We know that a square matrix (for example in $C$), can have eigenvalues equal to 0 or 1 simultaneously. For example: $$ A= \begin{pmatrix} a & b\\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} $$
has a characteristic equation $p_A(\lambda)=det(\lambda I - A)= (\lambda-1)\lambda=0$ with roots $\lambda_{1}=0$ and $\lambda_2=1$.
However, I have stumbled upon the following "proof" of the opposite (that a matrix in C can't have a pair of eigenvalues where one of them equals $0$, and another equals $1$).
let $\lambda_j \in \lambda(F) \iff p_F(\lambda_j)=0$
$$ 1\in \lambda(F) \iff 0 \notin \lambda(F) $$ Proof of necessity:
$$ 1\in \lambda(F) \implies \exists_{v\neq0} v=Fv\implies rk(v)=rk(Fv) \implies rk(F) = full \implies |F|\neq0\implies 0 \notin \lambda(F) $$
Where $v$ is a vector in $C$. I suspect the mistake I made was in raking the rank of both sides for the equation $v=Fv$, but I can't wrap my mind around why this is illegal.
So: Can I take the rank of two matrices that are equal? And if so, why? And if I can, where did I go wrong?
Thanks
The very helpful comments under the question show that the error is in assuming that the existence of a non-zero vector $v$ for which $v=Fv$ implies that $F$ is full rank - this is not true, because $F$ is full rank if and only if every non-zero input vector is mapped to some non-zero output vector. Any non-zero rank-defficient matrix maps some non-zero inputs to some non-zero outputs, so going from the presence of a unit eigenvalue, we have $\exists_{v\neq0}v=Fv\neq0$, and by taking the rank, we have $\exists_{v\neq0}rk(v)=rk(Fv)=1$, which doesn't deny the possibility of there being a zero eigenvalue $\exists_{w\neq0}Fw=0$.
Furthermore, by going from $\exists_{v\neq0}v=Fv\neq0$ to $\exists_{v\neq0}rk(v)=rk(Fv)=1$, we have actually lost information, because the first condition holds only for the unit eigenvectors of F, while the second holds for every non-zero vector in the span of F (the rank of a non-zero vector is one, and as long as $F$ doesn't take it to zero, so is the rank of $Fv$).