Form of anticommuting matrices. Suppose $F$ is a field where $1 \neq -1$ and $V$ is a $2n$ dimensional $F$-vector space. Also suppose that $M,N$ are involutions, i.e. $M^2 = I$ and $N^2 = I$, and that $M$ and $N$ anti-commute, i.e. $MN = -NM$.
I would like to show that
$$
M = \left[\begin{array}{cc}A & 0\\ 0 & -A \end{array} \right],~~ N  =\left[\begin{array}{cc}0 & B\\ B & 0 \end{array} \right]
$$
(these are given in block matrix notation, so that $A,B$ are matrices, not scalars).
I found this website
which makes the makes the same claim ($M,N$ involutions implies they are invertible, the website handles a slightly more general case) but I am not able to follow the argument.
First off, could anyone verify that this is true for an arbitrary $F$-vector space as described?
Also some help with the proof would be much appreciated, thank you.
 A: The right statement is that the matrices are similar to
$$\begin{pmatrix} \mathrm{Id}_n & 0 \\ 0 & - \mathrm{Id}_n \end{pmatrix} \quad 
\begin{pmatrix} 0 & \mathrm{Id}_n \\ \mathrm{Id}_n & 0 \end{pmatrix}.$$
Similar to means that we can choose a basis of $V$ so that the matrices are of this form. In coordinates, your matrices look like
$$S \begin{pmatrix} \mathrm{Id}_n & 0 \\ 0 & - \mathrm{Id}_n \end{pmatrix} S^{-1} \quad 
S \begin{pmatrix} 0 & \mathrm{Id}_n \\ \mathrm{Id}_n & 0 \end{pmatrix} S^{-1}$$
for some invertible $S$.
The subscript $n$ means that I am talking about the $n \times n$ identity matrix.

This looks like homework, so I'd rather not give a full solution. 

Since $M^2 = 1$, the matrix $M$ is diagonalizable with eigenvalues $1$ and $-1$. So we can choose a basis where 
$$M = \begin{pmatrix} \mathrm{Id}_k & 0 \\ 0 & - \mathrm{Id}_{2n-k} \end{pmatrix}.$$
Write $N$ in block form as $\left( \begin{smallmatrix} A & B \\ C & D \end{smallmatrix} \right)$. 
Now what can you deduce from the equation $MN=-NM$? And, once you've used that, what can you deduce from the equation $N^2=\mathrm{Id}_{2n}$?
