How many elements of $M$ are similar to the following matrix?

Question

Let k be the field with exactly $7$ elements. Let $M$ be the set of all $2\times 2$ matrices with entries in k. How many elements of $M$ are similar to the following matrix?

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

I know that two matrices in $M$ are similar if and only if their characteristic polynomial is equal, i.e., $A$ is similar to the given matrix if and only if $\operatorname{tr}(A)=1$ and $\det(A)=0$. We can take all matrices of the form: $$ \begin{pmatrix} a & b \cr c & 1-a\end{pmatrix} $$ with $a(1-a)=bc$ in k. But this method takes long time for calculation but is full-proof.

One of the comments here says: Do you know about eigenvalues and eigenvectors? If you do you can quite quickly count that there are $\left(\frac{p^2-1}{p-1}\right)\left(\frac{p^2-p}{p-1}\right)$ of them, where $p=7$. Can someone please explain the proof of how the above result comes? Thanks for the help.

Answer 1

Let $B$ denote the matrix $$ B = \pmatrix{0&0\\0&1}. $$ Regarding the last paragraph: we note that $A$ is similar to $B$ if and only if it is diagonalizable with eigenvalues $0,1$. With that established, we can completely specify a matrix $B$ by selecting a one-dimensional eigenspace associated with $0$ and a one-dimensional subspace associated with $1$. Because the total number of one-dimensional subspaces of $K^2$ is $n = \frac{p^2 - 1}{p - 1} = 8$, the total number of ways that a matrix $B$ can be constructed is $n(n-1) = 56$, which is the desired answer.

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Regarding your method: it is indeed possible to count the matrices of the form $$ A = \pmatrix{a & b\\c & 1-a}. $$ We divide these matrices into two cases.

Case 1: $a = 1$ or $a = 0$. We must have $bc = 0$, so $b = 0$ or $c = 0$. Each matrix that satisfies this description can be built by

1. Selecting $a = 0$ or $a = 1$ ($2$ possibilities),
2. Selecting $b,c$ with one equal to $0$ ($7^2 - 6^2 = 13$ possibilities).

Case 2: We can build the remaining matrices as follows:

1. Select $a$ ($5$ possibilities)
2. Select a non-zero $b$ ($6$ possibilities)

From there, we are forced to select $c = \frac{a(1-a)}{b}$.

In total, we have $$ 2 \times 13 + 5 \times 6 = 56 $$ matrices.