# 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.

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Are you sure about this result? Are you not missing "similar to". Take $M=\mathrm{diag}(1,-1,1,-1)$ and $N=\begin{array} 70& 0 & 0 & 1\\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & \\ 1 & 0 & 0 & 0 \end{array}$. – Nathan Portland Feb 7 '12 at 7:48
I guess the result is not true then. So what do the solutions look like then? – nullUser Feb 7 '12 at 13:36

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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How can you conclude from $M^2 = 1$ that $M$ is diagonalizable? I see that if $M$ has an eigenvalue it must be $\pm 1$, but what if $M$ doesn't? If I knew I could diagonalize I would have solved this problem long ago :( – nullUser Feb 7 '12 at 16:34
Do you know the theorem that, if the minimal polynomial of $M$ has all its roots in $K$, and has no double roots, then $M$ is diagonalizable over $K$? – David Speyer Feb 7 '12 at 16:37
If not, hint: Induction on $\dim M$. We have $(M-\mathrm{Id}) (M+\mathrm{Id})=0$, so at least one of these factors has a kernel. Then what? – David Speyer Feb 7 '12 at 16:38
Minimal polynomial and characteristic polynomial aren't the same by any chance are they? From context I would guess that in this case $M^2-1$ is the minimal polynomial. Also, what is meant by $\dim M$? I have heard of $\dim \ker M$ and $\dim \operatorname{im} M$, but not of $\dim M$. – nullUser Feb 7 '12 at 16:50
Oh, sorry, just the size of the matrix $M$. In general, minimal polynomial and characteristic polynomial are not the same. For example, if $M$ is the $n \times n$ identity matrix, then its minimal polynomial is $x-1$ and the characteristic polynomial is $(x-1)^n$. There are a lot of ways to show that (over a field of characteristic not $2$), a matrix satisfying $M^2=\mathrm{Id}$ is diagonalizable, and I didn't mean to focus on one particular one. If you are near the beginning of a rigorous linear algebra course, then this is a good challenge for you. – David Speyer Feb 7 '12 at 17:07