Rudin Functional Analysis, Chapter $10$, exercise $12$ Here is the exercise:
$12$. Find the spectrum of the operator $T\in {\mathcal B}(\ell^2)$
given by
$$T(x_1, x_2, x_3, x_4,...)=(-x_2, x_1, -x_4, x_3,...).$$
I am confused by ${\mathcal B}(\ell^2)$. What does ${\mathcal B}(\ell^2)$ mean? Bounded and linear functions
on $\ell^2$, right? I know this, but what does exactly $\ell^2$
mean?
Thank you for your help.
UPDATE $1$: based on the comments and helps, this is my answer. Please tell me if it is true or not:
We should find every $\lambda \in {\mathbb C}$ such that
$\lambda I-T$ is not invertible in ${\mathcal B}(\ell^2)$.
We can describe the operator $\lambda I-T$ as follows:
\begin{eqnarray*}
(\lambda I-T)(x)&=&\lambda x-Tx=(\lambda x_1, \lambda x_2,
...)-(-x_2, x_1, -x_4, x_3,...)\\
&=&(\lambda x_1+x_2, \lambda x_2-x_1, \lambda x_3+x_4, \lambda
x_4-x_3,...)
\end{eqnarray*}
It is obvious that this operator is a member of ${\mathcal
B}(\ell^2)$. Because $R(T)\subset\ell^2$. However I guess its
inverse
$$(\lambda I-T)^{-1}(x)=\big({\lambda x_1-x_2\over\lambda^2+1},{\lambda x_2+x_1\over\lambda^2+1},...\big)\ \ \ \ \ \ \ \ (*)$$
for $\lambda\not=0$ is not a member of ${\mathcal B}(\ell^2)$.
Because
$$\big|{\lambda x_1-x_2\over\lambda^2+1}\big|^2+\big|{\lambda x_2+x_1\over\lambda^2+1}\big|^2+...\nless \infty$$
which means $(\lambda I-T)(x)$ for $\lambda\not=0$ does not have inverse in ${\mathcal B}(\ell^2)$.
So $\sigma(T)={\mathbb C}-\{0\}$. Is it correct?
UPDATE $2$: Thanks for all for their helpful comments, I realized my
mistake in the UPDATE $1$. As @Mindlack and @Martin Argerami have pointed out and $(*)$ shows,
$(\lambda I-T)^{-1}$ as a member of ${\mathcal B}(\ell^2)$ is
meaningful for any $\lambda$, unless $\lambda=\pm i$, the roots of
$\lambda^2+1=0$.
So $\sigma(T)=\{+i, -i\}$.
 A: The space $\ell^2$ is defined as :
$$\ell^2 = \bigg\{ x = (x_n)_{n \geq 1} : \sum_{n=1}^\infty |x_n|^2 < \infty\bigg\}$$
In word, it is oftenly called the set of all the sequences that are square summable. The standard norm of $\ell^2$ is defined as :
$$\|x\|_2 = \sqrt{\sum_{n=1}^\infty |x_n|^2}$$
Finally, truly $\mathcal{B}(\ell^2)$ is the set of all the bounded linear operators from $\ell^2$ to $\ell^2$.
A: $\ell^2$ is the space of all sequences $a=(a_n)_{n \geq 1}$ such that $\|a\|_2^2 := \sum_{i \geq 1}{|a_i|^2} < \infty$. 
And I would say that your guess for $\mathcal{B}(\ell^2)$ is right.
As for the solution: note that $T^2=-Id$, so $\ell^2=Ker(T-iId) \oplus Ker (T+iId)$, and both spaces are nonzero (you can check it), and if $\lambda$ is a complex number not $\pm i$, then $-(T-\lambda)(T+\lambda)=(\lambda^2+1)Id$ is invertible, hence so is $T-\lambda$. 
A: "Bounded" means that $\|Tx\|\leq c\|x\|$ for some number $c$ and all $x$. It is not enough that $R(T)\subset\ell^2$ to say that $T$ is bounded. 
As for the spectrum, it is always compact, so it is always bounded (provided that your space is complete, which it is in this case). 
You say correctly that the inverse of $\lambda I-T$ is 
$$\left({\lambda x_1-x_2\over\lambda^2+1},{\lambda x_2+x_1\over\lambda^2+1},...\right)
=\left( \tfrac\lambda{\lambda^2+1}\,I+T\right)x.
$$
The bounded operators are a Banach space, which means that linear combinations (and limits, but we don't need that here) are again bounded. So the only obstruction for the inverse to exist, is that $\lambda^2+1=0$. That's the case when $\lambda=\pm i$. So 
$$
\sigma(T)=\{i,-i\}. 
$$
