Let $T$ be an operator on a $\Bbbk$-linear space $V$. The notions of commutant and bicommutant of $T$ seem to be important but I don't have any conceptual intuition for them. In hopes of mending this I have a few questions. As background and motivation, below are some relevant results.
Elementary Linear Algebra.
Proposition. If $T$ is cyclic, then its commutant consists of the polynomial operators in $T$.
Lemma. Suppose $V\cong \bigoplus _i U_i$ is an internal direct sum decomposition. Then each $U_i$ is $T$-invariant iff each projection $\pi_i$ commutes with $T$.
Corollary. If $S\in \mathrm C^2(T)$ then each $T$-invariant direct sum decomposition of $V$ is also $S$-invariant.
Theorem. If $V$ is finite dimensional then its bicommutant consists precisely of the polynomial operators in $T$.
I can't summarize the proof well, but the finite dimension is used via existence of cyclic decomposition which reduces to relating the cyclic cases already attended to.
Hilbert Spaces.
Definition. Let $(V, \left\langle \cdot \right\rangle )$ be an inner product space and $T\in \mathrm{End}_\Bbbk(V)$. A subspace of $V$ reduces $T$ if both it and its orthogonal complement are $T$-invariant.
By the above lemma, a subspace of a Hilbert space reduces $T$ iff $T$ commutes with its orthogonal projection.
In this MSE question the OP asks for intuition for the bicommutant in the context of Hilbert space theory. Two parts of the answer are:
- The following observation for a Hilbert space. "A normal operator $S$ is in the bicommutant of $T$ if and only if every reducing subspace of $T$ is also a reducing subspace of $S$".
- The bicommutant $\mathrm C^2(T)$ of $T$ is parallel to the sigma algebra $\sigma (X)$ generated by a random variable $X$. Indeed the first paragraph of the answer draws analogy between [something involving normal operators and generated algebras] in functional analysis and the Doob-Dynkin lemma.
Questions.
- "Why" is the bicommutant of an operator parallel to the sigma algebra generated by a random variable? Where does the commutant fit in, and how to think of it?
- The Doob-Dynkin lemma is intuitive, but I don't see the intuition behind the assertion made for a normal operator in the linked answer. Why should we expect it?
