Integral Operator in $L^2$ I was trying to do this exercise and I'm wondering if I figured it out well:
I have $\mathcal{H} := L^2(0,1)$ and $T$ the operator with integral kernel $K(x,y) = \min\{x,y\}$, $x,y \in [0,1]$. I have to show that $T$ is compact and self-adjoint.
To show that is compact I was thinking to say that because $\min\{x,y\} \in [0,1]$ then
\begin{equation}
\dim(\operatorname{Im}T) = 1
\end{equation}
(The self adjointness I think is trivial..)So T belongs to finite rank operators and so it is compact. (Is this correct?) Then it asks me to find eigenvalues and eigenvectors of $T$ and here I really don't know how to proceed...

We have
$$
\int_{(0,1)^2} |k(x,y)|^2\ \mathsf d(x\times y) = \int_0^1\int_0^1 (x\wedge y)^2\ \mathsf dx\ \mathsf dy 
\leqslant \int_0^1\int_0^1\ \mathsf dx\ \mathsf dy = 1 <\infty,
$$
so T is a Hilbert-Schmidt operator and hence is compact.
