2
$\begingroup$

I'm dealing with a problem related to linear transformation.

Problem: Let $f \in L\left( V \right)$, where $L\left( V \right)$ is the set of all linear operators on $V$. Prove that if $fg = gf$ for all $g \in L\left( V \right)$, then $f = ai$, where $a$ is a scalar and $i$ is identity map.

I can't go on. Is there any hint?

$\endgroup$

marked as duplicate by Gerry Myerson, Eric Stucky, daw, rschwieb, drhab Oct 9 '14 at 10:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Are you assuming $\dim V < \infty$? $\endgroup$ – Travis Oct 9 '14 at 7:56
  • $\begingroup$ There isn't any restriction. $\endgroup$ – egrtomath Oct 9 '14 at 8:04
  • $\begingroup$ I'm sure someone asked about this just the other day (but it isn't easy to find it when everyone uses subjects like "problem about linear transformations"). $\endgroup$ – Gerry Myerson Oct 9 '14 at 8:49
  • $\begingroup$ The finite dimensional case is done at math.stackexchange.com/questions/27808/… and Robert Israel says his answer works in the infinite-dimensional case, given the Axiom of Choice. $\endgroup$ – Gerry Myerson Oct 9 '14 at 8:53

Browse other questions tagged or ask your own question.