How would you go about showing that there cannot be a set of four 2 by 2 matrices that satisfy the anticommutative relation $AB + BA = 0 $ or $2I$ if $A=B$? i.e minimum order has to be 4.
I know that for such matrices A, B, C and D (of order 4), $A^2 = I$, $det(A) = +1$ or $-1$. I have also tried writing out these properties using $a_{ij}$ notation and see if some unsolvable equations would come out, unsuccessfully.
Just need a hint to get started, thanks.
There is an easy answer, anyway, for formally real fields like the real numbers. Assume this, and that $n\geq 4$.
I make extensive use of the fact that if $\{A_1,\dots,A_n\}$ is a collection of pairwise anticommuting things which each square to identity $I$, then $(\sum\lambda_i A_i)^2=(\sum\lambda_i^2)I$.
We claim that the $A_i$ are linearly independent. Suppose there exists $\lambda_i$'s such that $\sum\lambda_iA_i=0$. Squaring, we find that $(\lambda_i^2)I=0$, but this implies all the $\lambda_i=0$, and thus the $A_i$ are linearly independent. But since $M_2(F)$ is only four dimensional, this must mean $n=4$ that they are a basis for $M_2(F)$.
But $M_2(F)$ has things that square to zero, like $B=\begin{bmatrix}0&1\\0&0\end{bmatrix}$. Since the $A_i$ form a basis, you can find coefficients such that $\sum\alpha_iA_i=B$. But then squaring again, we find that all the $\alpha_i$ are zero, so that $B=0$, a contradiction.