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.



Does a suitable famlly of eigenvectors of non self-adjoint operators, sufficiently close to an adjoint one, form a basis?
In one space dimension, the answer is yes, and the eigenvalues are real and simple (Sturm-Liouville theory).


In higher space dimension, the answer is negative, because the operator needs not be diagonalisable. If you let the data $(A,C)$ depend upon a parameter, you can pass from a situation where all the eigenvalues are real to one where there is a pair of complex conjugate eigenvalues. At the transition point, the operator is not semi-simple, admitting a double eigenvalue whose eigenspace is only one-dimensional.


An explicit construction is given in this [answer](https://mathoverflow.net/q/376056). Just rewrite the operator $A:D^2$ as ${\rm div}(AD)+C\cdot D$ where $-C$ is the row-wise divergence of $A$.


