Suppose I have a linear dynamical system that is described by the following differential equation
$$\dot{\bf{x}} = \bf{A}\bf{x}$$
I wish to represent the right-hand side of this system as a gradient of a function $U({\bf x})$. Now, if $\bf{A}$ is symmetric, I can construct a quadratic form of $\bf{A}$, so that:
$$ U({\bf x}) = \frac12 \bf{x}^\top \bf{A} \bf{x} $$
and
$$ \dot{{\bf x}} = {\bf I} \dfrac{dU}{d{\bf x}} $$
where $\bf I$ is identity matrix of a proper order. However, how to construct $U({\bf x})$ when $\bf{A}$ is not symmetric? Note that for the problem I am solving I can't have $\bf I$ something else rather than identity.
One idea I was trying to explore was decomposing $\bf{A}$ as a sum of symmetric and skew-symmetric matrices, but it led me nowhere.
If $Ax$ was the gradient of $U$ we'd have $$A_{ij} = \dfrac{\partial}{\partial x_i} (A x)_j = \dfrac{\partial^2}{\partial x_i \partial x_j} U = \dfrac{\partial^2}{\partial x_j \partial x_i} U = \dfrac{\partial}{\partial x_j} (A x)_i = A_{ji}$$ so $A$ would have to be symmetric.