Through some facts, when $A$ is invertible, I found out that the eigenvalue can't be $0$, since if the eigenvalue is $0$, then $\det(A)=0$, which means that is is not invertible. Since it is over $\mathbb{Z}_2$, then eigenvalue is $1$.
Since the eigenvalue $(\lambda)$ is $1$, and $Av = \lambda v$, then $Av = v$ when $A=I$.
However, I didn't quite get how it can be diagonalizable if and only if $A=I$, or did I get it wrong somewhere? please help :)
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Since $0$ and $1$ are the only scalars available, and a diagonal entry$~0$ in a diagonal matrix makes it singular (for instance because there is then a zero column), the only invertible diagonal matrix is the identity matrices$~I_n$. A diagonalisable matrix is by definition similar to some diagonal matrix, and invertibility is unchanged under similarity. Hence an invertible matrix over $\Bbb Z/2\Bbb Z$ that is diagonalisable (over that field) is similar to$~I_n$, but only the identity matrix itself is similar to$~I_n$.
If $A$ is diagonalizable with eigenvalues $\lambda_1, \dots, \lambda_n$ (counted with multiplicities) then $\det(A) = \prod_{i=1}^n \lambda_i$. If in addition $A$ is invertible then all the eigenvalues (which are in $\mathbb{Z}_2$ must be equal to one which implies that $A = I$ (since if $v_1, \dots, v_n$ are the eigenvectors and $v = \sum_{i=1}^n c_i v_i$ is a general vector, then $Av = \sum_{i=1}^n c_i Av_i = \sum_{i=1}^n c_i v_i = v$).