Matrix Vector Product, as a Derivative?

Question

Let $p\in\mathbb{R}^d$ and $A\in \mathbb{R}^{d\times d}$, can the product $Ap$ be the derivative of some function?

In particular given a matrix $A$, how can I construct $H:\mathbb{R}^d\to \mathbb{R}$, such that $\nabla H(p)=Ap$. Note below I can do this for specific $A$ but I want it for general $A$ (say positive semi-definite, but possibly singular).

$\textbf{My Work :}$

• If $A$ is the identity matrix then we have $\nabla H(p)=\nabla \frac{1}{2}\|p\|^2=p=Ap$.

• If $A$ can be written as $B^TB$ for some matrix $B\in\mathbb{R}^{d\times d}$, then take $\nabla H(p)=\nabla \frac{1}{2}\|Bp\|^2=B^TBp=Ap$.

What about more general $A$?

Answer 1

If $A$ is symetric you can set $H(p) = \frac{1}{2}p^TAp $.

If you then calculate the derivative using the chain rule you get $\nabla H(p) = Ap$.

If $A$ is not symetric you can't find such a function $H$.

For example let $a_{12} \neq a_{21}$. You can write down the product of $Ap$ and start integrating the first element of the vector with respect to $p_1$:

Then

$H(p) = \frac{1}{2}a_{11}p_1^2 + p_1a_{12}p_2 + p_1a_{13}p_3 +...$

Similarly when you integrate the second element with respect to $p_2$ you get:

$H(p) = p_1a_{21}p_2 + \frac{1}{2}a_{22}p_1^2 +...$

So we get a contradiction. You can do a similar argument for any non symetrical matrix.