# Does a suitable famlly of eigenvectors of non self-adjoint operators, sufficiently close to an adjoint one, form a basis?

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.

• I think that this old post should contains the answer (which is positive) mathoverflow.net/questions/348931/…. However I do not know any simple perturbation argument yielding the result and I would be very interested at it. Apr 29, 2021 at 7:11
• By the way, if $A, C$ are constant matrices, then the proof is simple. Apr 29, 2021 at 8:19
• @GiorgioMetafune Is it? That's a good news. Yes, the question is for constant coefficients, but you are right, the wider question is the non constant case, hopefully as a perturbation of the constant case first, for smooth coefficients, and then the rough coefficient case. But first things first: why is the constant case simple? Apr 29, 2021 at 8:27
• Consider $A=I$ (otherwise you make a linear change of variables) and, more generally, a first order term $\nabla \phi \cdot \nabla u$. Then you can write the operator as $e^{-\phi} div (e^{\phi} \nabla u)$ which is self-adjoint with respect to the (equivalent) measure $e^{\phi}\, dx$. Apr 29, 2021 at 8:32
• @GiorgioMetafune Thank you! I referred to your comment in the post, and stated the question in general. I also proposed a "next case" problem, but yes, I would like to know if such a perturbation result exists, which would avoid to consider a catalog of cases one by one. Apr 29, 2021 at 9:24

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. Just rewrite the operator $$A:D^2$$ as $${\rm div}(AD)+C\cdot D$$ where $$-C$$ is the row-wise divergence of $$A$$.