# (When) is the bicommutant of a linear operator equal to the set of “simpler” operators?

Let $$V$$ be a $$\Bbbk$$-linear space and $$T\in \mathrm{End}_\Bbbk(V)$$.

Definition. Say an operator $$S\in \mathrm{End}_\Bbbk(V)$$ is simpler than $$T$$ if every $$T$$-invariant (internal) direct sum decomposition of $$V$$ is also $$S$$-invariant.

Lemma. An internal direct sum decomposition of $$V$$ is $$T$$-invariant iff all of its projections commute with $$T$$.

Corollary. The elements of the bicommutant $$\mathrm C^2(T)$$ are operators simpler than $$T$$.

Question 1. Suppose $$V$$ is finite dimensional. (When) is the bicommutant equal to the set of simpler operators?

Beginning of argument. Suppose $$S$$ is simpler than $$T$$. This means that for every $$T$$-invariant decomposition with projections $$\{\pi_i\}$$ we have $$\{\pi_i\}\subset \mathrm C(S)$$. We wish to prove that $$S\in \mathrm C^2(T)$$, i.e $$L\in \mathrm C(T)\implies L\in \mathrm C(S)$$. The only visible way forward seems to be if $$L$$ is in the subalgebra generated by the projections $$\{\pi_i\}$$...

Theorem. The bicommutant of an operator $$T$$ on a finite dimensional vector space consists of the polynomials in $$T$$.

Question 2. What's an example of an operator which is simpler than $$T$$ but not polynomial in $$T$$? (If possible, suppose $$V$$ is finite dimensional.)

Let $$S=\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$$ and $$T=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$$. Then neither $$S$$ nor $$T$$ has any nontrivial invariant direct sum decompositions, so they are equally simple, but neither is in the bicommutant of the other (or is a polynomial of the other).
Note that if $$V$$ is finite-dimensional and $$S$$ is diagonalizable, then its bicommutant is spanned by the projections onto its eigenspaces. So, it is natural to consider non-diagonalizable matrices in looking for counterexamples.
• Well, $S$ could be diagonalizable--for instance, in my example you could replace $S$ with any diagonal matrix with distinct diagonal entries, and $S$ would still be simpler than $T$, but not in its bicommutant. – Eric Wofsey May 12 at 15:11