# Is this a good proof of Schur's Lemma?

kent davidge
There are plenty of proofs of Schur's lemma on the internet, but I find them hard to follow. Then I came up with my own result, but I'm not sure if it's good enough.

Consider ##A v = \kappa v## and ##A v=\kappa v ##. Operating with ##D(g)## the equation then becomes ##D(g)A v = \kappa D(g) v## or ##D(g)(A - \kappa I) v = 0##. But ##A## and ##D(g)## commute, so ##(A - \kappa I)D(g) v = 0##.

Then there would be two possibilites. The first is that the parenthetic quantity vanishes and the second ##D(g)## anihilates ##v##. But this is not true in general for any ##g##. Then the first case would hold, i.e., ##A = \kappa I##.

kent davidge
Oops I should have said that ##D(g)## is irreducible representation. I guess it's completely possible for a reducible representation to map ##v## to the zero vector for any group element ##g##, that is anihilate it, that is let the space consisting of the zero vector invariant.

Mentor
2021 Award
Oops I should have said that ##D(g)## is irreducible representation
You should also have stated which version of Schur's Lemma you are talking about, what ##A,D,v,g,\kappa## are, why ##A## commutes with ##D(g)## and why you can conclude that ##A-\kappa I = 0## only because it vanishes for a certain element ##w:=D(g)(v)## which you have specified by ##Av=\kappa v\,.##

jim mcnamara