Question: Assume $T:V\rightarrow V$ is a linear transformation on an $n$-dimensional vector space $V$ over a field $F$ whose characteristic is $2$, such that $T^2=I$. Put $W=\{v\in V | T(v)=v\}$. Prove that $\dim W\ge \frac{n}{2} $.
PS: My effort for this question has been written as a separate answer below.
On
My solution:
$W=ker (T-I)$. Moreover, according to the given information $T^2=I$, it's obtained that $(T-I)(T+I)=0$. Hence, $Im(T+I)\subseteq ker (T-I)$. On the other hand, since transformation $T$ is defined on a field whose characteristic is $2$, we have: $$-1=1\Rightarrow T-I=T+I.$$ Regarding the rank-nullity theorem:$$rank(T-I)+nullity(T-I)=n\Rightarrow rank(T+I)+nullity(T-I)=n\Rightarrow dim(W)\geq\frac{n}{2}.$$ The second conclusion holds because: $$Im(T+I)\subseteq ker (T-I)\Rightarrow rank(T+I)=dim(Im(T+I))\leq nullity(T-I).$$
PS: I had written a solution attempt in my question. Although my attempt was misleading to some extent, it, along with the answer above, contained everything needed to solve the question. After all, I deleted some parts of my question. I hope firstly this new solution contains no error and secondly what I did isn't against Stack Exchange rules.
Rank-nullity-theorem gives,
$n=\operatorname{Nullity}(T-I)+\operatorname{Range}(T-I)\dots(1)$.
Now try to show $\operatorname{Ker}(T-I)\subseteq W$ and $\operatorname{Im}(T-I)\subseteq W$. From which $(1)$ will give $n\leq2\operatorname{dim}(W)$.
Note: To show $\operatorname{Im}(T-I)\subseteq W$, you will need the fact that $F$ has characteristics $2$, that is, $1=-1$ in $F$.