Spectrum of $ T(x_1, x_2, x_3, x_4, \ldots )=(-x_2, x_1, -x_4, x_3, \ldots). $ Find the spectrum of the operator $T: l^p\to l^p$, $1\leq p\leq \infty$
$$
T(x_1, x_2, x_3, x_4, \ldots )=(-x_2, x_1, -x_4, x_3, \ldots).
$$
My attempt
I have that $Tx = \lambda x$
$$ \implies \lambda (x_1, x_2, x_3, x_4, \ldots )=(-x_2, x_1, -x_4, x_3, \ldots)$$
$$ \implies \lambda (x_1, x_2, x_3, x_4, \ldots ) - (-x_2, x_1, -x_4, x_3, \ldots) = 0$$
$$ \implies   (\lambda x_1 + x_2, \lambda x_2 - x_1, \lambda x_3 + x_4, \lambda x_4  - x_3 , \ldots ) = 0$$
 $$ \implies x_1 = \lambda x_2, x_2 = - \lambda x_1$$ and similarly for $x_3, x_4 \cdots \cdots$. we then have
$$ \implies  \lambda = +i, -i.$$
Then what is the spectrum? Does it only consist of the eigen spectrum?
 A: You have
$$x_{2k+1} =\lambda x_{2k+2} = -\lambda^2 x_{2k+1}\implies x_{2k+1} = 0\text{ or }\lambda^2 = -1.$$
I.e., $\lambda = \pm i.$
If the base field is $\Bbb C$ then the spectrum is $\{i,-i\}$. If the base field is $\Bbb R$ then the spectrum is $\emptyset$.
A: Using this method you will find the eigenvalues, $\pm i$ as Martín-Blas Pérez Pinilla says, which are therefore in the spectrum.  However, you have given no reason why the spectrum of this operator consists only of eigenvalues.  In general, the spectrum of an operator on $\ell^p$ can contain many elements that are not eigenvalues.  For example, it might be an entire disk of points, none of which is an eigenvalue.  
In this case, your operator is essentially an infinite direct sum of copies of the map $\begin{bmatrix}x\\y\end{bmatrix}\mapsto A \begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}0&-1\\1&0\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}$ on $\mathbb C^2$, whose spectrum consists only of $\pm i$.  If $\lambda\not\in\{i,-i\},$ then $A-\lambda I_2$ is invertible, and $(T-\lambda I)^{-1}$ is an infinite direct sum of copies of the map $(A-\lambda I_2)^{-1}$.
(If this were on $\ell^2$ you could note that $T$ is a skew-adjoint unitary operator, hence $\pm i$ are automatically the only possible elements of the spectrum.)
