Suppose we have a matrix $A$ with entries in $\mathbb F_2$. Let $p_1(\lambda)$ be the characteristic polynomial of $A$ and $p_2(\lambda)$ be the characteristic polynomial of $A^2$, both of which are polynomials over $\mathbb Z_2[x]$.
For any polynomial $p(\lambda)$ in $\mathbb Z_2[x]$, it is true that $p(\lambda^2)=p(\lambda)^2$. Thus, we have the following relationship between $p_1$ and $p_2$:
$$p_2(\lambda)^2=p_2(\lambda^2)=\det(A^2+\lambda^2 I)=\det((A+\lambda I)^2)=\det(A+\lambda I)^2=p_1(\lambda)^2$$ so we have $p_2^2=p_1^2$ and $p_2=p_1$.
But it seems absurd that the characteristic polynomial of a matrix over $\mathbb F_2$ should always equal the characteristic polynomial of its square. Is this really true, or have I made a mistake?
Some consequences of this would be:
- the vectors fixed by left-multiplication by $A$ are exactly those fixed by left-multiplication by $A^2$
- the iterated map $\vec{v}\mapsto A\vec{v}$ cannot have any $2$-cycles, regardless of the matrix $A$
- the null spaces of $A$ and $A^2$ are equal
These seem impossible. What is amiss here?