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 $v$, $v_2 := ((v_2)1, (v_2)2, . . . , (v_2)n)^T$.