How to find the spectrum? Let the Hilbert space $H=l_2$ over the complex field.
How to find the point spectrum $\sigma_p(A)$ of $A$
$Ax=(x_1,ix_2,-x_3,-ix_4,x_5,....)$
Any help is very appreciated. thanks :)
 A: By definition of eigenvalues if $Ax=\lambda x$ then $x_1=\lambda x_1$, $ix_2=\lambda x_2$, $-x_3=\lambda x_3$, $-ix_4=\lambda x_1$, $\dots$. If not all $x_i$ are zero the only solutions are: $\lambda=1$ and $x_1,x_5,\dots$ are non-zero;  $\lambda=i$ and $x_2,x_6,\dots$ are non-zero; $\lambda=-1$ and $x_3,x_7,\dots$ are non-zero; and $\lambda=-i$ and $x_4,x_8,\dots$ are non-zero. So $\sigma_p(A)=\{1,i,-1,-i\}$.
A: For $x=(x_{1},x_{2},x_{3},\cdots)$, define
$$
\begin{align}
    P_{1}x & =(x_{1},0,0,0,x_{5},0,0,0,x_{9},\cdots)\\
    P_{2}x & =(0,x_{2},0,0,0,x_{6},0,0,0,x_{10},\cdots)\\
    P_{3}x & =(0,0,x_{3},0,0,0,x_{7},0,0,0,x_{11},\cdots)\\
    P_{4}x & =(0,0,0,x_{4},0,0,0,x_{8},0,0,0,x_{12},\cdots).
\end{align}
$$
The $P_{n}$ are mutually disjoint, orthogonal projections that sum to $I$. And $AP_{n}=i^{n-1}P_{n}$ for $n=1,2,3,4$, which proves $\{1,i,-1,-i\}\subseteq\sigma_{P}(A)$. Clearly $A=\sum_{n=0}^{3}i^{n-1}P_{n}$ (this is a spectral decomposition.) Let $d(\lambda)=\mbox{dist}(\lambda,\{1,i,-1,-i\})$. Then
$$
\begin{align}
    \|(A-\lambda I)x\|^{2}
   & =\|\sum_{n=0}^{3}i^{n-1}P_{n}x-\lambda\sum_{n=0}^{3}P_{n}x\|^{2} \\
   & = \sum_{n=0}^{3}|i^{n-1}-\lambda|^{2}\|P_{n}x\|^{2}\ge d(\lambda)^{2}\sum_{n=0}^{3}\|P_{n}x\|^{2} = d(\lambda)^{2}\|x\|^{2}.
\end{align}
$$
So $(A-\lambda I)x=0$ and $x\ne 0$ together imply $d(\lambda)=0$, which finishes the proof that $\sigma_{P}=\{1,i,-1,-i\}$.
A: I think just intuitively that it is a diagonal matrix (infinite of course) so the spectrum ought to be $\{1,-1,i,-i\}$ Essentially it is just the same four dimensional action repeated infinitely many times. 
