Is the Lie bracket of a Lie Algebra of a Matrix Lie Group always the commutator?

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!

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.

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 $$\mathfrak{g}=T_1G$$): $$[X,Y]=ad_XY=\left.\frac{d}{dt}\right|_{t=0}Ad_{\exp(tX)}Y=$$ $$\left.\frac{d}{dt}\right|_{t=0}dR_{\exp(tX)^{-1}}dL_{\exp(tX)}Y_1=$$ $$\left.\frac{d}{dt}\right|_{t=0}dR_{\exp(-tX)}dL_{\exp(tX)}Y_1=$$ $$\left.\frac{d}{dt}\right|_{t=0}\exp(tX)Y\exp(-tX)=$$ $$\left.\frac{d}{dt}\right|_{t=0}e^{tX}Ye^{-tX}=XY-YX.$$