Jacobian of $x^{\top} \cdot A$ - question about Jacobian of a row vector

If $$x \in \mathbb{R}^m$$ and $$A \in \mathbb{R}^{m \times n}$$, and we let $$A = [a_1,a_2,\dots,a_n]$$ where each where each of the $$a_i \in \mathbb{R}^m , 1 \leq i \leq n$$, and \begin{align} a_i &= \begin{bmatrix} a_{1,i} \\ a_{2,i} \\ \vdots \\ a_{m,i} \end{bmatrix} \end{align} then we have that $$x^T A=x^T\big[a_1,a_2,\dots,a_n\big]=\begin{bmatrix} x^T \cdot a_1, & x^T \cdot a_2, & \dots & x^T \cdot a_m \end{bmatrix}\\$$, so $$x^TA$$ is a 1 by m row vector. How do you take the Jacobian of this?

I was always under the impression that the Jacobian is typically applied to column vectors, so for example, the Jacobian of $$[x^TA]^T=\begin{bmatrix} x^T \cdot a_1\\ x^T \cdot a_2\\ \vdots \\ x^T \cdot a_m \end{bmatrix}\\$$ would be $$A^T$$ (which is A transpose). However, how does this work for just $$x^TA$$ since it is a row vector?

This may be a dumb question, but I am having trouble with this (maybe its all in the notation!)

• $x^TA$ is a 1 by n row vector. Commented Jan 22, 2020 at 15:43

The function $$\mathbb{R}^m \to \mathbb{R}^n, x \mapsto x^TA$$ maps each vector $$x=(x_1,...,x_m)$$ to the vector $$(\sum_{k=1}^{m}x_ka_{k,1},...,\sum_{k=1}^{m}x_ka_{k,n})$$.
Therefore the (i,j) entry of the jacobian matrix is $$\frac{\partial \sum_{k=1}^{m}x_ka_{k,i}}{\partial x_j}=a_{j,i}$$, which is the (i,j) entry of the matrix $$A^T$$