MHB Find the eigenvalue of a linear map

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The linear map A defined on R^3 is its own inverse, indicated by A(A(v)) = v. This property implies that A satisfies the polynomial equation x^2 - 1. Consequently, the minimal polynomial of A can only be x^2 - 1, x - 1, or x + 1. Since A is not equal to the identity or negative identity, the minimal polynomial must be x^2 - 1, leading to the conclusion that the eigenvalues of A are -1 and 1. Understanding these eigenvalues is crucial for analyzing the behavior of the linear map A.
Barioth
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Hi everyone,
I have this linear map $$A:R^3 \rightarrow R^3$$

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

I know that $$A(A(v))=v$$ this telling me that A is it's own inverse.
From there, how can I find the eigenvalue of A?
Thanks
 
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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 $\{-1,1\}$.
 
Thanks!
 
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