# Eigenvalue of a matrix

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? –  Gerry Myerson Jul 23 '12 at 7:36
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. –  Kelly Jul 23 '12 at 7:43
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 $x^2-1 = (x+1)(x-1)$. –  copper.hat Jul 23 '12 at 8:03

Hint: If $-1$ is not an eigenvalue of $A$, then $A+I$ is invertible. What is $(A^2-I)(A+I)^{-1}$?

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If $A^2=I$, then $A$ satisfies the polynomial $t^2-1=(t-1)(t+1)$. Therefore, the minimal polynomial of $A$ divides $(t-1)(t+1)$.

If the minimal polynomial is $t-1$, then that means that $A-I=0$, so $A=I$; but we are assuming that $A\neq I$, so this is not the case.

That means that the minimal polynomial is divisible by $t+1$; since every irreducible factor of the minimal polynomial divides the characteristic polynomial, it follows that $t+1$ divides the characteristic polynomial of $A$, and hence that $-1$ is an eigenvalue of $A$, as desired.

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HINT: What you have so far is that if $\lambda$ is an eigenvalue for $A$, then $\lambda=\pm1$. This is correct.

Now assume that the eigenvalue $\lambda=-1$ does not occur and try to figure out what the Jordan canonical form of $A$ might be.

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Now you have λ=1 or −1. Now assume λ=1 then you get Ax=x (since Ax=λx) now we know x≠0 that means A=I, but we know that A≠I, contradiction so λ=-1 is the eigenvalue

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Welcome to MSE! It helps readability to format using MathJax (see FAQ). Regards –  Amzoti Aug 22 '13 at 13:34