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

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 i^{th}and k

^{th}column, so the matrix is symmetric.