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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?

roger
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2 Answers2

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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 $$ \left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&1&0 \\ 0&1&1&0\\ 1&0&0&1\end {array} \right] \ \text{and}\ \left[ \begin {array}{cccc} 1&0&-\sin \left( t \right) &\cos \left( t \right) \\ 0&1&\cos \left( t \right) &\sin \left( t \right) \\ -\sin \left( t \right) &\cos \left( t \right) &1&0\\ \cos \left( t \right) &\sin \left( t \right) &0&1\end {array} \right] $$

Robert Israel
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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.

Chris Godsil
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    How do you show that a diagonal entry of the resolvent is the characteristic polynomial divided by n? – roger Dec 22 '15 at 01:19
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    Don't you need more than just strongly regular for the characteristic polynomials to be the same? – Robert Israel Dec 22 '15 at 05:59
  • @roger: morally, the $i$-th diagonal entry of the resolvent is the generating function for the closed walks in the graph that start at $i$. But the diagonal entries of $A^m$, for a strongly regular graph, are all equal. Equivalently srgs are "walk regular". To prove this, verify that the three idempotents in the spectral decomposition of $A$ have constant diagonal. – Chris Godsil Dec 22 '15 at 12:20
  • ok I see, So there exist non-isomorphic strongly regular graphs with the same number of closed walks of any given length starting from any vertex. Thanks, that helped! – roger Dec 22 '15 at 13:00
  • @roger: look up Latin square graphs, e.g., in http://www.cs.yale.edu/homes/spielman/561/2009/lect23-09.pdf – Chris Godsil Dec 22 '15 at 13:12
  • @Chris Aren't all latin square graphs of size $n^2$ isomorphic? – roger Dec 22 '15 at 13:53
  • @roger: not at all, in particular the graphs corresponding to the multiplication tables of the two groups of order four are not isomorphic. – Chris Godsil Dec 22 '15 at 14:00
  • perfect! thanks – roger Dec 22 '15 at 14:05