Showing that a symmetric matrix $\mathbb{R}^{n\times n}$ vector multiplied will yield a conservative vector field Hi I am trying to check if a Force field
$$
F(x) = Ax \hspace{1cm} A\in \mathbb{R}^{n\times n} \hspace{0.5cm}
$$
with A being a symmetric matrix, is conservative.
Because no explicit function is given, you cannot apply the integrability condition or similar methods. Thus my and the textbooks approach is to try to find a potential G(x) for which the following is true:
$$F(x) = \nabla G(x)$$
The problem starts here. Namely the textbook plainly states that  $G(x) = \frac{1}{2}x \cdot Ax$, unfortunately I am unable to even show that this is the case.
Here is what I tried:
$$ 
G(x) = \frac{1}{2}x \cdot Ax = \frac{1}{2} \begin{pmatrix}x_1\\x_2\\...\\x_n\end{pmatrix} \cdot \begin{pmatrix} a_{11} & a_{12} &... & a_{1n} \\a_{21} & a_{22} &... & a_{2n} \\ ... & ... &... & ... \\a_{n1} & a_{n2} &... & a_{nn} \end{pmatrix}\begin{pmatrix}x_1\\x_2\\...\\x_n\end{pmatrix}
$$
This yield in the following expression:
$$
G(x) = \frac{1}{2} a_{11} x_1^2  +  \frac{1}{2}a_{12}x_2x_1  +...+  \frac{1}{2} a_{1n}x_{n}x_1 
\hspace{0.25cm} + \hspace{0.25cm} \frac{1}{2} a_{21} x_1x_2  +  \frac{1}{2}a_{22}x_2^2  +...+  \frac{1}{2} a_{2n}x_{n}x_2 +\hspace{0.25cm}...
$$
Now the issue is that if I build the first derivative w.r.t $x$ then only in the very first term in which $x$ is squared the coefficient $\frac{1}{2}$ won't be there. Meaning that if I reason correctly the two matrices $A$ cannot be equal to each other...
I would be very thankful if you could explain to me how I can show that $F(x)$ is conservative and why $G(x)$ in this form is a Potential of $F$.
All help is appreciated, thanks!
 A: There aren't nearly as many $\frac 1 2$ terms as it seems. Because $A$ is symmetric, $a_{ij} = a_{ji}$ for all $i,j$. Thus all the off-diagonal terms in $G(x)$ combine with one another. In other words, the terms of the form $\frac 1 2 a_{ij}x_ix_j$ and $\frac 1 2 a_{ji}x_jx_i$ for $i\neq j$ combine together into $a_{ij}x_ix_j$. This leaves $\frac 1 2$ coefficients only on the diagonal terms (where $i=j$), where they cancel with the factor of $2$ that comes down when differentiating.
For a more formal proof, we can write
\begin{align*}
G(x) &= \frac 1 2\sum_{i=1}^n a_{ii}x_i^2 +\frac 1 2 \sum_{\substack{1\leq i,j\leq n\\i\neq j}} a_{ij}x_ix_j \\
&= \frac 1 2\sum_{i=1}^n a_{ii}x_i^2 + \frac 1 2 \left( \sum_{1\leq i<j\leq n} a_{ij}x_ix_j + \sum_{1\leq i<j\leq n} a_{ji}x_ix_j \right) \\
&= \frac 1 2\sum_{i=1}^n a_{ii}x_i^2 + \frac 1 2 \left( 2\sum_{1\leq i<j\leq n} a_{ij}x_ix_j \right) \\
&= \frac 1 2\sum_{i=1}^n a_{ii}x_i^2 + \sum_{1\leq i<j\leq n} a_{ij}x_ix_j .
\end{align*}
Here the second equality follows from that fact that if $i\neq j$ then $i<j$ or $j<i$, and the third equality follows from the fact that $A$ is symmetric.
Not sure if this is as formal as you'd like, but perhaps this could be something to start with...
