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I have a dumb question:

I have proven that there is one isomorphism class for all two-dimensional non-abelian Lie algebras, with basis $\{x,y\}$ and bracket $[x,y]=x$.

and it was written in an answer that $$\{x,y\}=\left\{\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}1&0\\0&0\end{bmatrix}\right\}$$ gives us one of these Lie algebras.

But my problem is $[x,y]=xy-yx=-x=-[x,y]=[-x,y]$. What is happening here?

So many hats
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Simply that with such definitions you have $[x,y]=-x$ (so that the equality $-x=-[x,y]$ is false). You can easily find at this point an isomorphism in terms of matrices to the standard form you prefer.

N. Ciccoli
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  • I don't understand sorry. The matrix Lie bracket is $[x,y]=xy-yx$ correct, so then do my matrices above not actually give a basis that works for this Lie algebra? – So many hats Aug 20 '15 at 08:06
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    They give you a basis in which $[x,y]=-x$. If you change to a new basis $X=x$, $Y=-y$ then you have $[X,Y]=[x,-y]=-[x,y]=x=X$; so to have a basis with the structural coefficients you'd like you simply need to change the sign in the non zero element of the second matrix. – N. Ciccoli Aug 20 '15 at 08:58