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What is the total number of matrices that are similar to \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} over $\mathbb{Z}_7$, that is, the finite field on $7$ elements $\{0,1,2,3,4,5,6\}$?

I know that total number of invertible matrices of order $2$ over $\mathbb{Z}_7$ is $2016$. From this how to proceed?

Rodrigo de Azevedo
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3 Answers3

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Here are the matrices satisfying the conditions in Dietrich Burde's answer.

The count is $(11\times 4) + (5\times 2) + (2\times 1) = 56$

Depending whether there can be $2$ or $1$ or $0$ exchange along the two diagonals (i.e. when numbers are not equal).


  • Matrices having $0,1$ in main diagonal

$\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\quad\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\quad$

$\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}\quad\begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix}\quad\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\quad\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}\quad$

$\begin{pmatrix} 0 & 2 \\ 0 & 1 \end{pmatrix}\quad\begin{pmatrix} 0 & 0 \\ 2 & 1 \end{pmatrix}\quad\begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix}\quad\begin{pmatrix} 1 & 0 \\ 2 & 0 \end{pmatrix}\quad$

$\begin{pmatrix} 0 & 3 \\ 0 & 1 \end{pmatrix}\quad\begin{pmatrix} 0 & 0 \\ 3 & 1 \end{pmatrix}\quad\begin{pmatrix} 1 & 3 \\ 0 & 0 \end{pmatrix}\quad\begin{pmatrix} 1 & 0 \\ 3 & 0 \end{pmatrix}\quad$

$\begin{pmatrix} 0 & 4 \\ 0 & 1 \end{pmatrix}\quad\begin{pmatrix} 0 & 0 \\ 4 & 1 \end{pmatrix}\quad\begin{pmatrix} 1 & 4 \\ 0 & 0 \end{pmatrix}\quad\begin{pmatrix} 1 & 0 \\ 4 & 0 \end{pmatrix}\quad$

$\begin{pmatrix} 0 & 5 \\ 0 & 1 \end{pmatrix}\quad\begin{pmatrix} 0 & 0 \\ 5 & 1 \end{pmatrix}\quad\begin{pmatrix} 1 & 5 \\ 0 & 0 \end{pmatrix}\quad\begin{pmatrix} 1 & 0 \\ 5 & 0 \end{pmatrix}\quad$

$\begin{pmatrix} 0 & 6 \\ 0 & 1 \end{pmatrix}\quad\begin{pmatrix} 0 & 0 \\ 6 & 1 \end{pmatrix}\quad\begin{pmatrix} 1 & 6 \\ 0 & 0 \end{pmatrix}\quad\begin{pmatrix} 1 & 0 \\ 6 & 0 \end{pmatrix}\quad$


  • Matrices having $2,6$ in main diagonal

$\begin{pmatrix} 2 & 5 \\ 1 & 6 \end{pmatrix}\quad\begin{pmatrix} 2 & 1 \\ 5 & 6 \end{pmatrix}\quad\begin{pmatrix} 6 & 5 \\ 1 & 2 \end{pmatrix}\quad\begin{pmatrix} 6 & 1 \\ 5 & 2 \end{pmatrix}\quad$

$\begin{pmatrix} 2 & 6 \\ 2 & 6 \end{pmatrix}\quad\begin{pmatrix} 2 & 2 \\ 6 & 6 \end{pmatrix}\quad\begin{pmatrix} 6 & 6 \\ 2 & 2 \end{pmatrix}\quad\begin{pmatrix} 6 & 2 \\ 6 & 2 \end{pmatrix}\quad$

$\begin{pmatrix} 2 & 4 \\ 3 & 6 \end{pmatrix}\quad\begin{pmatrix} 2 & 3 \\ 4 & 6 \end{pmatrix}\quad\begin{pmatrix} 6 & 4 \\ 3 & 2 \end{pmatrix}\quad\begin{pmatrix} 6 & 3 \\ 4 & 2 \end{pmatrix}\quad$


  • Matrices having $3,5$ in main diagonal

$\begin{pmatrix} 3 & 1 \\ 1 & 5 \end{pmatrix}\quad\begin{pmatrix} 5 & 1 \\ 1 & 3 \end{pmatrix}\quad$

$\begin{pmatrix} 3 & 4 \\ 2 & 5 \end{pmatrix}\quad\begin{pmatrix} 3 & 2 \\ 4 & 5 \end{pmatrix}\quad\begin{pmatrix} 5 & 4 \\ 2 & 3 \end{pmatrix}\quad\begin{pmatrix} 5 & 2 \\ 4 & 3 \end{pmatrix}\quad$

