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$A$ is a square matrix, why does $A^2=I$ imply $nullity(A)=0$?

This is the key step in the solution, which I can't get it. Please help

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    $\begingroup$ Thanks for this post. With little modification I just generated a nice problem for my lin algebra students... $\endgroup$ – imranfat May 14 '14 at 14:49
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Suppose that $Ax = 0$ for some $x$. Then also $A^2 x = 0$, but $A^2 x = I x = x$, so $x = 0$.

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  • $\begingroup$ As simple and short as that. +1 $\endgroup$ – DonAntonio May 14 '14 at 13:08
  • $\begingroup$ The proof writes itself! +1 $\endgroup$ – rschwieb May 14 '14 at 13:11
  • $\begingroup$ Nice proof, but after all, the problem was also pretty simple. $\endgroup$ – mathse May 14 '14 at 13:13
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If you know that invertible matrices have nullity $\{0\}$, then you could just observe that $A$ is invertible.

If you don't know this, then fuglede's excellent elemenary answer suffices.

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    $\begingroup$ Excellent observation. It is invertible because there exists a matrix $B$ such that $AB=BA=I$. In fact, $B=A$. $\endgroup$ – mathse May 14 '14 at 13:16

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