I found somewhere the following
If $\lambda_1,\dots,\lambda_n$ are the eigenvalues of the symmetric matrix $A$, show that $\sum_i \lambda_i v_i(1)^2=A_{11},$
where $v_i(1)$ is the first coordinate of the $i$-th normalized eigenvector.
I tested it with $2\times 2$ matrices and the identity is indeed true, however I am not sure about a formal proof. Using the eigendecomposition was not useful so I guess there should be another trick involved. Any hints are appreciated.
I would use the eigenvalue decomposition $A = V \Lambda V^T$.
This format is basically a shorthand for $A = \sum \limits_{i=1}^n \lambda_i v_i v_i^T$, where $v_i^T$ denotes the transpose of $v_i$, a row vector. This is a summation of dyadic outer products.
Evaluating $A_{i,j}$ comes down to choosing an entry of the dyadic product in every summation by choosing the right indices in $v_i$ and $v_i^T$.
From there, it is easy to get your desired result.