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Let $A$ be a square matrix with real entries, and let $\lambda$ be its complex eigenvalue. Suppose $v=(v_1,v_2,...,v_n)^T$ is a corresponding eigenvector, $Av=\lambda v$.

Prove that the $\lambda_2$ is an eigenvalue of $A$ and $Av_2 = $lambda_2 $v_2$. Here $v_2$ is the complex conjugate of the vector $v$, $v_2 := ((v_2)1, (v_2)2, . . . , (v_2)n)^T$.

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    $\begingroup$ Draw a line over $Av$ and then another one over $\lambda v$. $\endgroup$
    – Git Gud
    May 22, 2016 at 0:05
  • $\begingroup$ You say that "v2 is the complex conjugate of the vector v" but you don't say what $\lambda 2$" is. My guess would be that you mean it also to be the complex conjugate of $\lambda$ rather than $\lambda$ squared. Since A has all real entries, its characteristic equation has all real coefficients. Complex roots of such an equation must be in complex conjugate pairs. $\endgroup$
    – user247327
    May 22, 2016 at 0:10
  • $\begingroup$ @GitGud lol! Great answer! $\endgroup$
    – user5389726598465
    Jun 6, 2017 at 3:44

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