Let $k$ be a field of characteristic $2$. Let $V$ be a finite dimension $k$-vector space and
$$\rho: \mathbb{Z}/2\mathbb{Z} \to{\rm Aut}_k(V)$$
a representation of $\mathbb{Z}/2\mathbb{Z}=\{1,\sigma\}$ over $k$. Show that $V$ can be decomposed as $W_1 \oplus W_2$ where $W_1$ and $W_2$ are subspaces of $V$, moreover $W_2$ can be decomposed as a direct sum of subspaces $U_i$ of dimension $2$ so that $\sigma$ acts on $W_1$ as indentity and acts on $U_i$ as follows: there exists a basis $x_i,y_i$ of $U_i$ so that $\sigma(x_i)=x_i$ and $\sigma(y_i)=y_i+x_i$.
My wrong attempt (thanks @Omnomnomnom for the comment): We have $\rho(\sigma)^2=\rho(\sigma^2)=Id$, i.e $\rho(\sigma)=f$ is an idempotent element in ${\rm End}(V)$. Hence, we can decompose $V={\rm im}(f)\oplus {\rm im}(1-f)$. However, I cannot prove the latter part of the problem.
P/s: By Maschke's theorem: Every representation of a finite group $G$ over a field $F$ with characteristic not dividing the order of $G$ is a direct sum of irreducible representations. However, the hypothesis of the problem does not satisfy the condition of this theorem.
Best Answer
Hint: The statement to be proven is equivalent to saying that there exists a basis relative to which the matrix of $\rho(\sigma)$ is given by $$ M = \pmatrix{I\\ &A\\ && \ddots \\ &&& A} $$ where $I$ denotes an identity matrix of some size, and $$ A = \pmatrix{1&0\\1&1}. $$ Note in particular that $M$ is in Jordan normal form.
You might find it informative to consider the representation $\rho:G \to k^n$ for which $\rho(\sigma) = M$. Note that in this case, we would have $x_i = e_{2j-1},y_i = e_{2j}$ for some index $j$, where $e_1,\dots,e_n$ denotes the standard basis of $k^n$.
Hint: Note that $\rho(\sigma)^2 + I = 0$. Thus, the minimal polynomial $m \in k[x]$ of $\rho(\sigma)$ must divide $$ x^2 + 1 = (x - 1)^2. $$ What does this let you deduce about the Jordan form of $\rho(\sigma)$? It now suffices to show that there exist a choice of basis of $V$ over $k$ (rather than over the algebraic completion $\bar k$) relative to which $\rho(\sigma)$ has the form of $M$.
Alternatively, we can skirt the issue of algebraic completions if we stick to the Frobenius normal form. In particular, note that the matrix $M$ above is similar to $$ \tilde M = \pmatrix{I \\ & C \\ && \ddots \\ &&& C} $$ where $A$ is as defined above, and $C$ is the companion matrix to the polynomial $x^2 + 1$. This holds because $A$ is similar to $C$.
The fact that $\rho(\sigma)$ must have the matrix $\tilde M$ relative to some basis is a direct consequence of the structure theorem.