Computing spectrum of operator via matrix

Question

I saw this question online and didn't understand the given solution.

Question: Define $T : L^2(\mathbb{T}) \to L^2(\mathbb{T}) : u(x) \mapsto \displaystyle\int_{-\pi}^\pi \sin(x-y) u(y) \; dy$, where $\mathbb{T} = \mathbb{R} \setminus (2 \pi \mathbb{Z} ) $ is the torus (we identify it with $[-\pi,\pi]$ here). Compute the spectrum $\sigma_p(T), \ \sigma_c(T), \ \sigma_r(T)$.

Solution: Note $T$ is compact since it's a finite rank operator $$ [Tu](x) = \sin x \int_{-\pi}^{\pi} \cos(y) u(y) \; dy - \cos(x) \int_{-\pi}^{\pi} \sin(y) u(y) \; dy $$

Next, they define orthonormal unit vectors $v_1(x) = \dfrac{ \sin x }{\sqrt{\pi}}$ and $v_2(x) = \dfrac{ \cos(x)}{\sqrt{\pi}}$ so that $T u = \pi v_1 \langle v_2, u \rangle - \pi v_2 \langle v_1, u \rangle$. Now here's where I start getting confused. (More or less verbatim from here on).

Set $G = \text{span}\{v_1,v_2\}$ and note $G, G^\perp$ are invariant subspaces of $T$. Then the restriction of $T$ to $G$ has the matrix $T = \begin{pmatrix} 0 & \pi \\ -\pi & 0 \end{pmatrix}$, and $T$ has eigenvalues $\pm i \pi$. The restriction of $T$ to $G^\perp$ is zero. Therefore $\sigma(T) = \sigma_p(T) = \{0,i\pi,-i\pi\}$.

Remarks: I have never seen spectrum calculated this way before . . . My textbook certainly doesn't cover it. I don't see where the invariant subspaces come into play, nor the matrix $T$.

I would think $T$ is $ \begin{pmatrix} \pi & 0 \\ 0 &-\pi \end{pmatrix}$ since then $\begin{pmatrix} v_1 \\ v_2 \end{pmatrix} \mapsto \begin{pmatrix} \pi v_1 \\ - \pi v_2 \end{pmatrix}$ which seems closer to their decomposition of $T$.

I believe that since $T$ is normal, we know $\sigma_r(T)$ is empty. $0$ is in the spectrum since $T$ is compact, but how do they immediately conclude $0 \notin \sigma_c(T)$?

If you could explain this problem and this technique to me, that would be great. Thank you.

Answer 1

We have $T v_1 = -\pi v_2$ and $T v_2 = \pi v_1$ so the matrix representation is: $$ T [v_1 \ \ v_2]= [v_1 \ \ v_2] \left( \begin{matrix} 0 & \pi \\ -\pi & 0 \end{matrix} \right)$$

$G$ is two dimension and invariant so can only consists of eigenvalues, which is then simply the eigenvalues of the matrix. $G^\perp$ is closed and it maps to zero. So it also consists of point spectrum only (every vector in it is an eigenvector of eigenvalue zero). The full space is a direct sum of the two subspaces so our spectral description is complete. In the present exercise you don't have to worry about residual and continuous spectrum.

Answer 2

You can verify directly that $Tv_1=\pi v_2$ and $Tv_2=-\pi v_1$. Thus indeed, the restriction of $T$ to the subspace $G$ has that matrix. Since the matrix has two distinct eigenvalues, these certainly belong to the point spectrum. You can calculate the corresponding eigenvalues of this matrix and then consider them as a combination of $v_1$ and $v_2$, these will be the corresponding eigenvectors (not $v_1$ and $v_2$ themselves as you seem to suggest).

The next key step is figure out why $T$ is zero on $G^{\perp}$. Notice that $G^{\perp}$ is a huge space so certainly there is a nonzero-vector $v_3\in G^{\perp}$ such that $Tv_3=0$. Hence $0$ also belongs to the point spectrum. Thus $\left\{0,i\pi,-i\pi\right\}\subset \sigma_p(T)$. Since $G\oplus G^{\perp}=L^2(\mathbb{T})$, there are no other eigenvalues or elements in the point spectrum, thus $\sigma_p(T)=\left\{0,i\pi,-i\pi\right\}$. It remains to figure out why there is no continuous and residual spectrum.