Diagonalizable binary matrices over $\mathbb{Z}_4$ I'm trying to figure out the following question:
Are symmetric, binary $n\times n$ matrices with zeros on the diagonal, are diagonalizable over $\mathbb{Z}_4$?
I know that it isn't true that $\textit{any}$ binary matrix is invertible over $\mathbb{Z}_2$, but I'm not sure about what happens in $\mathbb{Z}_4$, in particular in the case where the diagonal is all-zero.
Thanks!
 A: First, notice that if $R$ is an invertible matrix with coefficients in $\mathbb{Z}_4$ then $\det(R)$ is an invertible element of $\mathbb{Z}_4$. So $\det(R)=1$  or $\det(R)=3$.
If $\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix} a & b\\ c & d\end{pmatrix}\begin{pmatrix} \lambda  & 0\\ 0 & \mu \end{pmatrix}$ then  $\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix} a & b\\ \lambda a & \mu b\end{pmatrix}$
Hence,  $\det\begin{pmatrix} a & b\\ c & d\end{pmatrix}=ad-bc=(\mu -\lambda)ab$.
Next, notice that
$$-1.(ad-bc)=\det\left(\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}\begin{pmatrix} a & b\\ c & d\end{pmatrix}\right)=\det\left(\begin{pmatrix} \lambda a & \mu b\\ \lambda c & \mu d\end{pmatrix}\right)=\lambda\mu (ad-bc).$$
If $ad-bc=1$ or $3$ the $\lambda\mu=-1$. Thus, $\mu-\lambda= 2$.
Therefore, $\det\begin{pmatrix} a & b\\ c & d\end{pmatrix}=2ab$.
This contradicts the assumption that $ad-bc=1$ or $3$.
So the matrix $\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}$ can not be written as  $RDR^{-1}$, where $R$ is an invertible matrix with coefficients in $\mathbb{Z}_4$ and $D$ is a diagonal matrix.
A: This paper sounds relevant, but I don't know anything about matrices over commutative rings.  I don't understand the second sentence of the abstract!
I tried to post this has a comment, but the link to the paper wouldn't fit.
