Why does $A^2=I$ imply $nullity(A)=0$?

Question

$A$ 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

Thanks for this post. With little modification I just generated a nice problem for my lin algebra students... — imranfat, May 14 '14 at 14:49

score 14 · Accepted Answer · answered May 14 '14 at 13:08

14

Suppose that $Ax = 0$ for some $x$. Then also $A^2 x = 0$, but $A^2 x = I x = x$, so $x = 0$.

answered May 14 '14 at 13:08

fuglede

As simple and short as that. +1 – DonAntonio May 14 '14 at 13:08
The proof writes itself! +1 – rschwieb May 14 '14 at 13:11
Nice proof, but after all, the problem was also pretty simple. – mathse May 14 '14 at 13:13

score 7 · Answer 2 · answered May 14 '14 at 13:12

7

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.

answered May 14 '14 at 13:12

rschwieb

2

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

2 Answers2