Let $T$ be a bounded normal operator on a Hilbert space $\mathcal H$ with $0 \in \sigma (T).$ Then $0 \in \sigma (A),$ where $A = T^* T.$
$\textbf {My Attempt} :$ Since $T$ is normal and $0 \in \sigma (T)$ it follows that $0 \in \sigma_{\text {app}} (T).$ So there exists a sequence of unit vectors $\{x_n\}_{n \geq 1}$ such that $Tx_n \to 0.$ Since $T^*$ is bounded it follows that $T^*Tx_n \to 0$ i.e. $Ax_n \to 0.$ This shows that $0 \in \sigma_{\text {app}} (A) = \sigma (A)\ (\because$ $A$ is self-adjoint and hence normal as well).
Is it fine what I did or something missing? Could anyone please check it?
Thanks for investing your valuable time.
Your proof is correct. Also, no need to point out that $A$ is normal as well since it suffices $0 \in \sigma_{\text{app}}(A) \subseteq\sigma(A)$.
Here is a more elementary proof, by contrapositive. Assume that $0 \notin \sigma(T^*T)$ so $T^*T$ is invertible.
Therefore we have $$I = (T^*T)^{-1}T^*T = ((T^*T)^{-1}T^*)T$$ and $$I = T^*T(T^*T)^{-1} \stackrel{T\text{ normal}}{=} TT^*(T^*T)^{-1} = T(T^*(T^*T)^{-1})$$ so we see that $T$ is left and right invertible. It follows that $T$ is invertible with $$T^{-1} = (T^*T)^{-1}T^* = T^*(T^*T)^{-1}$$ so $0 \notin \sigma(T)$.