The setup. Let $\mathbb{T^2}$ denote the two-dimensional torus, i.e. $$ \mathbb{T}^2 \simeq [-\pi,\pi)^2 $$ induced by identifying opposing faces of $[-\pi,\pi)^2$. Note that $$ L^2(\mathbb{T^2}) \simeq L^2_{per}([-\pi,\pi)^2).$$
The question. My question is:
Is it possible to find a basis $\lbrace e_n ~:~ n \in \mathbb{N} \rbrace$ of $L^2(\mathbb{T}^2)$ such that there are $\lambda_n \in \mathbb{C}$ satisfying $$ \Delta e_n = \lambda_n e_n ~~~\mbox{and}~~~ \lambda_n \neq 0 ~~~ \mbox{for all}~ n \in \mathbb{N}? $$
An example that does not work. From another question I know, that $$ u_{j,k}(x,y) = \exp(i (jx+ky))$$ is a basis of $L^2(\mathbb{T^2})$. This basis does not meet my requirements as $\Delta u_{0,0} =0$.
Other thoughts. For other spaces, namely $L^2([0,1])$, I know that the answer to my question would be affirmative. Here we have two bases $$ \lbrace \sqrt{2} \sin(k \pi x) ~:~ k \in \mathbb{N}\rbrace ~~~\mbox{and}~~~ \lbrace 1, \sqrt{2} \cos(k \pi x) ~:~ k \in \mathbb{N}\rbrace. $$ For the first one we the answer to my question is positive, for the last one it is negative. Yet: regardless of the Dirichlet/vonNeumann character of the bases the both span $L^2([0,1])$. This gives me hope that the answer to my original question can be positive as well.
Any help would be much appreciated!
It's not possible to find such a basis. The problem is that you have constant functions living in $L^2(\mathbb{T}^2)$, i.e. if $1 \in L^2(\mathbb{T}^2)$. Suppose, by way of contradiction, that you could find a basis satisfying your requirements. Then since $\lambda_n \neq 0$ for any $n$, we would have $$ \int_{\mathbb{T}^2} 1 e_n(x) dx = \int_{\mathbb{T}^2} \frac{\Delta e_n}{\lambda_n} = \frac{1}{\lambda_n} \int_{\mathbb{T}^2} \Delta e_n = 0,$$ where the last equality follows from the divergence theorem and periodicty. Since $\{e_n\}$ is a basis the only element orthogonal to every basis element is $0$, but this contradicts the fact that $1 \neq 0$.
Another point of view here is that your question is about the eigenvalues of the Laplacian on $\mathbb{T}^2$. One can show that they form an orthonormal basis of $L^2(\mathbb{T}^2)$, which is almost what you want. The hang up is that one of the eigenfunctions is constant. In particular, $e_0(x) = 1$ is an eigenfunction with $\lambda_0 = 0.$ The rest of the eigenvalues will be strictly positive, though.
Also, the reason you can find such a basis on $L^2([0,1])$ with Dirichlet boundary conditions is because the condition $e_n(0)=e_n(1)=0$ precludes the possibility of a constant eigenfunction.