Derivative of exponential function wrt a vector

Question

Let $\mathbf{A}$ be a $k\times k$ invertible matrix, let $\mathbf{x}$ be a $k\times 1$ vector and let $\mathbf{1}$ be a $k\times 1$ vector of ones. For a generic $k\times 1$ vector $\mathbf{z}$, let the function $\exp\left(\cdot\right)$ be defined as follows:

$\exp\left(\mathbf{z}\right)=\exp\left(\left[\begin{array}{c} z_{1}\\ z_{2}\\ \vdots\\ z_{k} \end{array}\right]\right)=\left[\begin{array}{c} e^{z_{1}}\\ e^{z_{2}}\\ \vdots\\ e^{z_{k}} \end{array}\right]$

Is the following equality true? If so, under what conditions?

$\frac{d}{d\mathbf{x}}\;\mathbf{1}'\mathbf{A}^{-1}\exp\left(\mathbf{Ax}\right)=\exp\left(\mathbf{Ax}\right)$

More in general, I am looking for a scalar function whose derivative with respect to vector $\mathbf{x}$ is $\exp\left(\mathbf{Ax}\right)$ (or its transpose).

Thanks a lot in advance!

Answer 1

OK, I'll give you a counterexample that your quest is impossible for a 2x2 matrix, for simplicity; that is, you assume there is a potential φ (pardon the physicsese...), s.t.
$$\frac{d}{d\mathbf{x}} \phi =\exp\left(\mathbf{Ax}\right) \qquad \Longrightarrow \\ \frac{d}{d x _1} \phi =\exp\left(A_{11}x_1+A_{12}x_2\right) , \qquad \frac{d}{d x _2} \phi =\exp\left(A_{21}x_1+A_{22}x_2\right) . $$

Now the integrability condition for this simplest linear system is, by above, $$ \left (\frac{d}{d x _2} \frac{d}{d x _1} - \frac{d}{d x _1} \frac{d}{d x _2}\right) \phi=0=A_{12}\exp\left(A_{11}x_1+A_{12}x_2\right) - A_{21} \exp\left(A_{21}x_1+A_{22}x_2\right) . $$ That is, $$ A_{12}e^{\left(A_{11}-A_{21}\right)x_1 } = A_{21} e^{\left(A_{22} -A_{12}\right)x_2}, $$ hopeless except in special circumstances, such as diagonal A, as proffered in @greg 's answer.

Answer 2

Define two new variables and their differentials $$\eqalign{ y &= Ax &\implies dy=A\,dx \cr e &= \exp(y) &\implies de = {\rm Diag}(e)\,dy = E\,dy \cr }$$ Now finding the differential and gradient of your function is straightforward $$\eqalign{ \phi &= 1:A^{-1}e \cr d\phi &= 1:A^{-1}de \cr &= 1:A^{-1}E\,dy \cr &= 1:A^{-1}EA\,dx \cr &= A^TEA^{-T}1:dx \cr \frac{\partial\phi}{\partial x} &= A^TEA^{-T}1 \cr &= A^T{\rm Diag}(e)A^{-T}1 \cr\cr }$$ In some of the steps above, a colon is denote the trace/Frobenius product, i.e. $$A:B = {\rm tr}(A^TB)$$ There are many ways to rearrange the arguments in a Frobenius product, which follow from the cyclic properties of the trace function.

For example, all of the following are equivalent $$\eqalign{ A:BC &= BC:A \cr &= A^T:(BC)^T \cr &= B^TA:C \cr &= AC^T:B \cr\cr }$$ For your general question, notice that if we use an invertible diagonal matrix $$A={\rm Diag}(a)$$ then (since diagonal matrices commute) the above result reduces to $$\frac{\partial\phi}{\partial x} = E1 = \exp(Ax) $$