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.

 

[fresh_42]

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 $Av=\kappa v\,.$