Show Lie Bracket of X & Y is Linear Comb. of Commuting Vector Fields

[WannabeNewton]

Homework Statement

Show that if the vector fields X and Y are linear combinations (not necessarily with constant coefficients) of m vector fields that all commute with one another, then the lie bracket of X and Y is a linear combination of the same m vector fields.

The Attempt at a Solution

I started off by denoting the m vector fields by V_{(a)} so that [V_{(c)}, V_{(d)}] = 0 for all V_{(a)}. I wrote the vector fields X and Y as X = \alpha ^{c}V_{(c)} and Y = \beta ^{d}V_{(d)} where \alpha ,\beta are scalar functions. Then, [X, Y] = [\alpha ^{c}V_{(c)}, \beta ^{d}V_{(d)}] right? I worked out the lie derivative in component form and put it back in abstract form to get [X, Y]= [\alpha ^{c}V_{(c)}, \beta ^{d}V_{(d)}] = \alpha ^{c}V_{(d)}(V_{(c)}\cdot \triangledown \beta ^{d}) - \beta ^{d}V_{(c)}(V_{(d)}\cdot \triangledown \alpha ^{c}) but I don't see how this helps me at all in showing that [X, Y] can be written as a linear combination of the V_{(a)}'s (the m vector fields). Help please =D.

EDIT: I forgot to mention that I am using the Einstein summation convention here so that any repeated letters with one on top and one on bottom indicates summation.

 

[WannabeNewton]

Apparently, [X, Y]= \alpha ^{c}V_{(d)}(V_{(c)}\cdot \triangledown \beta ^{d}) - \beta ^{d}V_{(c)}(V_{(d)}\cdot \triangledown \alpha ^{c}) is sufficient to conclude that the lie bracket is a linear combination of the m vector fields. I'm not convinced only because, while the expression does contain linear combinations of the m vector fields, the expression doesn't look neat in the slightest. If anyone wants to weigh in I would be very glad but I will assume that the conclusion is correct for now as it agrees with what the text has.