Lie bracket of linear combinations 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.