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Given a Matrix Lie Group, the Lie Bracket is of the associated Lie Algebra is given by the Lie Derivative. Is this always the commutator if we start from a Matrix Lie Group?

Cheers!

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Yes; one can prove that the Lie bracket in the Lie algebra of the general linear group is the matrix commutator. Any matrix group is a subgroup of this with corresponding subalgebra and the bracket of the subalgebra is just the restriction of the original bracket.

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Anthony's answer is the most elegant one that I can think. But I'll post one which is more constructive (and waaaaaay less elegant), because, sometimes people need to see algebra working to truly convince theirselves (and here I use some important technics to the theory).

If you already know that $[X,Y]=ad_XY$ and that in matrix group $dL_gX=gX$ and $dR_gX=Xg$ and the Lie exponential coincides with the matrix exponential, for every $X,Y\in\mathfrak{g}$ and for every $g\in G$, then (under the natural identification $\mathfrak{g}=T_1G$): $$[X,Y]=ad_XY=\left.\frac{d}{dt}\right|_{t=0}Ad_{\exp(tX)}Y=$$ $$\left.\frac{d}{dt}\right|_{t=0}dR_{\exp(tX)^{-1}}dL_{\exp(tX)}Y_1=$$ $$\left.\frac{d}{dt}\right|_{t=0}dR_{\exp(-tX)}dL_{\exp(tX)}Y_1=$$ $$\left.\frac{d}{dt}\right|_{t=0}\exp(tX)Y\exp(-tX)=$$ $$\left.\frac{d}{dt}\right|_{t=0}e^{tX}Ye^{-tX}=XY-YX.$$

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