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I got completely stumbled upon the following question. Let's suppose that I'm given a certain representation of a Lie algebra and want to find its decomposition into irreducible blocks (not only formulae like $2\otimes2=1\oplus3$, but also the explicit change of basis). I know what I should do in two other cases: a) in case of the finite groups I should have calculated the characters; b) in case of tensor products I should have found highest-weight states. But here I'm completely stuck and don't even understand what to do.

The question arose from the following problem i tried to solve unsuccessfully. 3 matrices $G^i$ (with $\sigma^i$ being Pauli $\sigma$-matrices): $$ G^1 = i \begin{pmatrix} 0 & -\sigma_3\\ \sigma_3& 0\end{pmatrix}, \quad G^2 = i \begin{pmatrix} 0 & \sigma_1\\ -\sigma_1& 0\end{pmatrix}, \quad G^3 = \begin{pmatrix} \sigma_2&0\\ 0&\sigma_2\end{pmatrix} $$

form an $su(2)$-representation after certain rescalings: $[G^i, G^j] = -2 i \epsilon^{ijk}G^k$. They answer (which I could not obtain) is that they, actually, form a reducible representation with the following form $$ U^\dagger G^i U = \begin{pmatrix} -\sigma^i & 0\\ 0& -\sigma^i\end{pmatrix} $$

achieved with the help of unitary transformation $$ U =\frac{1}{\sqrt{2}} \begin{pmatrix} 0 & -i & -1 & 0\\ 0 & 1 & i & 0\\ -1 & 0 & 0 & i \\ i & 0 & 0 & -1 \end{pmatrix}$$

I am more interested in understanding the general situation and whether there are some algorythms to do this, but will equally appreciate the explanation of what happened in the explicit example.

Update. Not sure if it's helpful, but in the particular example the following route seems possible. To determine the structure of representation: one could simply find the eigenvalues of the matrices $G^i$, which turn out to be $(1,1,-1,-1)$, from which it's pretty much clear that it's $2\oplus2$. Then I tried diagonalizing $G^3$, since it can correspond to $diag(-\sigma_3,-\sigma_3)$, after some juggling with rows permutation I almost obtained the required decomposition, but not yet: there still is a residual similarity transformation.

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