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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

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  • $\begingroup$ If $X=Y=0$, then $XY=-YX$. That is, they anti-commute. However, $X$ and $Y$ are not linearly independent. $\endgroup$ – José Carlos Santos Jul 3 '20 at 20:55
  • $\begingroup$ @JoséCarlosSantos my b, i forgot to mention that $X_i^{2} notEQ 0$ $\endgroup$ – John Rawls Jul 3 '20 at 20:57
  • $\begingroup$ Then I suggest that you edit your question. $\endgroup$ – José Carlos Santos Jul 3 '20 at 21:00
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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 $X_2^{\,2}=0$.

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Hint Try a proof by contrapositive: if $X,Y$ are linearly dependent, show that they commute (and therefore do not anticommute).

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  • $\begingroup$ @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. $\endgroup$ – Ben Grossmann Jul 3 '20 at 21:15
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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.

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  • $\begingroup$ The OP mentioned in the comments that $X_i^2 \neq 0$.I used that directly. $\endgroup$ – Lelouch Jul 3 '20 at 21:52

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