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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.

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    I guess the result is not true then. So what do the solutions look like then?2012-02-07

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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 $N^2=\mathrm{Id}_{2n}$?

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    @David I was wondering if you would be able to clarify something for me. I am interested in this form for anti-commuting matrices, however I am wondering if this is the form or the more general one given on [this site.](http://goo.gl/YKhG2) That general one seems to have zero entry rows and columns, whereas this case both matrices are full rank. Is what you are saying that if we decompose one of the matrices into a form of S*Diag(I,-I)*S^{-1}, then in that same basis, the other one will always be as you have written there, will the identity matrices in the skew diagonal?2012-08-25