Question: Suppose that $\displaystyle T$ is an invertible linear operator. Prove that $\displaystyle \lambda$ is an eigenvalue of $\displaystyle T$ if and only if $\displaystyle \lambda^{-1}$ is an eigenvalue of $\displaystyle T^{-1}$.

Let $\displaystyle A$ be defined as an $\displaystyle n \times n$ matrix such that $\displaystyle T(x) = Ax$. Then I tried approaching this by demonstrating that

$\displaystyle det(A - \lambda I_n) = det(A^{-1} - \lambda^{-1} I_n) = 0$

But the only information that I have is $\displaystyle AA^{-1}=A^{-1} A = I_n$ - how does that get me to the desired result?

I suppose that if we know $\displaystyle det(A - \lambda I_n) = 0$, when the rank of $\displaystyle A - \lambda I_n$ is not full (i.e. - less than $\displaystyle n$). And then we can try to prove that because of this, the rank of $\displaystyle A^{-1} - \lambda^{-1} I_n$ is also less than $\displaystyle n$...?