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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?

Gerry Myerson
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egrtomath
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  • Are you assuming $\dim V < \infty$? – Travis Willse Oct 09 '14 at 07:56
  • There isn't any restriction. – egrtomath Oct 09 '14 at 08:04
  • 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 09 '14 at 08:49
  • The finite dimensional case is done at http://math.stackexchange.com/questions/27808/a-linear-operator-commuting-with-all-such-operators-is-a-scalar-multiple-of-the and Robert Israel says his answer works in the infinite-dimensional case, given the Axiom of Choice. – Gerry Myerson Oct 09 '14 at 08:53

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