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Show that if $\lambda{}$ is an eigenvalue of $A$, then it is also an eigenvalue for $S^{-1}AS$ for any nonsingular matrix $S$.

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Try the definition of eigenvalue ($\lambda$) and corresponding eigenvector ($v$): $Av=\lambda v$. –  Berci Oct 24 '12 at 0:26
    
What have you tried? –  lhf Oct 24 '12 at 0:28
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We have $$Ax = \lambda x$$ Since $S$ is an invertible matrix, consider $x = S^{-1}y$. We then get that $$AS^{-1}y = \lambda S^{-1}y \implies SAS^{-1}y = \lambda y$$

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