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Suppose that there is a set of matrices.

1) $A^2 =0, C^2=0, E^2=0 .....$ and $AB+BA=0$, $CD+DC=0$... where $A,B,C..$ are matrices.

2) Matrices in the set either anticommute or commute.

3) $AD+BC \neq 0,... $

Is there any set of matrices that satisfies all of the aforementioned?

Edit: all matrices are not equal to each other.

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3) seems to contradict 1) if required for all $A,D,B,C$ from the set. –  Berci Oct 27 '12 at 22:14
    
@Berci how does it contradict? I am not gettting it. –  TTTY Oct 27 '12 at 22:18
1  
Take $A=B=C=D$ and you have a contradiction. –  kjetil b halvorsen Oct 27 '12 at 22:31

2 Answers 2

$ N:= \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}, \ \ \sigma:= \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}\ \ \ $

$\Longrightarrow \ \ \ N^2=0, \ \ \ \sigma^2=1,\ \ \ \sigma N=N=-N\sigma$

So define

$A:=a\ N, \ \ \ C:=c\ N, \ \ \ D:=d\ N,$

$B:=b\ N+\beta\ \sigma$

$\Longrightarrow$ products among $\{A,C,D,...\}$ are zero, while $B^2=\beta^2\ 1$ and e.g. $BA=ab\ \sigma$.

The relation $\sigma N=N=-N\sigma$ makes $N$ an "eigenvector" of $\sigma$ from left and right so as to fulfill the anti-commutation relation $AB+BA=0$, without $AB$ already being zero. This way the expression involving $B$ is the only one which survives in your last condition:

$AD+BC=ad\ N^2+bc\ \sigma\propto \sigma\ne 0.$

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$\mathrm\LaTeX$ tip: It's much easier to type (and remember) "\implies" (respectively "\impliedby") than "\Longrightarrow" (respectively "\Longleftarrow"). –  kahen Oct 28 '12 at 2:33

Finding $n$ matrices $A_1, ... A_n$ satisfying the commutation relations

$$A_i A_j + A_j A_i = 0 \forall 1 \le i, j \le n$$

is equivalent to finding a finite-dimensional module over the exterior algebra of an $n$-dimensional vector space. The exterior algebra has dimension $2^n$, hence is itself such a module (acting on itself by left multiplication). The exterior algebra acts faithfully on itself, hence the corresponding matrices satisfy only relations which follow from the relations above.

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