Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.

$S=(S_1,S_2)$ is called normal iff $S_1S_2=S_2S_1$ and both $S_1$ and $S_2$ are normal.

Assume that $F$ is an infinite-dimensional complex Hilbert space.

I look for an example of two normal operators $S_1$ and $S_2$ (which are not a scalar multiple of the identity) such that $S_1S_2=S_2S_1$ and $S_1\neq S_2$.

  • $\begingroup$ How about $S_1 = I$ and $S_2 = 2I$? $\endgroup$ Apr 4 '18 at 10:47
  • $\begingroup$ @Omnomnomnom You are right but I hope to get a non trivial example and thank you for your help $\endgroup$
    – Schüler
    Apr 4 '18 at 10:49
  • $\begingroup$ My point is that you should be clear about what "non-trivial" means in this context $\endgroup$ Apr 4 '18 at 10:50
  • $\begingroup$ Notably, every finite dimensional example can be thought of (up to a change of basis) as a pair of diagonal matrices. Would versions of these examples be considered "trivial"? $\endgroup$ Apr 4 '18 at 10:51
  • $\begingroup$ @Omnomnomnom Please see my edit. Thank you. $\endgroup$
    – Schüler
    Apr 4 '18 at 10:52

Here's a family of examples that you might find interesting. Take any two bounded sequences $(a_n),(b_n)$. Define the maps $S_i:\ell_2 \to \ell_2$ by $$ (T_1x)_n = a_nx_n, \qquad (T_2x)_n = b_n x_n $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.