First order derivative of inner product w.r.t. vector

Question

Let $\mathbf{x}\in\Bbb{R}^n$, $\mathbf{y}\in\Bbb{R}^m$, and $A\in\Bbb{R}^{m\times n}$. Also, let $f\colon\Bbb{R}^n\to\Bbb{R}$ given by $$ f(\mathbf{x}) = \big(A\mathbf{x}\big)\cdot\mathbf{y} = \big(A\mathbf{x}\big)^\top\mathbf{y} $$ What is the first derivative of $f$ (which will belong to $\Bbb{R}^n$) with respect to $\mathbf{x}$, i.e., $$ \frac{\partial f}{\partial\mathbf{x}}=\frac{\partial(A\mathbf{x})^\top\mathbf{y}}{\partial\mathbf{x}}=? $$

Will it be just $$ \frac{\partial(A\mathbf{x})^\top\mathbf{y}}{\partial\mathbf{x}} = A^\top\mathbf{y}, $$

or something else? Thanks a lot in advance!

Have you tried writing this out in coordinates to develop a sense of the answer? Then you'll probably see a more elegant proof. — Simon S, Nov 13 '14 at 17:17
Actually I did, but I would like some confirmation, as I am not sure... — nullgeppetto, Nov 13 '14 at 17:18
Yes, $A^Ty$ is the correct expression for the gradient of $f(x).;$ — greg, Mar 12 '23 at 03:27

Simon S · Accepted Answer · 2014-11-13T20:12:53.307

1

Looks right as for vectors over $\mathbb{R}^j$,

$$Ax \cdot y = y \cdot Ax = y^TAx = (A^Ty)^T x$$

edited Nov 13 '14 at 20:12

answered Nov 13 '14 at 17:26

Simon S

26,524

Could you explain your notation please? Because I can only guess what you want to say! – nullgeppetto Nov 13 '14 at 20:08
Updated notation so it's hopefully now more consistent with your original post – Simon S Nov 13 '14 at 20:13
Right, that makes perfect sense! I should have seen that too! Thanks a lot! – nullgeppetto Nov 13 '14 at 20:15

First order derivative of inner product w.r.t. vector

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