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This answer claims the matrix $A = \begin{bmatrix}8 & 2 \\ 0 & 1\end{bmatrix}$ is not similar over $\mathbb Z$ to its transpose, without telling why. By taking a generic $S = \begin{bmatrix}x & y \\ z & w\end{bmatrix}$ and expanding $SA = A^t S$, i obtained a system of equations: $\begin{cases} 8y = 2x+y \\ 8z = 2x+z \\ 2z+w = 2y+w.\end{cases}$

Solving it, i got that $S = \begin{bmatrix} 7k & 2k \\ 2k & w\end{bmatrix}$ for some $k,w \in \mathbb Z$. Since $S$ should be invertible, we need its determinant to be $\pm 1$. The determinant is $7kw-4k^2 = k(7w-4k)$, which clearly cannot be $\pm 1$ so $A$ and $A^t$ are not similar.

Did i do this right? And, more importantly, is there an easier and/or more enlightning way to show they are not similar? Them being non-similar means their behaviour must be different in some way, but a straightforward calculation like this doesn't tell how their behaviours differ, only that they do.

user26857
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Carla_
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    Yes, this is a correct (and fairly simple way) to do it. One comment: The determinant is $7 k w-4 k^2 = k (7w-4k)$. It is easy to check that this cant be $\pm 1$ as the $4$ possibilities $k=\pm 1$, $7w-4 k=\pm 1$ all lead to $z$ being non-integral. – Kasper Andersen Jun 25 '24 at 15:55

1 Answers1

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I poked around and found this answer from KCd which links to his very nice note Ideal classes and matrix conjugation over $\mathbb{Z}$ which clarifies a great deal and discusses the case of transposes specifically (starting at Example 3.7).

KCd works exclusively with the case that the characteristic polynomial is irreducible. In the case at hand the characteristic polynomial actually has integer roots which simplifies the discussion enormously as follows. $A$ and $A^T$ have integer eigenvalues $1, 8$, hence integer eigenvectors, which moreover are unique up to sign if we ask them to be in lowest terms (have coordinates with no common factors). For $A$ they are

$$x_1 = \left[ \begin{array}{c} -2 \\ 7 \end{array} \right], x_8 = \left[ \begin{array}{c} 1 \\ 0 \end{array} \right]$$

while for $A^T$ they are

$$y_1 = \left[ \begin{array}{c} 0 \\ 1 \end{array} \right], y_8 = \left[ \begin{array}{c} 7 \\ 2 \end{array} \right].$$

Interestingly, note that $\{ x_1, x_2 \}$ and $\{ y_1, y_2 \}$ both span proper sublattices of $\mathbb{Z}^2$ of index $7$. This is related to the appearance of $7$ in the calculation. Now, since the integer eigenvectors in lowest terms are unique up to sign, a matrix $S$ conjugating $A$ to $A^T$ must do so by sending $x_1$ to $\pm y_1$ and $x_8$ to $\pm y_8$, and this condition determines exactly $4$ possible values of $S$, which are

$$S_{++} = \left[ \begin{array}{cc} 0 & 7 \\ 1 & 2 \end{array} \right]\left[ \begin{array}{cc} -2 & 1 \\ 7 & 0 \end{array} \right]^{-1} = \left[ \begin{array}{cc} 7 & 2 \\ 2 & \frac{5}{7} \end{array} \right]$$ $$S_{+-} = \left[ \begin{array}{cc} 0 & -7 \\ 1 & -2 \end{array} \right]\left[ \begin{array}{cc} -2 & 1 \\ 7 & 0 \end{array} \right]^{-1} = \left[ \begin{array}{cc} -7 & -2 \\ -2 & - \frac{3}{7} \end{array} \right]$$

together with $S_{--} = -S_{++}$ and $S_{-+} = - S_{+-}$. These matrices conjugate $A$ to $A^T$ over $\mathbb{Z} \left[ \frac{1}{7} \right]$, and hence over $\mathbb{Q}$ and over $\mathbb{F}_p$ for all $p \neq 7$, but they do not have integer coefficients, and since these are the only possibilities, it follows that $A$ and $A^T$ are not conjugate over $\mathbb{Z}$.

This calculation also explains why your calculation gives a $2$-parameter family of choices for $S$ over a field; over a field $S$ can scale each eigenvector arbitrarily.

Edit: Okay, here's something that can be said that's actually phrased in terms of properties one action has that the other doesn't. For $A$, the two eigenvectors in lowest terms are related by the fact that $x_1 + 2x_8$ is divisible by $7$ (or rather that there is some choice of signs that makes this true), whereas the same is not true for $A^T$ (or rather, there is no choice of signs that works). For example $y_1 + 2y_8 = \left[ \begin{array}{c} 14 \\ 5 \end{array} \right]$ so trying to divide it by $7$ produces the $\frac{5}{7}$ we saw in $S_{++}$ above, and we can check that $\pm y_1 \pm 2y_8$ don't work either. Instead for $A^T$ the eigenvectors satisfy the "dual" condition that $y_8 - 2y_1$ is divisible by $7$ and this is, symmetrically, not true for $A$. This divisibility by $7$ is significant because it witnesses the fact that $\mathbb{Z}^2$ mod the lattice spanned by eigenvectors is $\mathbb{Z}/7$.

Qiaochu Yuan
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