Given a self adjoint operator $T$ in a separable Hilbert Space $H$ and a vector $v \in H$, how can we find a sequence of finite rank operators, say $\{F_k\}$, such that $H + F_k$ has a finite dimensional reducing subspace $M_k$ for all $k \in \mathbb{N}$ and $dist[v, M_k] \rightarrow 0$? (Edit: Not yet answered)
This question is from *Banach Algebra Technique in Operator Algebras written by Ronald Douglas, Problem 5.17. Thanks for Martin's remainder, I should state the source of this question:
This question is for proving that, in a separable Hilbert Space $H$, given a self-adjoint operator $T$ there exists a compact self-adjoint operator $K$ such that $H+K$ has an orthonormal basis consisting of eigenvectors, namely diagonizable. Assume $\{f_i\}_{i \geq 1}$ is a orthonormal basis of $Ker[T]^{\perp}$ and $\{e_j\}_{j \geq 1}$ a orthonormal basis of $Ker[T]$. I have no control on $T f_j$ so can not proceed.
(Edit:Answered)In addition to the question above, for a general self-adjoint operator $T$ in a Hilbert Space $K$ (not necessarily separable), is there a sufficient condition (ideally also necessary) for $T$ to has a invariant subspace which is a proper subspace of $Ker[T]^{\perp}$? And is there such a condition for the same $T$ NOT TO have a invariant subspace that is a proper subspace of $Ker[T]^{\perp}$ (like $T f_j$ is not perpendicular to any vectors from any basis of $Ker[T]^{\perp})$?
What you are trying to prove is the Weyl-von Neumann Theorem. I don't think there's a straightforward proof for it.
As for your second question, the necessary and sufficient condition is that $T$ is not rank-one. By the Spectral Theorem, you can see $T$ as a multiplication operator $M_g\in L^2(X)$ for some measure space $X$. If $T$ is not rank-one, then $g$ is not supported on an atom; letting $E\subset \{g\ne0\}$ be a proper subset, we get that $1_E\,L^2(X)$ is a proper invariant subspace for $M_g$.