Let $\mathcal{H}$ be an infinite-dimensional Hilbert space and let $S$ be any operator on $\mathcal{H}$ that has no eigenvalues [for example, take $\mathcal{H} = L^2[0, 1]$ and let $S$ be the operator on $\mathcal{H}$ defined by $(Sf)(x) = xf(x)$.] Now define an operator $T$ on $\mathcal{H} \times \mathcal{H}$ by
$$T(f, g) = (0, Sg).$$ Then $0$ is the only eigenvalue of $T$, but $T$ is not nilpotent.
If you're working strictly within the context of an inner product space or Hilbert space, the statement you have learned is false. Even if you could identify certain "states" that you would call "eigenstates," those objects are not vectors in the space. They are some unknown, unspecified object in some unknown and extended space.
For example, start with the position operator $M$ on an interval $[0,1]$. This operator is multiplication by $x$, meaning $(Mf)(x)=xf(x)$. This is a perfectly legitimate selfadjoint linear operator on the Hilbert space $L^2[0,1]$, but it has no eigenvectors, meaning that there are no non-trivial functions $f \in L^2[0,1]$ for which $Mf = \lambda f$, regardless of the choice of $\lambda$. The operator $M$ has "continuous spectrum" but no eigenvectors or eigenvalues. There just are no such "eigenstate" objects.
Physicists typically introduce the $\delta$ function into such a space (which is logically inconsistent) and claim that $(M \delta_{y})(x) = x\delta(x-y)=y\delta_{y}$. For many reasons, on many levels, that is Mathematical non-sense, and cannot be salvaged within $L^2[0,1]$ in a logically consistent way. The axioms of Quantum use Hilbert spaces, but $\delta$ functions cannot live in spaces such as $L^2[0,1]$ because two functions that are equal a.e. in $L^2[0,1]$ are identical, which means that pointwise values have no meaning for elements of $L^2[0,1]$. Dirac knew this, but he liked the intuition.
John von Neumann, who was a contemporary of P.A.M. Dirac, created consistent ways of dealing with the selfadjoint operators of Physics through the Spectral Theorem, and this was available to Dirac at the time; but Dirac chose an intuitive presentation over logical consistency. And it wasn't because rigorous Mathematics was unavailable at the time to handle everything; the correct Math was there.
All of the Hilbert spaces of Quantum must be separable, meaning that the space contains a countable dense subset. The reason for this has to do with constructibility, and being able to bootstrap to general answers through finite approximations. $L^2[0,1]$ is a separable space. One of the consequences of having a separable space, is that an orthonormal basis of such a space is always finite or countably infinite. You cannot have mutually orthogonal objects that are indexed by an interval of the real line, for example. That's impossible in separable spaces. And, if you have a selfadjoint operator $M$ on a Hilbert space, and you have $Mf=\lambda f$ and $Mg = \mu g$ for $\lambda\ne \mu$, then $(f,g)=0$ must hold. So, a selfadjoint operator on a Hilbert space that is allowed in Quantum Mechanics cannot have more than a countable number of actual eigenvalues. That's a consequence of correct axiomatic systems for Quantum. You can have continuous spectrum, but that's different than having a continuum of eigenvalues. If you enlarge the space to allow for such things, then you lose the ability to approximate in a finite way, which disconnects the theory from a setting where finite approximation makes sense.
Sorry to disappoint you. The theorem you have--as stated--is not true. Correct spectral theory gives you essentially the same thing, but not exactly that.
Best Answer
An invertible operator can't have a non zero null space since the image of the null space is zero therefore a non zero null space implies that the operator is not injective.
Consider the shift operator defined by $T(e_n)=e_{n+1}$ in $Vect(e_0,..,e_n,...)$ its image does not contains $e_0$, so it is not surjective.