## Proof That Any Vector Field Represented by a Symmetric Jacobian Matrix is Conservative

Theorem
Any vector field
$\mathbf{F}$
represented by a symmetric Jacobian matrix on a region
$B$
then
$\mathbf{F}$
is a conservative field.
Proof
Take any two points
$x_0 , \: x_1$
in
$B$
and construct a rectangle with edges parallel to the coordinate axis in
$B$
. The
$K$
be the set of all paths from
$x_0$
to
$x_1$

Any element of
$K$
consists of a path which we may represent as by the directions and lengths of the path segment at each vertex on the path eg
$x_{a_1}e_{a_1} x_{a_2}e_{a_2} ... x_{a_m}e_{a_m}$
with
$\mathbf{e_{a_l}$
being the alth basis vector for the ith axis. Take any two paths
$\gamma_p : x_{p_1}e_{p_1} x_{p_2}e_{p_2} ... x_{p_m}e_{p_m}$

$\gamma_q : x_{q_1}e_{q_1} x_{q_2}e_{q_2} ... x_{q_m}e_{q_m}$

Each path segment occur once in each of the two paths since each is the side of a triangle, opposite sides being the same length, so each path can be transformed into the other by a sequence of interchanging of adjacent vectors. Consider the closed path
$C$
consisiting of the segments
$x_{p_s}e_{p_s} x_{p_t}e_{p_t}$
formed into a reacngle.
$\oint_C \mathbf{F} d \mathbf{x} = \oint_C F_p dx_p + F_q dx_q$

Green's Theorem tells us
$\oint_C \mathbf{F} d \mathbf{x} = \int_S ( \frac{\partial F_q}{\partial x_p} - \frac{\partial F_p}{\partial x_q}) dx_p \: x_q =0$

Since the Jacobian matrix is symmetric, so
$\frac{\partial F_q}{\partial x_p} = \frac{\partial F_p}{\partial x_q}$
.
This mean a change of path by swapping any two adjacent path segements does not change the integral. Since pthe paths can be changed into each other by a sequence of such operation, each integral gives the same result and the field is conservative.