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Assume A is invertible, Let $\lambda$ be an eigenvalue of $A$. Prove that $\lambda^{-1}$ is an eigenvalue of $A^{-1}$.

The answer goes:

given that $A$ is invertible, $Ax=\lambda x$, $A$ is invertible, and $\lambda\neq 0$, we have
$$Ax=\lambda x\implies A^{-1}Ax=A^{-1}\lambda x\implies x=\lambda A^{-1}x\implies \frac1\lambda x=A^{-1}x.$$

My question is why are we assuimg $\lambda \neq 0$. And for a $\lambda \neq 0$, can we always claim that it has an inverse?

  • 2
    $\begingroup$ If $\lambda =0$ them $\lambda^{-1}$ does not exist ; otherwise it exists. $\endgroup$ – Kavi Rama Murthy Feb 20 '20 at 6:03

We can assume $\lambda \neq 0$ since $A$ is invertible if and only if $0$ is no eigenvalue of $A$. See this question for more details.

If $\lambda \neq 0$, then $\lambda$ always has a multiplicative inverse in the corresponding field, so $1/\lambda$ exists.


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