TITLE: Vector derivative w.r.t its transpose $\frac{d(Ax)}{d(x^T)}$
QUESTION [46 upvotes]: Given a matrix $A$ and column vector $x$, what is the derivative of $Ax$ with respect to $x^T$ i.e. $\frac{d(Ax)}{d(x^T)}$, where $x^T$ is the transpose of $x$?
Side note - my goal is to get the known derivative formula $\frac{d(x^TAx)}{dx} = x^T(A^T + A)$ from the above rule and the chain rule.

REPLY [6 votes]: Mathematicians kill each other about derivatives and gradients. Do not be surprised if the students do not understand one word about this subject. The previous havocs are partly caused by the Matrix Cookbook, a book that should be blacklisted. Everyone has their own definition. $\dfrac{d(f(x))}{dx}$ means either a derivative or a gradient (scandalous). We could write $D_xf$ as the derivative and $\nabla _xf$ as the gradient. The derivative is a linear application and the gradient is a vector if we accept the following definition: let $f:E\rightarrow \mathbb{R}$ where $E$ is an euclidean space. Then, for every $h\in E$, $D_xf(h)=<\nabla_x(f),h>$. In particular $x\rightarrow  x^TAx$ has a gradient but $x\rightarrow Ax$ has not ! Using the previous definitions, one has (up to unintentional mistakes):
Let $f:x\rightarrow Ax$ where $A\in M_n$ ; then $D_xf=A$ (no problem). On the other hand $x\rightarrow x^T$ is a bijection (a simple change of variable !) ; then we can give meaning to the derivative of $Ax$ with respect to $x^T$: consider the function $g:x^T\rightarrow A(x^T)^T$ ; the required function is $D_{x^T}g:h^T\rightarrow Ah$ where $h$ is a vector ;  note that $D_{x^T}g$ is a constant. EDIT: if we choose the bases $e_1^T,\cdots,e_n^T$ and $e_1,\cdots,e_n$ (the second one is the canonical basis), then the matrix associated to $D_{x^T}g$ is $A$ again.
Let $\phi:x\rightarrow x^TAx$ ; $D_x\phi:h\rightarrow h^TAx+x^TAh=x^T(A+A^T)h$. Moreover $<\nabla_x(f),h>=x^T(A+A^T)h$, that is ${\nabla_x(f)}^Th=x^T(A+A^T)h$. By identification, $\nabla_x(f)=(A+A^T)x$, a vector (formula (89) in the detestable matrix Cookbook !) ; in particular, the solution above $x^T(A+A^T)$ is not a vector !