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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.

Rodrigo de Azevedo
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Dots_and_Arrows
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2 Answers2

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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.

J. W. Tanner
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Romulus Augustulus
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  • Why do we need to divide by 48 ? I really can't see. Can anyone please elaborate. – Another_Ramanujan_Fan Sep 16 '21 at 04:18
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    $26\cdot 24=624$ is the number of couples of linearly independent vectors in $(\mathbf{Z}/(3))^3$. Every such couple spans one two-dimensional subspace, but different couples can span the same space. The number of couples of linearly independent vectors that span the same 2-dim subspace is the number of its bases, that is $48$. So you must divide $624$ by $48$ – Romulus Augustulus Sep 30 '21 at 11:53
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I know this is an old post, but here is another, more combinatorial and group-theoretic, way to answer such questions.

The group $\mathrm{GL}_3(\Bbb{Z}_3)$ acts on itself by conjugation, and an orbit of this action is the same thing as a conjugacy class. The Orbit-Stabilizer Theorem from group theory says that the size of an orbit is equal to the index of the stabilizer subgroup. That is, for an element $x$ in a set $X$ on which a group $G$ acts, if $\mathcal{O}(x)$ is its orbit and $G_x$ is the stabilizer of $x$, then $\left| \mathcal{O}(x) \right| = \frac{|G|}{|G_x|}$.

In this instance, the stabilizer is the same thing as the centralizer; i.e. the set of matrices which commute with the given one. So the answer to your question is equal to the size of $\mathrm{GL}_3(\Bbb{Z}_3)$ (which you have pointed out is 11,232) divided by the number of matrices which commute with the given diagonal matrix.

Let a given $3 \times 3$ matrix be denoted by $$ \begin{pmatrix} a & b & c \\ x & y & z \\ p & q & r \end{pmatrix} $$ Writing down the conditions for this matrix to commute with the diagonal matrix with entries $2,2,1$ is equivalent to saying that $c=z=p=q=0$. In other words, the matrices which commute with the diagonal one are the invertible matrices of the form $$ \begin{pmatrix} a&b&0 \\ x&y&0 \\ 0&0&r \end{pmatrix} $$ For this to be invertible, we must have $r \neq 0$ (so $r=1$ or $r=2$), and the $2 \times 2$ block $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ must be invertible. One quickly sees that there are 96 such matrices.

So the final answer is $\frac{11232}{96} = 117$.

Nick
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