Spectrum of a linear operator Bounded linear operator is given by $A(x_1,x_2,\dots) = (x_3,x_2,x_1,x_6,x_5,x_4,\dots)$
Compute spectrum,point spectrum and residual spectrum.
So I wrote down $A$ in matrix form and turned out that $A=A^*$. So we know that if $A=A^*$ then residual sperctum is an empty set,also we know that if $A=A^*$ the spectrum of $A$ is an improper subset of $[-||A||,||A||]$ so in our case this will be $[-1,1]$. I also did calculations and found out that continuous spectrum is empty set. How to proceed next? I think that since eigenvalues are $-1,1$ spectrum will be exactly set containing both of them that is $\{-1,1\}$. However how do I show it?
 A: It remains to find the point spectrum.  Suppose that $\lambda \in \sigma_p(A)$, so $(\lambda I - A)x = 0$ for some $x$.  Then
\begin{cases}
x_3 = \lambda x_1 \\
x_2 = \lambda x_2 \\
x_1 = \lambda x_3
\end{cases}


*

*$x_2 = 0$, so we can't have $x_3 = 0$.  (Otherwise $x_1 = 0$)  We have $x_1 = \lambda^2 x_1$, so $\lambda^2 = 1 \iff \lambda = \pm1$.

*$x_2 \ne 0$, so $\lambda = 1$.


Therefore, $\sigma_p(A) = \{-1,1\}$.

(Edited in response to OP's comment.)
When $\lambda \ne \pm1$, $\lambda^2 \ne 1$, so the matrix
$$ B =
\begin{bmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{bmatrix}
- \lambda I = 
\begin{bmatrix}
-\lambda & 0 & 1 \\
0 & 1-\lambda & 0 \\
1 & 0 & -\lambda
\end{bmatrix}
$$
has nonzero determinant $(1-\lambda)(\lambda^2-1)$.
\begin{align}
B^{-1} &= \begin{bmatrix}
-\lambda(\lambda^2-1)^{-1} & 0 & -(\lambda^2-1)^{-1} \\
0 & (1-\lambda)^{-1} & 0 \\
-(\lambda^2-1)^{-1} & 0 & -\lambda(\lambda^2-1)^{-1}
\end{bmatrix} \\
&= \begin{bmatrix}
\lambda(1-\lambda^2)^{-1} & 0 & (1-\lambda^2)^{-1} \\
0 & (1-\lambda)^{-1} & 0 \\
(1-\lambda^2)^{-1} & 0 & \lambda(1-\lambda^2)^{-1}
\end{bmatrix}
\end{align}
From this, we observe that
$$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \end{bmatrix} \mapsto
\begin{bmatrix} \lambda(1-\lambda^2)^{-1} x_1 + (1-\lambda^2)^{-1} x_3 \\ (1-\lambda)^{-1} x_2 \\ (1-\lambda^2)^{-1} x_1 + \lambda(1-\lambda^2)^{-1} x_3 \\ \vdots \end{bmatrix}$$
gives the inverse of $A-\lambda I$.
Therefore, $\sigma_c(A) = \varnothing$
