Spectrum of a linear operator

Question

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?

Answer 1

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}

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

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

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(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, $\sigma_c(A) = \varnothing$