I am currently starting to learn about Banach Algebras/Spectral theory. And I have the following question: Take $\mathcal{A}:=B(l^2(\mathbb N))$ as a Banach Algebra. And consider the right shift operator
$$
T( (x_n)_{n \in \mathbb N}) =
\begin{cases}
0 ,\quad & n = 1 \\
x_{n-1} , \quad & n > 1
\end{cases}
$$
We need to prove that $\sigma(T^{\ast} T) \neq \sigma(T T^{\ast})$, where $\sigma$ denotes the spectrum. (Usually this is defined only for elements in the Banach Algebra but by abuse of notation, we need to find an element in the algebra, such that the spectra differ from one-another).
My progress: Clearly $T^{\ast} : l^2(\mathbb N) \to l^2(\mathbb N)$ is the left shift-operator. Thus consider $x \in \mathcal{A}$, then
$$
T^{\ast} T x = (x_1,x_2,\dots) \quad T T^{\ast} x = (0,x_2,x_3,\dots)
$$
This reflects that we have a Hilbert-Space structure i.e. the adjoint is the inverse operator. So for $x=0 \in \mathcal A$, I believe that $0 \notin \sigma(T^{\ast} T)$ but $0 \in \sigma(T T^{\ast})=B_1(0)$. However I am not 100% sure that one can deduct the last equality from knowing the spectrum of the right shift? Moreover, can we deduce more information about the Banach Algebra if we know that $\sigma(T^{\ast} T) \neq \sigma(T T^{\ast})$?
Thanks in advance.
Best Answer
You have that $T^*T=I$, the identity operator, and it is trivial to check that $\sigma(I)=\{1\}$. Meanwhile, $TT^*=I-P$, where $P$ is the projection onto the first coordinate. Then $\sigma(TT^*)=\{0,1\}$, as such is the spectrum of any nontrivial idempotent.