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Find the spectrum of the operator $T: l^p\to l^p$, $1\leq p\leq \infty$ $$ T(x_1, x_2, x_3, x_4, \ldots )=(-x_2, x_1, -x_4, x_3, \ldots). $$

My attempt

I have that $Tx = \lambda x$ $$ \implies \lambda (x_1, x_2, x_3, x_4, \ldots )=(-x_2, x_1, -x_4, x_3, \ldots)$$ $$ \implies \lambda (x_1, x_2, x_3, x_4, \ldots ) - (-x_2, x_1, -x_4, x_3, \ldots) = 0$$ $$ \implies (\lambda x_1 + x_2, \lambda x_2 - x_1, \lambda x_3 + x_4, \lambda x_4 - x_3 , \ldots ) = 0$$ $$ \implies x_1 = \lambda x_2, x_2 = - \lambda x_1$$ and similarly for $x_3, x_4 \cdots \cdots$. we then have $$ \implies \lambda = +i, -i.$$ Then what is the spectrum? Does it only consist of the eigen spectrum?

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  • $\begingroup$ Are you assuming that every element of the spectrum is an eigenvalue, or have you already proved it for this operator? $\lambda$ being an eigenvalue is a sufficient but not necessary condition for $\lambda$ being in the spectrum of an operator on $\ell^p$. $\endgroup$ – Jonas Meyer Apr 20 '17 at 18:10
  • $\begingroup$ Please don't remove your question by editing it out. To delete the question would require intervention by moderators or closing and deleting by high rep users, but there is no apparent reason to do so. $\endgroup$ – Jonas Meyer Apr 27 '17 at 20:04
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You have $$x_{2k+1} =\lambda x_{2k+2} = -\lambda^2 x_{2k+1}\implies x_{2k+1} = 0\text{ or }\lambda^2 = -1.$$ I.e., $\lambda = \pm i.$ If the base field is $\Bbb C$ then the spectrum is $\{i,-i\}$. If the base field is $\Bbb R$ then the spectrum is $\emptyset$.

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Using this method you will find the eigenvalues, $\pm i$ as Martín-Blas Pérez Pinilla says, which are therefore in the spectrum. However, you have given no reason why the spectrum of this operator consists only of eigenvalues. In general, the spectrum of an operator on $\ell^p$ can contain many elements that are not eigenvalues. For example, it might be an entire disk of points, none of which is an eigenvalue.

In this case, your operator is essentially an infinite direct sum of copies of the map $\begin{bmatrix}x\\y\end{bmatrix}\mapsto A \begin{bmatrix}x\\y\end{bmatrix}= \begin{bmatrix}0&-1\\1&0\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}$ on $\mathbb C^2$, whose spectrum consists only of $\pm i$. If $\lambda\not\in\{i,-i\},$ then $A-\lambda I_2$ is invertible, and $(T-\lambda I)^{-1}$ is an infinite direct sum of copies of the map $(A-\lambda I_2)^{-1}$.

(If this were on $\ell^2$ you could note that $T$ is a skew-adjoint unitary operator, hence $\pm i$ are automatically the only possible elements of the spectrum.)

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  • $\begingroup$ @manhattan: That is why I posted this answer. It is about what should be done, after the warning in the first paragraph that it wasn't done yet. $\endgroup$ – Jonas Meyer Apr 20 '17 at 18:34
  • $\begingroup$ It is the set of $\lambda\in \mathbb C$ such that $T-\lambda I$ is not invertible. Can you see what specific set it is for this $T$? $\endgroup$ – Jonas Meyer Apr 20 '17 at 18:39
  • $\begingroup$ Do you understand the second paragraph of this answer? That is what I have to offer you regarding showing exactly what the set is. By the way, it might be easier to think of the infinite direct sums in terms of block matrices, just a bunch of copies of the same $2$-by-$2$ matrix down the diagonal. $\endgroup$ – Jonas Meyer Apr 20 '17 at 18:43
  • $\begingroup$ @manhattan: Yes, a.k.a. $\{i,-i\}$. $\endgroup$ – Jonas Meyer Apr 20 '17 at 18:53

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