# Lie algebra isomorphic to $\mathfrak{so}(3)$

I'm trying to show that the Lie algebra $\cal{L}$ generated by the matrices $$t_1 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & +1 & 0 \end{pmatrix} \quad t_2 = \begin{pmatrix} 0 & 0 & x \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix} \quad t_3 = \begin{pmatrix} 0 & -x & 0 \\ +1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\\$$ with $x \in \mathbb{R}$ and $x \ne 1$ is isomorphic to $\mathfrak{so}(3)$. The corresponding commutation relations are $$[t_1, t_2] = t_3 \quad [t_2, t_3] = x \, t_1 \quad [t_3, t_1] = t_2$$ Specifically, I'm looking for a matrix $A$ such that $$\epsilon_{ijk}=\sum_{n,m,s} A_{in}\,A_{jm}\,(A^{-1})_{sk}\,f_{nms}$$ with $f_{nms}$ denoting ${\cal L}$'s structure constants $$f_{2,3,1} = x = -f_{3,2,1} \qquad f_{3,1,2} = f_{1,2,3} = 1 = -f_{1,3,2} = -f_{2,1,3}$$ and the Levi-Civita symbol $\epsilon_{ijk}$, the structure constants of $\mathfrak{so}(3)$. How do I construct $A$?

Hint: try correcting the relations rather than looking for an explicit isomorphism. More specifically, the first and last relation (which are the same as for $\mathfrak{so}(3)$) remain unchanged if you multiply $t_2$ and $t_3$ by the same factor. Now choose that factor in such a way that you "correct" the middle relation.
• Thanks, your hint helped me to clear up my confusion. It turns out, that $A$ is the diagonal matrix $diag(1,1/\sqrt{x},1/\sqrt{x})$. – user71769 Mar 19 '18 at 20:14