Find the eigenvalue of a linear map

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SUMMARY

The discussion centers on finding the eigenvalues of the linear map \( A: \mathbb{R}^3 \rightarrow \mathbb{R}^3 \) defined by \( A(v) = v - 2(v \cdot \hat{o})\hat{o} \), where \( \hat{o} \in \mathbb{R}^3 \) and \( ||\hat{o}|| = 1 \). It is established that \( A \) is its own inverse, leading to the conclusion that \( A^2 = I \). Consequently, the minimal polynomial of \( A \) is determined to be \( x^2 - 1 \), yielding the eigenvalues \( \{-1, 1\} \).

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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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