# If $f$ commutes with every linear transformation, it's a scalar multiple of the identity [duplicate]

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?

## marked as duplicate by Gerry Myerson, Eric Stucky, daw, rschwieb, drhabOct 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.

• Are you assuming $\dim V < \infty$? – Travis Oct 9 '14 at 7:56
• There isn't any restriction. – egrtomath Oct 9 '14 at 8:04
• 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"). – Gerry Myerson Oct 9 '14 at 8:49
• 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. – Gerry Myerson Oct 9 '14 at 8:53