Determine the number of distinct characteristic polynomial in $M_{2 \times 2} ( Z_{2})$ $?$
The question has already been asked here before. But that didn't clear my doubt.
The answer is $4$.
$t^{2}, t(t+1), t^{2} +1, t^{2} + t +1$
Can't we find more polynomials.
For example the matrix with all four entries $1$ is in $M_{2 \times 2} ( Z_{2})$ $?$ and its characteristic polynomial is $t^{2} -2t$.
Best Answer
Here I answer the more general question of the number of characteristic polynomials of matrices in $M_{n \times n}(\Bbb Z_2)$, which was originally posed by our OP Mathsaddict before editing. Obviously, $n = 2$ is a special case of this general result.
First, an example: let
$A \in M_{3 \times 3}(\Bbb Z_2), \; A = I; \tag 1$
that is,
$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}; \tag 2$
then
$A - tI = \begin{bmatrix} 1 - t & 0 & 0 \\ 0 & 1- t & 0 \\ 0 & 0 & 1 - t \end{bmatrix}, \tag 3$
whence
$\det (A - tI) = (1 - t)^3 = 1 - 3t + 3t^2 - t^3 \equiv t^3 + t^2 + t + 1 \mod 2. \tag 4$
The above example was constructed in light of the original (unedited) version of the question in which the class of matrices under consideration was stated to be $M_{n \times n}(\Bbb Z_2)$; it is meant to show the existence of characteristic polynomials of degree other than $2$ for matrices in this class.
These things being said, we proceed with the general case of $M_{n \times n}(\Bbb Z_2)$:
Now let
$p(t) \in \Bbb Z_2[t], \; \deg p(t) = n, \tag 5$
viz.
$p(t) = \displaystyle \sum_0^n p_i t^i, \; p_i \in \Bbb Z_2, \; 0 \le i \le n; \tag 6$
we observe that $p(t)$ is monic, that is,
$p_n = 1 \in \Bbb Z_2; \tag 7$
consider the matrix
$C \in M_{n \times n}(\Bbb Z_2), \tag 8$
$C = \begin{bmatrix} 0 & 0 & \ldots & 0 & p_0 \\ 1 & 0 & \ldots & 0 & p_1 \\ 0 & 1 & \ldots & 0 & p_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \ldots & 1 & p_{n - 1} \end{bmatrix}; \tag 9$
we have
$\det(C - tI) = \det \left ( \begin{bmatrix} -t & 0 & \ldots & 0 & p_0 \\ 1 & -t & \ldots & 0 & p_1 \\ 0 & 1 & \ldots & 0 & p_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \ldots & 1 & p_{n - 1} - t \end{bmatrix} \right ); \tag{10}$
when this determinant is expanded in minors along the last ($n$-th) column we obtain
$\det(C - tI) = \displaystyle \sum_0^n p_i t^i = p(t) \in \Bbb Z_2[t], \tag{11}$
where we have used the fact that
$-p_i = p_i, \; p_i \in \Bbb Z_2. \tag{12}$
The preceding discussion reveals that every polynomial as in (5)-(7) is in fact the characteristic polynomial of some $n \times n$ matrix over $\Bbb Z_2$. How many such polynomials are there? Since each such polynomial is monic, there are $n$ free coefficients $p_0$, $p_1$, . . ., $p_{n - 1}$, each of which may arbitrarily take on the values $0, 1 \in \Bbb Z_2$; thus their number in toto is $2^n$; this is the number of characteristic polynomials of matrices $C \in M_{n \times n}(\Bbb Z_2)$; the case $n = 2$ thus yields $2^2 = 4$ possible polynomials over $\Bbb Z_2$, corroborating the work of our OP.