Among $3 \times 3$ invertible matrices with entries from the field $ \mathbb{Z/3Z}$, how many matrices are similar to the following matrix?
\begin{pmatrix}
2 & 0 &0 \\
0&2 &0 \\
0&0 &1
\end{pmatrix}
Things I'm familiar with: I know similar matrices have the same determinant and trace. The number of invertible matrices over $ \mathbb{Z/3Z}$ is $(3^3-1)(3^3-3)(3^3-3^2)$
Please give me a hint to proceed from here.
Best Answer
A matrix similar to that one is determined by its eigenspaces.
The eigenspace of $2$ has dimension two. To count the two-dimensional subspaces of $(\mathbf{Z}/3\mathbf{Z})^3$ you can choose a non zero vector in $26$ ways, and a linearly independent vector in $24$ ways. Now you have to divide by the number of bases of a two dimensional vector space on $\mathbf{Z}/3\mathbf{Z}$, that is $(3^2-1)(3^2-3)=48$. So $(\mathbf{Z}/3\mathbf{Z})^3$ has $(26\cdot 24)/48 = 13$ two-dimensional subspaces.
Now you have to multiply for the number of one-dimensional subspaces that are the possible eigenspaces of $1$. You can choose a vector in $3^3-3^2=18$ ways, but you must divide by the number of non zero elements of a one-dimensional vector space over $(\mathbf{Z}/3\mathbf{Z})^3$, that is two.
So you get $13\cdot 9=117$ matrices.