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

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 a_{l}

^{th}basis vector for the i

^{th}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.