I've been working on this question for awhile, and am not sure how to finish the argument. It feels like I'm on the right track though. I've demonstrated my approach below:
Question: Let $E$ be a Hilbert space with orthonormal basis $\{e_i\}$. Let $T:E \to E$ be a bounded, linear, strictly upper triangular operator with respect to $\{e_i\},$ i.e. $\left< Te_j,e_i \right> = 0$ if $i \leq j$. Find all eigenvalues of $T$.
My Current Approach: Let $x \in E,$ be a unit eigenvector for $T$. Then
$$\lambda \left<x, x\right> = \left<Tx, x\right> = \left<\sum_j \alpha_j T e_j, x\right> $$ $$ = \sum_j \alpha_j \left<T e_j, x\right> $$ $$ = \sum_j \alpha_j \sum_i \bar{\alpha_i} \left<T e_j, e_i\right> $$ $$ = \sum_j \sum_{i > j} \alpha_j \bar{\alpha_i} \left<T e_j, e_i\right> $$
My intuition tells me that either all of the eigenvalues will be equal to zero, or there will be no eigenvalue eigenvector pairs at all. In my attempt, I'm trying to show that the right hand side reduces to zero, proving that $\lambda = 0.$