If is an eigenvalue of then there exists such that which happens iff (since both and are linear) and since is invertible it implies that can't be an eigenvalue so
Question: Suppose that is an invertible linear operator. Prove that is an eigenvalue of if and only if is an eigenvalue of .
Let be defined as an matrix such that . Then I tried approaching this by demonstrating that
But the only information that I have is - how does that get me to the desired result?
I suppose that if we know , when the rank of is not full (i.e. - less than ). And then we can try to prove that because of this, the rank of is also less than ...?