Is a symmetric matrix characterized by the diagonal of its resolvent? The resolvent of a square matrix $A$ is defined by $R(s) = (A-sI)^{-1}$ for $s \notin \operatorname{spect}(A)$. 
Is knowing the diagonal of $R(s)$ for all $s$ sufficient to recover $A$ when $A$ is symmetric?
edit: a counter-example of two matrices $A,B$ whose resolvent have the same diagonal has been found by Robert Israel. In the counter example, $A = P B P^T$ for some permutation matrix $P$. Now the question is, it is possible to recover $A$ up to permutations of rows and columns?
 A: No.  For example, consider $$\left[ \begin {array}{cccc} 1&0&1&1\\ 0&0&1&1
\\ 1&1&0&0\\ 1&1&0&1\end {array}
 \right] \ \text{and}\
 \left[ \begin {array}{cccc} 1&1&0&1\\ 1&0&1&0
\\ 0&1&0&1\\ 1&0&1&1\end {array}
 \right] $$
which both have diagonal of $R(s)$
$$ (s^4 - 2 s^3 - 3 s^2 + 4 s - 1)  \left[ \begin {array}{c} -{s}^{3}+{s}^{2}+2\,s-1\\ 
-{s}^{3}+2\,{s}^{2}+s-1\\ -{s}^{3}+2\,{s}^{2}+s-1
\\ -{s}^{3}+{s}^{2}+2\,s-1\end {array} \right] 
 $$
EDIT: For the second question, try 
$$ \left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&1&0
\\ 0&1&1&0\\ 1&0&0&1\end {array}
 \right] 
\ \text{and}\ \left[ \begin {array}{cccc} 1&0&-\sin \left( t \right) &\cos \left( t
 \right) \\ 0&1&\cos \left( t \right) &\sin \left( t
 \right) \\ -\sin \left( t \right) &\cos \left( t
 \right) &1&0\\ \cos \left( t \right) &\sin \left( t
 \right) &0&1\end {array} \right] 
$$
A: If $A$ and $B$ are the adjacency matrices of non-isomorphic strongly regular graphs on $n$ vertices with the same parameters, then the diagonal entries of the resolvent are all equal to $1/n$ times the common characteristic polynomial. Such graphs on 16 vertices can be constructed from the two $4\times4$ Latin squares. Since the graphs are not isomorphic, the two matrices are not permutation equivalent.
