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Let $A$ be an $n\times n$ matrix and let $I$ be the $n\times n$ identity matrix. Show that if $A^{2} = I$, and $A \neq I$, then $\lambda =-1$ is an eigenvalue of $A$. This problem doesn't seem that too hard to solve, but I am stuck near the end. Here is what I have done so far.

Since $A^{2}=I$, then by definition $Ax=\lambda x$, where $x$ is an eigenvector of $A$ and $\lambda$ is an eigenvalue of $A$. It follows that $x=Ix=A^{2}x=A(Ax)=A(\lambda x)= \lambda(Ax)=\lambda^{2}x$. (Now I was going to use the fact that since $A \neq I$, that $x\neq 0$, so we get that $\lambda = 1$ or $-1$). This is where I am stuck.

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    Do you know about Jordan normal form?2012-07-23
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    This is what I am currently having a hard time with. I don't quite understand the whole idea of it, just bits and pieces.2012-07-23
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    You have shown that all eigenvalues satisfy $\lambda = \pm 1$. To finish, you must show that at least one eigenvalue is $-1$. Proceed by contradiction; assume all eigenvalues are $+1$ and see what that means about $A$. @Robert Israel's suggestion is a good one. Remember that $x^2-1 = (x+1)(x-1)$.2012-07-23

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