# Operator in the commutant which is small on a given vector

Suppose $$x$$ is a non-zero vector in a Banach space, and $$T$$ is a fixed operator. Is the following true:

For any $$\varepsilon, \delta$$, there exists $$S$$ in the commutant of $$T$$ such that $$1\leq\|S\|<1+\delta$$ and $$\|Sx\|<\varepsilon$$.

This is true for some $$T$$ (for example the identity), but is it true for all $$T$$? Is there anything known in this direction, even for $$\ell_2$$?

## 1 Answer

Not even true for $$2\times 2$$ matrices. Let $$T$$ be the nilpotent $$\begin{pmatrix}0&1\\0&0\end{pmatrix}$$ and $$x$$ be the vector $$(0,1)$$. Then anything in the commutant of $$T$$ has form $$\lambda+\mu T$$. So if $$S$$ is in the commutant of $$T$$, then $$\|Sx\| = \sqrt{\lambda^2 + \mu^2}$$, so if $$\|Sx\| < \varepsilon$$, then $$\max(|\lambda|, |\mu|) < \varepsilon$$ and $$\|S\| < \sqrt{\frac {3 + \sqrt 5} 2}\varepsilon$$.

• Please provide additional details in your answer. As it's currently written, it's hard to understand your solution.
– Community Bot
Sep 2 at 10:53
• The estimate comes from ||S|| = sqrt (||S*S||).--just find highest eigenvalue of this positive definite matrix. Sep 2 at 13:12