0
$\begingroup$

Let $\mathbf{x}\in \Bbb{R}^n, \mathbf{y}\in \Bbb{R}^m$, and $\mathbf{A} \in \Bbb{R}^{n \times m}$. What is $\nabla_{\mathbf{x}}\mathbf{x}^{T}\mathbf{Ay}$?

I understand that Ay is a linear combination of the columns of A, where the coefficients are the entries of y:

[a1]y1 + [a2]y2 + ... + [a_m][y_m]

Thus, it results in a column vector of dim nx1. Let's call this column vector s = [s1 ... s_n].

I also understand that x.T dimension is 1xn, so this multiplication would result in a sum like:

x1*s1 + .... + x_n*s_n

I'm confused when it comes to the actual gradient, because if I'm taking it with respect to x I'm not sure how to express this in terms of the variables A and y. I don't know anything about the order of the members of x. If I knew they were all order 1, I'd be able to express this just as a sum of the terms in the s vector, but I think that x's members can be arbitrary order. Basically I have no idea where to go from here.

$\endgroup$
0
$\begingroup$

Let $f (\mathrm x) := \mathrm c^\top \mathrm x$. The directional derivative of $f$ in the direction of $\rm v$ at $\rm x$ is given by

$$\lim_{h \to 0} \frac{f (\mathrm x + h \, \mathrm v) - f (\mathrm x)}{h} = \mathrm c^\top \mathrm v = \langle \mathrm c, \mathrm v \rangle$$

and, thus, the gradient of $f$ is $$\nabla f (\mathrm x) = \mathrm c$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.