Existance of an invertible matrix.

Question

Show that there exists a $3\times 3$ invertible matrix $M\neq I_{3}$ with entries in the field $F_{2}$ such that $M^{7} = I$.

Answer 1

If $M^7=1$, then the minimal polynomial of $M$ divides $x^7-1$. Note that $$ x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1).$$ Over $\mathbb{F}_2$ we have $x-1=x+1$ and $(x^6+x^5+x^4+x^3+x^2+x+1)=(x^3+x^2+1)(x^3+x+1),$ thus $$ x^7-1=(x+1)(x^3+x^2+1)(x^3+x+1) \hspace{0.4cm} \text{over} \hspace{0.4cm} \mathbb{F}_2.$$ So, the minimal polynomial of $M$ must be $x^3+x^2+1$ or $x^3+x+1$.

For example, the companion matrix of $x^3+x^2+1$ is $$ \begin{bmatrix} 0 & 0 & -1 \\ 1 & 0 & 0 \\ 0 & 1 & -1 \\ \end{bmatrix},$$ and the companion matrix of $x^3+x+1$ is $$ \begin{bmatrix} 0 & 0 & -1 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \\ \end{bmatrix},$$ so this both matrices satisfy the requirements.