write a finite rank operator as sum of a nilpotent and diagonal operators

Question

In the highlighted part, I don't see why we can write a finite rank operator as sum of a nilpotent and diagonal operators.

I know that if the space is of finite dimension, then we can write any operator as a upper triangular matrix. But here, the space could be infinite dimensional...

Thank you for your help.

Answer 1

The operator $B_n$ is finite-rank. So, as an operator $(\ker B_n)^\perp\to\operatorname{Ran}B_n$, it is an invertible operator between finite-dimensional Hilbert spaces of the same dimension. Which allows you to see $B_n$ as an $m\times m$ matrix, where $m=\dim\operatorname{Ran} B_n$.