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When $E^{2} = I$ (where $I = n×n$ identity matrix), show that if $\lambda$ is an eigenvalue of $E$ then λ $\in$ {$\pm1$}.

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    $\begingroup$ $\lambda$ is an eigenvalue of $E$ iff there exists a nonzero vector $v$ with $Ev = \lambda v$. So apply $E$ to both sides. $\endgroup$
    – Jane Doé
    May 6, 2019 at 21:03
  • $\begingroup$ What have you tried? $\endgroup$
    – Spencer
    May 6, 2019 at 21:05
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    $\begingroup$ More generally: if $\lambda$ is an eigenvalue of $E$ and $p$ is a polynomial, then $p(\lambda)$ is an eigenvalue of $p(E)$. $\endgroup$
    – GEdgar
    May 6, 2019 at 21:06

2 Answers 2

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$E^{2} = I$ implies that $E^{2} x=Ix=x$. So we have to solve $E=ax$ such that $a^2=1$ since

$E^{2} x=Ix=x=EEx=Eax=aax$. And the only values for $a$ are one and negative one

($a^2=1$). IF $a$ is an eigenvalue of E

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If $x\neq 0$ and $Ex=\lambda x$ then $x=E^2x=\lambda^2 x$, so $(\lambda^2-1)x=0$.

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  • $\begingroup$ For this level of question, I would emphasize why $x$ is not zero. $\endgroup$ May 6, 2019 at 21:14

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