Find and classify the spectrum of the operator: $T(x_1, x_2, x_3 \ldots) = (2x_2 - 3x_1, 2x_3 - 3x_2, 2x_4 - 3x_3, \ldots)$

Question

Let $T:l_p \to l_p$ be an operator defined as: $$T(x_1, x_2, x_3 \ldots) = (2x_2 - 3x_1, 2x_3 - 3x_2, 2x_4 - 3x_3, \ldots)$$

I need to find and classify its spectrum.

First, I noticed that $T = 2S_l - 3I$, when $S_l$ is the shift left operator.

Using this I can say that $||T|| \le 2||S_l|| + 3||I|| = 2 + 3 = 5$. Which means that for $| \lambda | > 5$, $\lambda$ is a regular point.

Then, I tried to find its point spectrum. If $Tx = \lambda x$, then: $$(2x_2 - 3x_1, 2x_3 - 3x_2, 2x_4 - 3x_3, \ldots) = (\lambda x_1, \lambda x_2, \lambda x_3, \ldots)$$ which means that: $x_2 = \frac{3 + \lambda}{2} x_1, x_3 = \frac{3 + \lambda}{2} x_2, \ldots$

This gives $x = (\frac{3 + \lambda}{2} x_1, (\frac{3 + \lambda}{2})^2 x_1, (\frac{3 + \lambda}{2})^3 x_1, \ldots)$

This vector is in $l_p$ if $\frac{|3+\lambda|}{2} < 1$.

From here, I am not sure how to continue, and I still need to find the continuous spectrum and residual spectrum of $T$.

Help would be appreciated.

Answer 1

It's easier to concentrate on $S_l$ first.

You've found that it has eigenvectors $(1,\lambda,\lambda^2,\ldots)$ with eigenvalue $\lambda$, for any $|\lambda|<1$. Moreover its norm is $1$, so any $|\lambda|>1$ is a regular point.

This means that its point spectrum $\sigma_p(S_l)$ is the open unit ball, while $\sigma(S_l)$ is a subset of the closed unit ball. Moreover, the whole spectrum $\sigma(S_l)$ is a closed set bounded by the norm $\|S_l\|=1$. This implies that $\sigma(S_l)$ is precisely the closed unit ball.

By the spectral mapping theorem, the spectrum of $T=2S_l-3I$ is "$\sigma(T)=2\sigma(S_l)-3$", that is, it is the closed disk of radius $2$ centered at $-3$.

To classify the spectrum requires one more step. The adjoint of $S_l$ is $S_r$ so $$T^*=2S_r-3I$$ Repeating your exercise for the right-shift operator shows that it has no eigenvectors at all.
Now for any operator $A$, $$\sigma_r(A)\subseteq\sigma_p(A^*)$$ For $S=S_l$, this gives $\sigma_r(S_l)=\emptyset$. Hence the remainder of the spectrum of $S_l$ must be the continuous spectrum.
These results pass on to $T$ since it is just a translation of $S_l$, i.e., $\sigma_r(T)\subseteq\sigma_p(T^*)=\sigma_p(2S_r-3I)=\emptyset$. (If $\lambda\in\sigma_p(2S_r-3I)$ then $2S_r-(3+\lambda)I$ is not injective, so $(3+\lambda)/2$ would be an eigenvalue of $S_r$.)