I was working on the following proof and have completed it but I am trying to understand why
char$(K)\neq2$ is demanded.
Let $V$ be a finitely dimensional vector space over $K$ with char$(K)\neq2$ and let $f \in End_K(V)$. Show that $$f^2=Id_V$$ $$\iff$$
There exists an ordered basis $B$ of $V$ and $i, k \in \mathbb{N_0}$ such that
$$c_{B,B}(f)=\begin{bmatrix} E_i & 0 \\ 0 & -E_k \end{bmatrix}$$
Is there an example of an $f \in End_K(V)$ with char$(K)=2$ such that the equivalency does not hold? I have used char$(K)\neq2$ to show that $V=ker(f-Id_V) \oplus ker(f+Id_V)$, so with that there exist bases of those subspaces that form an ordered base of $V$ with $f=Id_V$ for the first and $f=-Id_V$ for the second kernel which proves "$\implies$".
Are there any counterexamples when char$(K)=2$?
Note that in characteristic $2$, the equation $f^2-Id_V=0_V$ is equivalent to $(f-Id_V)^2=0_V$, which suggests that we should take a Jordan block of size $2$ with eigenvalue $1$:
$$\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}.$$