Yes. There is a compact operator with spectrum is the the singleton set {0}. I think it is the Volterra integral operator. Refer Limaye: Functional Analysis book.

A quasinilpotent would do.

Yes.

Example: The Volterra Operator

Consider the Hilbert space H=L^2([0,1]), the space of square-integrable functions on the interval [0,1]. Define the Volterra operator T:L^2([0,1])→L^2([0,1]) as

(Tf)(x)=∫f(t) dt (integration between 0 to x) for all f∈L^2([0,1]).

The Volterra operator T is not normal. To see why, recall that T is compact and strictly triangular (its matrix representation relative to a suitable orthonormal basis is upper triangular with zeros on the diagonal), so it cannot be normal unless it is zero.

Since T is compact, its spectrum σ(T) consists of 0 together with the set of eigenvalues of T. However, since the Volterra operator has no eigenvalues (since it is injective but not surjective), the spectrum of T is just {0}. Technically, 0 is a boundary point of the spectrum, so if we want the spectrum to be contained in the set of positive real numbers (excluding zero), we might need to adjust the example. But since 0 is an isolated point in the spectrum, some interpretations of "spectrum in positive reals" might include 0.

If you want the spectrum to be a subset of the positive real numbers and avoid 0, consider the shifted Volterra operator T+λI, where λ>0. The spectrum of T is {0}, so the spectrum of T+λI is shifted to {λ}. This shifted operator is still not normal, but its spectrum is now entirely contained in the positive real numbers.

Why go into all this rigmarole: if a Hilbert space quasinilpotent is normal, then it is the zero operator. Indeed, any non-normal nilpotent (i.e. not equal to 0 operator) is an example.

Pick any pair of vectors f, g that linearly independent and such that =1 (for instance, f, h that are not multiples of each other with =z not zero, and alter g=1/z h). Consider the rank one operator T=< , f>g (i.e., Th=g).

Then it is easy that T^2=T, so that the spectrum of T is { 0 ; 1 }. But T is not normal (an idempotent that is normal is an orthogonal projection, or compute):

T*=< , g>f, T*T=< . f>f, T*T=< , g>g.