# Find basis for irreps

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?

Thanks!

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 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! – Faser Oct 12 '17 at 14:15
• 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... – Micah Oct 13 '17 at 3:06