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

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.

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 $$S_r$$.)

• Could you clarify more on why the spectrum of $T$ is a closed disk of radius $2$ centered at $-3$? I don't exactly understand and I couldn't find the spectral mapping theorem you talked about. I also still don't see what is $\sigma _{p}(T)$ Commented Dec 29, 2020 at 21:32
• @GabiG I've modified the answer to include more details. Commented Dec 30, 2020 at 6:33
• Thanks! Its much clearer right now. The only thing that I still don't exactly get is why $\sigma _r (A) \subset \sigma _p (A^*)$. I see why if $\lambda \in \sigma _r (A)$ then $\bar \lambda \in \sigma _p (A^*)$, but why is $\lambda \in \sigma _p (A^*)$ ? Commented Dec 30, 2020 at 10:48
• @GabiG That depends on how the adjoint $A^*$ is defined, whether in the Banach space sense $A^*f=f\circ A$ or as the Hermitian adjoint for Hilbert spaces. In this case, since $p$ is not necessarily $2$, I took the former. So it should be $\lambda\in\sigma_p(A^*)$. But not every author takes the same convention. In any case, here it does not matter since $\sigma(T)$ is symmetric about the real line. Commented Dec 30, 2020 at 11:26