr(e^{uX}X'e^{(1-u)X}A)du=$

$\int_0^1 tr(e^{(1-u)X}Ae^{uX}X')du=tr(\int_0^1 e^{(1-u)X}Ae^{uX}du.X')$.

Thus $\nabla(f)_X=(\int_0^1 e^{(1-u)X}Ae^{uX}du)^T$