I need to calculate the spectrum of the operator $T$ for $f \in L^2([0,1])$ defined by:
\begin{equation}
(Tf)(x) = \int_0^1 (x+y)f(y)dy.
\end{equation}
I know that $T$ is compact and self-adjoint so the residual spectrum is empty and the eigenvalues are real and a closed subset of $[-||T||, ||T||]$.
So I let $\lambda$ be an eigenvalue so I know that $\lambda f(x) = (Tf)(x)$. By differentiating twice, I found that $\lambda f''(x) = 0$ but I don't really know how I can continue.
Best Answer
Notice that the range of $T$ is the span of $\{1,x\}$, that is is the space of linear functions on $[0,1]$. So if $f(y)=ay+b$ is an eigenvector of $T$ corresponding to $\lambda\in\mathbb{C}\setminus\{0\}$, $$ (\frac12 a+ b)x + (\frac13 a + \frac12b) = \lambda a x + \lambda b $$ Thus $$ \begin{pmatrix} \frac12-\lambda & 1\\ \frac13 & \frac12-\lambda \end{pmatrix}\begin{pmatrix} a\\ b \end{pmatrix}=\begin{pmatrix} 0\\0\end{pmatrix} $$