Proving that the matrix representing the 5 point method for pdes is symmetric

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Consider this

For any ordering of the grid points, the matrix A of the system \begin{align*} u_{i-1, j}+u_{i+1, j}+u_{i, j-1}+u_{i, j+1}-4 u_{i, > j}=h^{2} f_{i, j}. \end{align*} is symmetric and negative definite.

The given proof is:

The above equation implies that if $a_{i j} \neq 0$ for $i \neq j$, then the $i$-th and $j$-th points of the grid $(p h, q h)$, are nearest neighbours. Hence $a_{i j} \neq 0$ implies $a_{i j}=a_{j > i}=1$, which proves the symmetry of $A$. Therefore $A$ has real eigenvalues and eigenvectors.

It remains to prove that all the eigenvalues are negative ...

I really do not understand this. I can picture the grid. However, surely I can label the nondiagonal entries with $-4$ and get some sort of contradiction? I think there is some structure to $A$ or something that I am missing. In regards to the argument, I have $0$ understanding what the author.

Could someone clarify this for me?