Normal $T\in B(H)$ has a nontrivial invariant subspace I am wondering if the following is true:

Every normal $T\in B(H)$ has a nontrivial invariant subspace if $\dim(H)>1$?


Let $ T \in B(\mathcal{H}) $ be a normal operator. Let $ \sigma(T) $ denote the spectrum of $ T $. We then have two cases to consider: (i) $ \sigma(T) $ is a singleton set, and (ii) $ \sigma(T) $ contains at least two points.

Case (i): Suppose that $ \sigma(T) = \{ \lambda \} $ for some $ \lambda \in \mathbb{C} $. Let $ \text{id}_{\lambda} $ and $ 1_{\lambda} $ denote, respectively, the identity function on $ \{ \lambda \} $ and the constant function on $ \{ \lambda \} $ with value $ 1 $. By the Continuous Functional Calculus, we can apply these two functions to $ T $. As $ \text{id}_{\lambda} = \lambda \cdot 1_{\lambda} $, we obtain $ T = \lambda I $, where $ I: \mathcal{H} \to \mathcal{H} $ is the identity operator. Clearly, $ I $ has non-trivial subspaces (this is only true if we assume that $ \dim(\mathcal{H}) > 1 $), so $ T $ has non-trivial subspaces as well.

Case (ii): Suppose that $ \sigma(T) $ contains two distinct points $ \lambda_{1} $ and $ \lambda_{2} $ (note that this is not possible if $ \dim(\mathcal{H}) = 1 $). Let $ U_{1} $ and $ U_{2} $ be disjoint open neighborhoods (contained in $ \sigma(T) $) of $ \lambda_{1} $ and $ \lambda_{2} $ respectively. If $ \mathbf{E} $ denotes the resolution of the identity corresponding to $ T $, then we have non-zero projection operators $ P_{1} := \mathbf{E}(U_{1}) $ and $ P_{2} := \mathbf{E}(U_{2}) $ satisfying the following two properties:


*

*$ P_{1} P_{2} = \mathbf{0}_{B(\mathcal{H})} = P_{2} P_{1} $ and

*$ T P_{1} = P_{1} T $ and $ T P_{2} = P_{2} T $, i.e., $ P_{1} $ and $ P_{2} $ commute with $ T $.
Property (1) says that $ P_{1} $ is not the identity operator; otherwise $ P_{1} P_{2} = P_{2} \neq \mathbf{0}_{B(\mathcal{H})} $, which is a contradiction. Next, Property (2) says
$$
T[{P_{1}}[\mathcal{H}]] = {P_{1}}[T[\mathcal{H}]] \subseteq {P_{1}}[\mathcal{H}],
$$
which shows that $ {P_{1}}[\mathcal{H}] $ is a non-trivial invariant subspace of $ T $.

Conclusion: Every normal operator $ T \in B(\mathcal{H}) $ has a non-trivial invariant subspace if $ \dim(\mathcal{H}) > 1 $.