$\begin{pmatrix} 3 & 5 \\ 3 & 5 \end{pmatrix}\quad\begin{pmatrix} 3 & 3 \\ 5 & 5 \end{pmatrix}\quad\begin{pmatrix} 5 & 5 \\ 3 & 3 \end{pmatrix}\quad\begin{pmatrix} 5 & 3 \\ 5 & 3 \end{pmatrix}\quad$

$\begin{pmatrix} 3 & 6 \\ 6 & 5 \end{pmatrix}\quad\begin{pmatrix} 5 & 6 \\ 6 & 3 \end{pmatrix}\quad$


* Matrices having $4,4$ in main diagonal

$\begin{pmatrix} 4 & 2 \\ 1 & 4 \end{pmatrix}\quad\begin{pmatrix} 4 & 1 \\ 2 & 4 \end{pmatrix}\quad$

$\begin{pmatrix} 4 & 3 \\ 3 & 4 \end{pmatrix}\quad$

$\begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix}\quad$

$\begin{pmatrix} 4 & 6 \\ 5 & 4 \end{pmatrix}\quad\begin{pmatrix} 4 & 5 \\ 6 & 4 \end{pmatrix}\quad$

zwim
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Two matrices in $M_2(\Bbb F_7)$ are similar if and only if they have equal minimal and equal characteristic polynomial, i.e., $A$ is similar to the given matrix if and only if $\operatorname{tr}(A)=1$ and $\det(A)=0$. Now you can count easily the total number of such $A$. Take all matrices $$ \begin{pmatrix} a & b \cr c & 1-a\end{pmatrix} $$ with $a(1-a)=bc$ in $\Bbb F_7$. One has to avoid double counts.

Dietrich Burde
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  • Is the answer $49$? –  Oct 19 '19 at 17:18
  • yes. I have count them. I think there will be $49$ such matrices –  Oct 20 '19 at 05:41
  • I have more than $49$. Call the left upper corner element $a$. For $a=0$ I have $13$ different matrices, for $a=1$ I have $13$ further matrices, for $a=2$ I have $6$ new ones etc. How many matrices do you count, say, for $a=0$? – Dietrich Burde Oct 21 '19 at 19:01
  • Sorry. The correct answer will be $56$. Isn't it? – vqw7Ad Oct 22 '19 at 10:35
  • @J.Doe Let JNS do this, right? – Dietrich Burde Oct 31 '19 at 13:04
  • Dietrich, you have managed to confuse at least one user (see the comment under this answer) by opening with too bold a claim in the first half of your first sentence. I know the imprecision does not matter in the context of this question, but you might consider going for a slightly more careful formulation. – Marc van Leeuwen Sep 07 '22 at 10:21
  • @MarcvanLeeuwen I am sorry. Can you help me to clarify? I thought that for matrices of size $n\le 3$, equal minimal and characteristic polynomial are enough to ensure similiarity. For $\Bbb C$ I found a confirmation here. Is this correct for finite fields? You have an answer here, but it doesn't mention the field. – Dietrich Burde Sep 07 '22 at 10:40
  • @DietrichBurde My apologies; I misread your answer before commenting to it, not noticing that it mentioned minimal polynomials at all. To my defence, the phase used to only contain one instance of "equal", making it somewhat unclear to be asking for $2$ equalities among $4$ polynomials (the two minis and the two chars); the phrase is perfectly clear now though. I confirm that these equalities ensure equivalence in dimensions up to$~3$, as the linked argument works over algebraically closed fields, and matrices are similar over a field if and only if they are so over a containing field. – Marc van Leeuwen Sep 07 '22 at 12:02
  • @MarcvanLeeuwen No need to apologize. It was really a typo and I have corrected it. Now it seems right. So thank you again. – Dietrich Burde Sep 07 '22 at 12:05
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Call your matrix $D$. Its two eigenvalues, $0$ and $1$, are distinct. Therefore a matrix $A$ is similar to $D$ if and only if it shares the same spectrum with $D$. In turn, it is similar to $D$ if and only if both its rank and trace are equal to $1$. This means $A=uv^T$ for some vectors $u$ and $v$ such that $v^Tu=1$. Normalise $u$ such that the first nonzero entry of $u$ is $1$. Then $A$ must take one of the following two forms, where $x,y,z$ are arbitrary scalars in $\mathbb F_7$: $$ \pmatrix{1\\ x}\pmatrix{1-xy&y}\text{ or } \pmatrix{0\\ 1}\pmatrix{z&1}. $$ These representations are unique and so it's a trivial matter to count them.

user1551
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