Here is the original problem: Let $S$ and $T$ be linear operators on a finite-dimensional vector space $V$. Show that $TS$ and $ST$ have the same eigenvalues.
I can prove it. However, my question is: Is it also true for infinite dimensional vector space $V$? What is the counter-example for this case?
Thank you.
It is always true that two linear operators share the same nonzero eigenvalues, regardless of the dimension of the vector space. However, in the finite case, it is also true that $TS$ is invertible if and only if $ST$ is, thus also settling the case of the zero eigenvalue.
The standard counterexample for this in infinite dimensions is to let $V = \ell^2(\mathbb{N})$, the set of all sequences of the form $\{a_n\}$ such that $\sum a_n^2 < \infty$. Let $R$ be the right shift, and $L$ the left-shift. Then $LR = I$, but $RL$ is not invertible.