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.