I have a self-adjoint operator $d$ which acts on vector fields defined on $\mathbb{R}^n$. I am interested on its eigenvalues. That is, I study the equation $d(X)-\lambda X=0$.

I have found that if $\lambda\neq 0$ then $X=0$ is the only solution of $d(X)-\lambda X=0$. This would mean that if $\lambda$ is an eigenvalue, then it must be zero. However, I have also shown that the kernel of $d$ is trivial. This then implies, i think, that $d$ has no eigenvalues! Is this really possible? or did something go wrong?


Edit: It is a bit difficult to write the specific operator $d$ in a brief way, but let me try for $\mathbb{R}^3$. Let $S$ the vector field $S=z\partial_x-(y^2+x)\partial_y+0\partial_z$. We define the operator $f$ (acting on polynomial vector fields of degree $k$) as $f(U)=[S,U]$. Note that $f$ is not self adjoint as it maps polynomial vector fields of degree $k$ to polynomial vector fields of degree $k+1$ . Let $f^*$ be the adjoint of $f$ (there is a well defined inner product in the space of polynomial vector fields that allows us to do this). Then $d=ff^*$.

As a side note, I've seen here that integral operators may have no eigenvalues.

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    $\begingroup$ Could you please give $d$ explicitely? It shouldn't be possible. $\endgroup$ – xavierm02 Sep 4 '14 at 9:35

In general, it is fully acceptable for a bounded self-adjoint operator to not have any eigenvalues. Consider for instance the operater $M\in L^2([0,1])$ defined on a representative by $$ (Mf)(x)=xf(x). $$ It is not hard to see that the spectrum of this operator is $[0,1]$ (because it is a multiplication operator corresponding to a real continuous function with range $[0,1]$), while the eigenvalue equation $Mf=\lambda f$ implies $f=0$ almost everywhere.


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