Let $T$ be an invertible linear operator on a finite-dimensional vector space $V$. Prove that the eigenspace of $T$ corresponding to $\lambda$ is the same as the eigenspace of $T^{-1}$ corresponding to $\lambda^{-1}$.
Let $T$ be a linear operator on a finite-dimensional vector space $V$. Let $E_\lambda(T)$ denote the eigenspace of T. How can we conclude $E_\lambda(T)=E_{\lambda^{-1}}(T^{-1})$?
Let $\lambda$ be an eigenvalue of $T$ with corresponding eigenvector $v$. Then, per definition, $Tv = \lambda v$. Multiplying this equation by $T^{-1}$ yields $T^{-1}Tv = T^{-1} \lambda v$. Since $\lambda \neq 0$ by the invertibility of $T$, this is equivalent to
$$T^{-1}v = \lambda^{-1} v.$$
But this precisely means that $v$ is an eigenvector of $T^{-1}$ with corresponding eigenvalue $\lambda^{-1}$. Since $v$ was arbitrary, the eigenspaces are the same.