Let $T$ be a bounded normal operator on a complex Hilbert space $V$. Suppose $\alpha \in \mathbb{C}$ is in the spectrum of $T$, i.e. $T - \alpha I$ is not invertible. I'd like to show that $|\alpha|^2$ is in the spectrum of $T^*T$.
I think I've managed to prove this as follows.
Proof of claim: Note that if $T$ is normal, then it is easy enough to check that $T - \alpha I$ and $T^*T - |\alpha|^2I$ are normal for any $\alpha \in \mathbb{C}$.
We will make use of the fact that for normal operators, invertibility is equivalent to injectivity and having a closed image. Using this, we claim that $T^* T - |\alpha|^2I$ is invertible implies that $T - \alpha I$ is invertible, which would suffice to complete the proof. To that end, suppose $T^* T - |\alpha|^2I$ is invertible.
First, we show that $T - \alpha I$ is injective. If it were not, then there would exist a nonzero $f \in V$ with $Tf = \alpha f$. Since $T$ is normal, we also have $T^* f = \overline{\alpha} f$. A short calculation shows that $(T^*T - |\alpha|^2I)f = 0$, which contradicts the invertibility of $T^*T - |\alpha|^2I$. Hence, $T - \alpha I$ is injective.
Next, we show that $T - \alpha I$ has closed image. Suppose $(f_k) \subset V$ is a sequence such that $(T - \alpha I)f_k \to g$ converges to some $g \in V$. It would suffice to show that $f_k \to f$ for some $f \in V$, whereupon the continuity of $T - \alpha I$ would show that $(T - \alpha I)f_k \to (T - \alpha I)f$, i.e. showing that $T - \alpha I$ has closed image. Indeed, since $T^*T - |\alpha|^2I$ is invertible, there exists some $\beta \in (0, \infty)$ such that for all $f_j, f_k$, we have $$||f_j - f_k|| \leq \beta ||(T^*T - |\alpha|^2 I) (f_j - f_k)||.$$
Then, we have $$||f_j - f_k|| \leq \beta ||{ (T^*T - |\alpha|^2I) (f_j - f_k) }||$$
$$= \beta || (T^* (T - \alpha I) + \alpha(T - \alpha I)^*) (f_j - f_k)|| $$
$$\leq \beta \left( || T^* || ||(T - \alpha I) (f_j - f_k) || + |\alpha| || (T - \alpha I)^* (f_j - f_k) || \right) $$
$$= \beta \left( ||{T^*}|| + |\alpha| \right) ||{(T - \alpha I) (f_j - f_k)}||.$$
The first equality is an easy calculation, the second inequality is from the triangle inequality, and the last equality is from the fact that $T - \alpha I$ is normal. Since $(T - \alpha I) f_k$ converges, it is Cauchy, hence $(f_k)$ is Cauchy by the above calculation. The completeness of $V$ shows $f_k \to f$ for some $f \in V$. This completes the proof.
I have two questions:
- Is my proof correct?
- My proof is pretty unwieldy and I get the sense there should be a better way of proving this claim. Is there a shorter, more elegant proof?
Your argument looks fine to me. As for an easier argument, it depends on the background available on normal operators. I'll give you three arguments. The first one is more or less elementary, while the other two use more machinery.
The argument without lots of machinery is the following: because $T-\alpha I$ is normal, $\ker(T-\alpha I)=\ker(T^*-\overline\alpha I)$. From this it is kind of easy to derive that the residual spectrum of $T$ is empty. This means that $\sigma(T)$ consists entirely of approximate eigenvalues. So if $\alpha\in\sigma(T)$ there exists $\{x_n\}\subset V$ with $\|x_n\|=1$ for all $n$, and $Tx_n-\alpha x_n\to0$. Since $\|(T-\alpha I)x_n\|=\|(T-\alpha I)^*x_n\|$, we also have $T^*x_n-\overline\alpha x_n\to0$. Then $$ T^*Tx_n-|\alpha|^2x_n=T^*(Tx_n-\alpha x_n)+\alpha(T^*x_n-\overline\alpha x_n)\to0, $$ and then $\alpha\in\sigma(T^*T)$.
The other two arguments below rely on working on the unital C$^*$-algebra C$^*(T)$ generated by $T$. This C$^*$-algebra is abelian. And because we are dealing with C$^*$-algebras, the spectrum does not depend on the algebra.
The first alternative way is to use that in an abelian unital Banach algebra, the spectrum of an element is the image of the element under all characters (nonzero multiplicative linear maps). It is standard that characters are $*$-preserving. Hence, with $\Sigma$ denoting the character space of C$^*(T)$, \begin{align} \sigma(T^*T)&=\{\gamma(T^*T):\ \gamma\in\Sigma\} =\{|\gamma(T)|^2:\ \gamma\in\Sigma\}\\[0.3cm] &=\{|\lambda|^2:\ \lambda\in\sigma(T)\}. \end{align}
The second alternative is to use the Gelfand transform. The unital C$^*$-algebra generated by $T$ is isomorphic to $C(\sigma(T))$, with $T$ mapped to the identity function. Then $T^*T$ is mapped to the function $z\longmapsto |z|^2$. As the spectrum of $g\in C(\sigma(T))$ is $g(\sigma(T))$, we again get that \begin{align} \sigma(T^*T)=\{|\lambda|^2:\ \lambda\in\sigma(T)\}. \end{align}