# If matrix $X$ & $Y$ anti-commute then show that the two matrices are linearly independent

Show that if matrix $$X_1$$ & $$X_2$$ anti-commute then show that the two matrices are linearly independent and $$X_i ^{\,2}\ne0$$

I know $$X_1X_2=-X_2X_1$$ from the definition then I tried the following:

$$X_1^{-1}X_1X_2=-X_1^{-1}X_2X_1$$ $$X_2 = -X_1^{-1}X_2X_1 \ (1)$$

$$and$$ $$X_1X_2X_2^{-1}=-X_2X_1X_2^{-1}$$ $$X_1=-X_2X_1X_2^{-1} \ (2)$$

Then I'll substitute (1) into (2) to get:

$$X_1=X_1^{-1}X_2X_1X_1X_2^{-1}$$ $$X_1=-X_1^{-1}X_1X_2X_1X_2^{-2}$$ $$X_1=X_1X_2X_2^{-2}$$

But I'm not sure if this does anything

• If $X=Y=0$, then $XY=-YX$. That is, they anti-commute. However, $X$ and $Y$ are not linearly independent. – José Carlos Santos Jul 3 '20 at 20:55
• @JoséCarlosSantos my b, i forgot to mention that $X_i^{2} notEQ 0$ – John Rawls Jul 3 '20 at 20:57
• Then I suggest that you edit your question. – José Carlos Santos Jul 3 '20 at 21:00

If $$X_1=\lambda X_2$$, then $$0=X_1X_2+X_1X_1=2\lambda X_2^{\,2}$$. So, either $$\lambda=0$$ (in which case $$X_1=0$$) or $$X_2^{\,2}=0$$.
Hint Try a proof by contrapositive: if $$X,Y$$ are linearly dependent, show that they commute (and therefore do not anticommute).
• @RobertIsrael They can't because $XY$ is a multiple of $X^2$ or $Y^2$, which (per the comment on the question) is assumed to be non-zero. – Ben Grossmann Jul 3 '20 at 21:15
We would say that $$X_1$$ and $$X_2$$ are linearly independent if for scalars (from the underlying field of the respective matrix space) $$a,b$$ we have, $$aX_1 + bX_2 =0 \implies a=b=0$$
So, we start by assuming $$aX_1+bX_2=0$$. Multiply once from the left and once from the right by $$X_2$$ and add. We then get, $$a(X_1X_2 + X_2X_1) + 2bX_2^2 = 2bX_2^2=0$$ This clearly implies $$b=0$$ and back-substituting in the assumption, we also get $$a=0$$. Hence they are linearly independent.
• The OP mentioned in the comments that $X_i^2 \neq 0$.I used that directly. – Lelouch Jul 3 '20 at 21:52