I set my self a practice problem of finding the $2D$ irreducible representation, $R_2$, of the group $D_3$. So far I have used projection matrices to find the generators in $R_2 \oplus R_2$ to be: $$R_2 \oplus R_2(a)=\begin{pmatrix}0& 1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0 \end{pmatrix}$$ and $$R_2 \oplus R_2(b)=\begin{pmatrix}0& 0&1&0\\-1&0&0&-1\\1&0&0&0\\0&-1&-1&0 \end{pmatrix}$$ I now need to simultaneously block diagonalize these matrices, but as far as I can tell there is no standard procedure to doing this. Does anyone know how it can be done in this relatively simple case?
Note: this source actually gives the $R_2(g)$ irreducible rep: https://www.its.caltech.edu/~xcchen/img/Ph129b2017/lecture/lecture0124.pdf
I have found how to simultaneously block diagonalize these matrices out of an educated guess looking at the eigenvectors of $R_2\oplus R_2(a)$ and working my way from there.
If we define: $$\tilde I=\begin{pmatrix} I_2&I_2\\I_2&-I_2\end{pmatrix}$$ and $$P=\frac{1}{\sqrt{2}}\begin{pmatrix} 1&1&0&0\\0&0&1&-1\\0&0&1&1\\1&-1&0&0\end{pmatrix}$$ then the basis transformation defined thru: $$A'=\tilde I^{-1}P^{-1}A P \tilde I$$ diagonalizes both $R_2\oplus R_2(a)$ and $R_2\oplus R_2(b)$ to give: $$R_2\oplus R_2(a)\mapsto\begin{pmatrix} 1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&1\end{pmatrix}$$ $$R_2\oplus R_2(b)\mapsto\begin{pmatrix} -0.5&0.5&0&0\\1.5&0.5&0&0\\0&0&0.5&-0.5\\0&0&-1.5&-0.5\end{pmatrix}$$ from which we can read of values for $R_2(a)$ and $R_2(b)$. Note these will be different to those given in the link in the question - but are required to be isomorphic.