Motivated by the apparent lack of possible classification of integer matrices up to conjugation ([see here](https://mathoverflow.net/questions/51942/analogue-of-smith-normal-form-for-matrices-over-mathbb-zt)) and by a question about possible complete graph invariants ([see here](https://mathoverflow.net/questions/52141/invariants-that-might-determine-graph-up-to-isomorphism)), let me ask the following: 



> 
> **Question:** Is there an example of a pair of non-isomorphic simple finite graphs which have conjugate (over $\mathbb Z$) adjacency matrices?
> 
> 
> 


It is well-known that there are many graphs which have the same spectrum. This implies that their adjacency matrices are conjugate over $\mathbb C$.


In *Allen Schwenk, Almost all trees are cospectral. New directions in the theory of graphs (Proc. Third Ann Arbor Conf., Univ. Michigan, Ann Arbor, Mich., 1971), pp. 275–307. Academic Press, New York, 1973* it was shown that almost all trees have cospectral partners. Maybe $\mathbb Z$-conjugate graphs can be found among trees? 



Adjacency matrices of graphs
Yes.


Consider the adjacency matrices
$$ A = \left[\begin{array}{rrrrrrrrrrr}
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
\end{array}\right] $$
and
$$ B = \left[ \begin{array}{rrrrrrrrrrr}
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \\\\
0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
\end{array}\right]. $$
These are both the adjacency matrices of trees, and both have characteristic polynomial
$$\lambda^{11}-10\lambda^9+34\lambda^7-47\lambda^5+25\lambda^3-4\lambda.$$
Each tree
has exactly two vertices of degree 3, separated by a path of length 1 in the case of $A$ but length 2 in the case of $B$. In particular, the trees are not isomorphic.


Now consider the [EDIT: improved, much nicer] matrix
$$ C = \left[\begin{array}{rrrrrrrrrrr}
1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 \\\\
0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\\\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\\\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\\\
\end{array}\right] $$
with determinant $-1$.


Since $C^{-1}AC = B$, the two trees (on 11 vertices) are non-isomorphic but
have adjacency matrices that are conjugate over $\mathbb Z$.


Now to explain the where the example comes from. The pair of graphs
was constructed by a method, attributed to Schwenk, that I found in Doob's
chapter of *Topics in algebraic graph theory* (edited by Beineke and Wilson).
The first 9 rows and columns of $A$, in common with $B$, come from a particular tree on 9 vertices that has a pair of attachment points such that extending the tree in the same way from either point gives isomorphic spectra.
Adding a single pendant vertex cannot work for this problem, as I found using Brouwer and van Eijl's trick, mentioned
by Chris Godsil, of comparing the Smith normal forms of (very) small polynomials in $A$ and $B$, in this case $A+2I$ and $B+2I$. When a path of length two is added at either of the two special vertices, however, there doesn't seem to be any obstruction of this type.


I then set about trying to conjugate both $A$ and $B$, separately, to the companion matrix of their mutual characteristic polynomial, by looking for a random small integer vector $x$ for which the matrix $X\_A = [ x\ Ax\ A^2x\ \ldots\ A^{10}x]$ has determinant $\pm 1$, and similarly $y$ giving $Y\_B$. (The fact that I succeeded fairly easily may have something to do with the fact that $A+I$ is invertible over $\mathbb Z$.) The matrix
$X\_AY\_B^{-1}$ then acts like the $C$ above.


[EDIT: The actual matrix $C$ I found at random and first posted was not nearly so pretty, with a Frobenius norm nearly ten times the current example. But taking powers 0 to 10 of $A$ times $C$ gave a $\mathbb Q$-basis for the full space of conjugators, whose Smith normal form (as 11 vectors in $\mathbb R^{121}$) was all 1's down the diagonal, so in fact it was a $\mathbb Z$-basis. Performing an LLL reduction on this lattice basis then gave a list of smaller-norm matrices, the third of which is the more illuminating $C$ given above, of determinant $-1$. The other determinants from the reduced basis were all $0$ and $\pm 8$.]


Taking rational $x$ and not restricting the determinant of $X\_A$ gives a space of possible rational matrices $C$ of dimension 11, which are generically invertible; varying $y$ gives the same space [EDIT: as does multiplying on the left by powers (or in the more general case commutants) of $A$]. Since the spectrum of $A$ has no repeated roots, this is also the dimension of the commutant of $A$, and every matrix conjugating $A$ to $B$ lies in this space. Starting with a rational basis, it is not hard to find an exact basis for the integer sublattice, and taking the determinant of a general point in the integer lattice gives an integer polynomial in 11 variables which takes the value $1$ or $-1$ if and only if the matrices $A$ and $B$ are conjugate over $Z$. If there are repeated roots, you have to work a little harder; in general the full space has dimension the sum of the squares of the multiplicities, and is generated by multiplying on the left by a basis for the commutator space of $A$. A basis for the commutant can be produced (for a diagonalizable matrix) by first conjugating $A$ to a direct sum of companion matrices for the irreducible factors of the characteristic polynomial, and then one at a time, for each $k$-by-$k$ block corresponding to a $k$-times repeated factor of degree $m$, replacing each of the $k^2$ blocks with powers $0$ to $m-1$ of the companion matrix for that factor, with $0$ everywhere elsewhere.


