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.