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Find the eigenvalue of a linear map

Barioth

Member
Jan 17, 2013
52
Hi everyone,
I have this linear map \(\displaystyle A:R^3 \rightarrow R^3\)

I have that \(\displaystyle A(v)=v-2(v\dot ô)ô); v,ô\in R^3 ;||ô||=1\)

I know that \(\displaystyle A(A(v))=v\) this telling me that A is it's own inverse.
From there, how can I find the eigenvalue of A?
Thanks
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
IF $A^2 = I$, then $A^2 - I = 0$, that is, $A$ satisfies the polynomial $x^2 - 1$.

This means the minimal polynomial of $A$ can only be:

$x^2 - 1, x - 1$ or $x + 1$.

Since $A \neq \pm\ I$, it must be that the minimal polynomial of $A$ is $x^2 - 1$, so the eigenvalues of $A$ are $\{-1,1\}$.
 

Barioth

Member
Jan 17, 2013
52
Thanks!