Let $V$ be a finite-dimensional vector space over a field $\mathbb{F}$ of characteristic $\ne 2$. Let $X,Y: V \rightarrow V$ be linear transformations such that $X^2=Y^2=\rm Id$, where $\rm Id$ is the identity transformation.
Suppose that $XY=-YX$. Show that $\dim V$ is even, and that there is a basis of $V$ in which the matrices of $X$ and $Y$ are $$\begin{bmatrix} 0 & I_n \\ I_n & 0\end{bmatrix} \quad \& \quad \begin{bmatrix} I_n & 0 \\ 0 & -I_n\end{bmatrix}$$ respectively, where $I_n$ is the identity matrix and $\dim V=2n$.
Things I know:
I've showed that both $X$ and $Y$ are diagonalizable with eigenvalues $\lambda= \pm 1$.
I have also thought the since $X,Y$ are diagonalizable, we may use the Jordan canonical form. But from here, I'm not sure how to proceed.
Any help will be appreciated!
From the relation $XY=-YX$, you get that if $y$ is an eigenvector associated with eigenvalue $1$ of $Y$
$$X(Y(y))= X(y) =-YX(y)= -Y(X(y))$$ which means that $X(y))$ is an eigenvector associated with the eigenvalue $-1$ of $Y$. As we suppose that the characteristic of the field is not equal to $2$ ($1 \neq -1$), the two eigenvalues are distinct and the vectors $y, X(y)$ are linearly independent.
From this, you can build by induction a family of vectors $\{y_1, \dots, y_n\}$ that is a basis of the eigenspace $E_Y(1)$ associated to the eigenvalue $1$ of $Y$, and such that $\{y_1, \dots, y_n, X(y_1), \dots, X(y_n)\}$ is a basis of $V$. Therefore the dimension of the vector space is even. Let say $2n$.
Now let's have a look at the matrices of $X$ and $Y$ in the basis
$$e_1 = y_1, \dots, e_n = y_n, e_{n+1}= X(y_1), \dots, e_{2n}= X(y_n)$$
For $1 \le i \le n$ we have
$$\begin{cases} X(e_i) &= X(y_i) = e_{i+n}\\ X(e_{i+n}) &= X(X(y_{i})) =X^2(y_{i})=y_{i} = e_i\\ Y(e_i) &= Y(y_i)=y_i=e_i\\ Y(e_{i+n}) &= Y(X(y_i)) = -X(y_i)=-e_{i+n} \end{cases}$$
which is exactly the relations $$\begin{bmatrix} 0 & I_n \\ I_n & 0\end{bmatrix} \quad \& \quad \begin{bmatrix} I_n & 0 \\ 0 & -I_n\end{bmatrix}$$ we're looking for.