I am going through a longer proof of a theorem which states the following as an intermediate result:
Let $N$ be a bounded normal operator on a Hilbert space $H$ with $0 \in \sigma(N)$. Then the self-adjoint operator $A = NN^\ast$ has $0 \in \sigma(A)$.
I was able to find a proof of this assertion using the approximate point spectrum $\Pi(N)$ of $N$.
For a normal operator $N$ one can show that $\Pi(N) = \sigma(N)$ holds. With this statement, the claim follows: Since $0 \in \sigma(N)$, there is a sequence $(x_n)$ with $\| x_n\| = 1$ such that $\| Nx_n \| \longrightarrow 0$, therefore, $A x_n \longrightarrow 0$ and $A$ has $0 \in \sigma(A)$.
However, the proof of $\Pi(N) = \sigma(N)$ is somewhat long and it seems to me that the above statement could be proven in a much shorter and more elementary manner.
Question: Is there a way to show $0 \in \sigma(A)$ more easily and shorter and without using the arguments I cited above?
If $0\notin \sigma(A)$ then it means the operator $NN^*$ is invertible. So there is a bounded operator $T$ such that $NN^*T=TNN^*=I$. In particular this tells us that $N(N^*T)=I$, which means $N$ is invertible from the right side. Also, since $N$ is normal we get $I=TNN^*=TN^*N$, which means $N$ is also invertible from the left side. And it is a standard result that if an element is invertible from both sides then the one sided inverses must be equal, which means that the element is invertible. So we got that $N$ is invertible, which contradicts $0\in\sigma(N)$.