## Proof that the Jacobian Matrix Representing a Conservative Vector Field is Symmetric

Theorem
If
$\mathbf{F}$
is a continuously differentiable conservative vector field, then the Jacobian matrix representing
$\mathbf{F}$
is symmetric.
Proof
In general
$\mathbf{F}$
is a vector field with
$n$
components and
$n$
arguments.
We can write
$\mathbf{F} =(F_1 , F_2 , ...., F_n )$
and
$F_i = F_i (x_1 , x_2 , ..., x_n )$
for
$i = 1,2,3,..., n$
.
Since
$\mathbf{F}$
is conservative there is some function
$f=f(x_1 , x_2 , ..., x_n )$
such that
$F_i = \frac{\partial f}{\partial x_i } , \: i=1,2,3,...,n$

Differentiating the last expression with respect to
$x_k$
gives
$\frac{\partial F_i}{\partial x_k} = \frac{\partial^2 f}{\partial x_k \partial x_i } = \frac{\partial^2 f}{\partial x_i \partial x_k } = \frac{\partial F_k}{\partial x_i}, \: i,k=1,2,3,...,n$

$\frac{\partial F_i}{\partial x_k}$
is the entry in the ith and kth column, so the matrix is symmetric.