Originally Posted by

**davesface** I think you're right.

The solution to that ODE is $\displaystyle f(x)=e^{\lambda x}+c, c \in \mathbb{R}$, which is the general form of the eigenvectors of D. From this, and knowing that $\displaystyle \lambda$ must be a scalar in the space W, I would make the claim that the real numbers constitute all of the eigenvalues.

If I'm correct in both of the above claims, then it would seem as though it would be impossible to find a finite dimensional subspace of W on which D is diagonalizable because the subspace couldn't be both finite dimensional and closed under addition. How to show this though...I have no idea.