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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!)

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

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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 $A^T$

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