It is very well known that if $A\in L^\infty(B_1;\mathbb R ^{d\times d})$ is a positive definite symmetric matrix, the eigenvalue of the self adjoint operator $H^2(B_1)\cap H^1_0(B_1)\to L^2(B_1)$ $$T:u\to\text{div}(A Du)$$ are all real, and the corresponding eigenvectors can be chosen to form an orthonormal eigenbasis of $L^2(\Omega)$ and an $A$-orthogonal basis of $H^1_0(\Omega)$.

My question is whether it is known if this extends to the case when the operator is perturbed by a lower order term, namely $$T^\prime:u\to\text{div}(A Du)+C\cdot Du$$ with $C\in L^\infty(\Omega;\mathbb R^d)$? It could have complex eigenvalues and eigenvectors, but nevertheless would the span of the real and imaginary part of such vectors still be a generating family?

Giorgio Metafune has pointed out that the question is answered positively when $A$ and $C$ are constant, which was the formulation of this question before this edit. In that case, the eigenvalues and eigenvectors of $T^\prime$ are connected to that of $T$ thanks to a transformation. Indeed, note that $$ \text{div}(A Du)+C\cdot Du = \lambda u \Leftrightarrow \text{div}(A Dv)= \lambda v $$ where $v=u\exp(-\frac12 A^{-1}C \cdot x)$. It therefore immediately follows that the answer is positive. The very same argument carries over (as Giorgio Metafune pointed out) when $A$ and $C$ are non constant, but $$ A^{-1}C =D\phi $$ for some function $\phi$. The question remains in general. Decomposing the problem by this method, I suppose the next simple case is $$ T^\prime : u\to \Delta u+ \phi \cdot Du $$ where $\text{div} \phi =0$. But if a "lower order terms therefore perturbation therefore yes" argument exist, it will avoid this case by case dissection.