1
$\begingroup$

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 $S_1\neq S_2$.

$\endgroup$
5
  • $\begingroup$ How about $S_1 = I$ and $S_2 = 2I$? $\endgroup$ Apr 4, 2018 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, 2018 at 10:49
  • $\begingroup$ My point is that you should be clear about what "non-trivial" means in this context $\endgroup$ Apr 4, 2018 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, 2018 at 10:51
  • $\begingroup$ @Omnomnomnom Please see my edit. Thank you. $\endgroup$
    – Schüler
    Apr 4, 2018 at 10:52

1 Answer 1

3
$\begingroup$

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 $$ (T_1x)_n = a_nx_n, \qquad (T_2x)_n = b_n x_n $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .