The Eigenvalues of a Matrix over $\mathbb{F}_2$ Consider $\bf A$ is a non-singular binary matrix over the field $\mathbb{F}_2$. Assume that the characteristic polynomial of $\bf A$ over $\mathbb{F}_2$ is ${\bf f}(x)^k$ for some $k$ where ${\bf f}(x)$ is an irreducible polynomial over $\mathbb{F}_2$.
Suppose that ${\bf g}(x)\neq {\bf f}(x)$ is an irreducible polynomial over $\mathbb{F}_2$. Consider the matrix ${\bf B}={\bf g}({\bf A})$ over $\mathbb{F}_2$. From linear algebra we know that ${\bf B}$ is not a zero matrix over $\mathbb{F}_2$. But what can we say about singularity of the matrix ${\bf B}$ over $\mathbb{F}_2$?
My question: Is it true this claim that the matrix ${\bf B}$ is a non-singular matrix over $\mathbb{F}_2$? 
My try: Consider $\lambda_i$ for $1\leq i \leq n$, are eigenvalues of the matrix $\bf A$. Becuse of the matrix $\bf A$ is a non-singular matrix over $\mathbb{F}_2$ we conclude that $\lambda_i$'s are non-zero. Moreover, 
from linear algebra we know that $\det({\bf B})=\prod_{i=1}^n{\bf g}(\lambda_i)$.
Therefore, if the matrix ${\bf B}$ be a singular matrix over $\mathbb{F}_2$ then there is at least an $\lambda_i$ for some $1\leq j \leq n$ such that ${\bf g}(\lambda_j)=0$ which means the matrix ${\bf B}$ has an eigenvalue equal to zero. I don't know how this issue helps to our discussion.
In fact, my problems is that how to obtain the eigenvalues of the  matrix $\bf A$ over $\mathbb{F}_2$?
I would very much appreciate any assistance you can offer me in this question.  
 A: EDIT. It's true because, in $F_2$ , $1$ is the only invertible.
The $(\lambda_i)$ are in an algebraic extension of $F_2$ and $f$ is their common unitary minimal polynomial. If $g(\lambda_j)=0$, then $g$ is a multiple of the minimal polynomial of $\lambda_j$, that is $g=f$. 
Answer 1 to @user0410. The minimal polynomial of $\lambda\in K$ over $F_2$, ($K$ an algebraic extension of $F_2$) is $h$, the unitary polynomial of minimal degree ($>0$) among the polynomials that admit $\lambda$ as a root. One can prove that $h$ is irreducible and, conversely, if a unitary irreducible polynomial admits $\lambda$ as a root, then it 's $h$. Here $f$ is the minimal polynomial of all its roots (the $(\lambda_i)$'s).
Answer 2 to @user0410. Your polynomial $f=x^4+x+1$ decomposes in $K$, an algebraic extension of degree $4$; $g=x^6+x+1$ factorizes in $K$ in the form $g=p(x)q(x)$ where $degree(p)=degree(q)=3$. Yet $g$ decomposes in $L\supset K$. Of course, $g(A)$ is invertible and I don't see any contradiction.
A: The eigenvalues $\lambda_i$ live in an extension of $\Bbb F_2$, namely in $\Bbb F_2[x]/(f)$ which is the finite field of size $2^{\deg f}$.
Since $f$ and $g$ are distinct irreducible polynomials, $\gcd(f, g)=1$, so they don't share a root even in a field extension.
It proves $g(\lambda_i)\ne0$, hence $\det B\ne 0$.
