Proof explanation: Spectral radius of operators

Question

I want to understand the following theorem from this paper

I don't understant the first equality.

Problem 71 is the following

Answer 1

First, by definition, the left-hand side is $2\sup_{\lambda\in\sigma(AB)}|\lambda|.$

Now, recall that the resolvent set $\rho(T)$ is the complement of the spectrum, and it's defined as $\lambda\in\rho(T)$ iff $(T-\lambda)^{-1}$ exists and is bounded. Let's call $$T=\begin{bmatrix} 2AB & 0 \\ B & 0\end{bmatrix}$$ and look at when $T-\lambda$ is invertible.

Note that $$T-\lambda=\begin{bmatrix} 2AB-\lambda & 0\\ B & -\lambda\end{bmatrix}.$$ Applying problem $71$ (observe that the invertibility assumption is satisfied now), we get that this is invertible iff $-(2AB-\lambda)\lambda$ is invertible, which occurs iff $\lambda\neq 0$ and $\lambda\in\rho(2AB).$ That is, $\lambda\in\sigma(T)$ iff $\lambda=0$ or $\lambda\in\sigma(2AB)=2\sigma(AB).$

So, the right-hand side will be$$\sup_{\lambda\in\sigma(T)}|\lambda|=\sup_{\lambda\in\{0\}\cup 2\sigma(AB)}|\lambda|=2\sup_{\lambda\in\{0\}\cup \sigma(AB)}|\lambda|,$$ which is equal to the left-hand side.

Essentially, the right-hand side will have the same spectrum with extra zeros, which won't affect the sup.