1. ## Eigenvalue Proof

I am suppose to prove that if R is a matrix, R^2 = I and lambda is an eigenvalue, then lamda is 1 or - 1

Do I somehow need to show that an eigenvector, say it is v that Rv and R^2 v ??? compared but I am lost!

Is it just that since R is squared then either 1 or -1 are values??? Frostking

2. Originally Posted by Frostking
I am suppose to prove that if R is a matrix, R^2 = I and lambda is an eigenvalue, then lamda is 1 or - 1

Do I somehow need to show that an eigenvector, say it is v that Rv and R^2 v ??? compared but I am lost!

Is it just that since R is squared then either 1 or -1 are values??? Frostking
$R^2=I\Longrightarrow (R-I)(R+I)=0\Longrightarrow \mbox{R is a zero of the polynomial }(x-1)(x+1)\Longrightarrow$ $\mbox{the minimal pol. of R divides }(x-1)(x+1)\Longrightarrow\,...$

Tonio

3. A slightly different proof, without using the minimal polynomial (directly):

If $\lambda$ is an eigenvalue of R then there exist non-zero v such that $Rv= \lambda v$. Apply R to both sides of that equation- $R^2v= R(\lambda v)= \lambda Rv= \lambda(\lambda v)= \lambda^2 v$. But it is given that $R^2= I$ so we are saying that $v= \lambda^2 v$ so that $(1- \lambda^2)v= 0$. Since v is not 0, we must have $1- \lambda^2= 0$ and, from that, $\lambda= 1$ or $\lambda= -1$.

4. Originally Posted by HallsofIvy
A slightly different proof, without using the minimal polynomial (directly):

If $\lambda$ is an eigenvalue of R then there exist non-zero v such that $Rv= \lambda v$. Apply R to both sides of that equation- $R^2v= R(\lambda v)= \lambda Rv= \lambda(\lambda v)= \lambda^2 v$. But it is given that $R^2= I$ so we are saying that $v= \lambda^2 v$ so that $(1- \lambda^2)v= 0$. Since $v\neq 0$, we must have $1- \lambda^2= 0$ and, from that, $\lambda= 1$ or $\lambda= -1$.
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