I have six-dimensional (complex) matrices which span a representation of $S_4$ that decomposes into the two three-dimensional irreducible representations of the group. I would like to find out the basis vectors such that my representation matrices will be block-diagonal, however I am failing to do so … Is there a trick or algorithm one could use?



You could compute the matrices $M_\chi=\frac{1}{24}\sum_{g \in S^4} \chi(1)\chi(g^{-1})M_g$ which correspond to the primitive central idempotents of the two representations. These will turn out to be projection maps to the two three-dimensional irreducible components, from which you can get the basis vectors you want.

I don't know how numerically stable all of this will be, though. It should work in the general case, but if the two components turn out to be nearly parallel you may need to work harder.

  • $\begingroup$ Hi Micah, thanks for your reply. Do you have a (introductory) reference where I could read more about this? I am a bit lost when I try googling it $\endgroup$ – Faser Oct 12 '17 at 8:20
  • $\begingroup$ I am having trouble with your suggestion. I computed $M_\chi$ for one irrep (am I right to take the characters of the irrep, and $M_g$ is my 6x6 matrix?) and I get $M^2 = M$, as I would expect. However I am not sure how I would figure the new basis vectors from that, or the block diagonal matrices? Computing $M_\chi M_g M_\chi$ does not work … Many thanks! $\endgroup$ – Faser Oct 12 '17 at 14:15
  • $\begingroup$ The best reference I can find online is here; specifically Theorem 3.15 is the thing we need. I'm not sure how elementary you would consider it, though. $\endgroup$ – Micah Oct 13 '17 at 3:02
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    $\begingroup$ If $M^2=M$, you're probably doing everything right. Hopefully also your $M$ has rank 3 (since it's a projection onto a 3-dimensional subspace). You want three of your basis vectors to form a basis for that subspace, so the easiest way to get them is probably to just take three linearly independent columns of $M$. Then you do it all over again with the other character to get the other three basis vectors... $\endgroup$ – Micah Oct 13 '17 at 3:06
  • $\begingroup$ Yeah, it works! Thanks a lot! I also found another reference: Lang's algebra, XVIII §4 :-) $\endgroup$ – Faser Oct 16 '17 at 12:44

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