# Normal compact operator commute with bounded self adjoint operator in Hilbert space.

Suppose $H$ is a Hilbert space and $A:H\rightarrow H$ is a normal compact operator such that $\ker(A)=0$. show that if $B$ is a bounded self adjoint operator that commutes with $A$ then the spaces in $\{\ker(B-\lambda Id):\lambda\in \mathbb{R}\}$ generate $H$.

## 3 Answers

Hint: the eigenspaces of $A$ are finite-dimensional and invariant under $B$, and their sum is dense in $H$.

Since apparently I have nothing better to do, I'll flesh out the hint for OP.

By assumption, $A$ has pairwise-orthogonal spectral projections $\{P_i\}$ such that

$$A = \sum_i \sigma_i P_i \;\; \mbox{in operator norm},$$

$$\forall i \; ran(P_i) < \infty ,$$ and, by injectivity of $A$,

$$I = \sum_i P_i \;\; \mbox{in SOT}.$$

Because $AB = BA$, we have (by continuous functional calculus, if you'd like) $BP_i = P_i B$ for all $i$. So apply the finite-dimensional spectral theorem to each $P_i B P_i$ and you're done.

A normal compact operator $A$ is a diagonalisable one, and so $ker(A)=0$ automatically. so the assumption $ker(A)=0$ must be omitted from the question.