Does representation irreducibility ensure non-zero determinant?

Does representation irreducibility ensure non-zero determinant?



If a set of matrix representation $M(g)$ for a group $G$ is irreducible, what can we say about their determinant for every $gin G$? Are they all of non-zero determinant?



Thank you very much!



Cheers,
Collin



P.S.: I'm a physics graduate student. So please use as little math terminology as possible, I would really appreciate that!




1 Answer
1



The inverse of $M(g)$ is $M(g^-1)$.



And of course, invertible matrices have non-zero determinant.



Note that this is true for all representations, not just irreducible ones.





$begingroup$
Oh, Yes! You're so right sir. I can't believe this. I just read this sentence yesterday. Thank you! ^. ^ @DanielMroz
$endgroup$
– Collin
Sep 15 '18 at 22:03



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