I am currently reading the book Spectral Methods in Automorphic Forms, and Iwaniec defines the trace operator in a different way than I am accustomed to. Throughout, assume that everything converges spectacularly - that's not important here.
In particular, if $K: F \times F \longrightarrow \mathbb{C}$ is a $C_0^\infty$ (that is, smooth and bounded) function and $L$ is the integral operator having $K$ as its kernel, i.e. $$ (Lf)(z) = \int_F K(z,w)f(w) d w,$$ then Iwaniec defines the trace of $L$ as the integral across the diagonal, $$ \text{Tr} L = \int_F K(z,z)dz. \tag{1}$$
I'm familiar with the trace of a more generic (linear operator $A$ over a Hilbert space by
$$ \text{Tr} A = \sum_j \langle Ae_j, e_j \rangle,\tag{2}$$
where the $e_j$ form an orthonormal basis of functions. Do these definitions agree? If we suppose in addition that the $e_j$ are eigenfunctions with eigenvalues $\lambda_j$, then I can see the equivalence in the following "wrong" way. Taking the spectral decomposition for $K(z,w)$, $$K(z,w) = \sum_j \lambda_j e_j(z) \overline{e_j(w)},$$ then since the $e_j$ are orthonormal, we have that $$\int_F K(z,z)dz = \sum_j \lambda_j \int_F e_j(z)\overline{e_j(z)}dz = \sum_j \lambda_j.$$ And Lidskii's Theorem says that $$\text{Tr} A = \sum_j \lambda_j,$$ where $\text{Tr} A$ is as in $(2)$. So I can conclude that $(1)$ and $(2)$ should agree, but I would like to see in a more fundamental, less roundabout way that they do actually agree.