Prove or disprove the statement:
If {v,w} is a linearly independent set of vectors in $R^3$, then {v,w,v cross w} is also linearly independent.
So far, this makes intuitive sense to be true. If u and v are independent, then they lie in the same plane, but not on top of one another. Then v cross w is orthogonal to both, thus it doesn't lie in the plane thus it seems to be true.
I just don't know how to go about writing a formal proof about this.
We assume at the outset $\{ u,v \}$ is a linearly independent set. Let's attack it directly. Consider, $$ c_1u+c_2v+c_3u\times v =0 $$ we seek to show $c_1=c_2=c_3=0$ hence establishing $\{ u,v, u \times v \}$ is a linearly independent set. Take the dot-product with $u$, $$ c_1u \cdot u +c_2v \cdot u+c_3(u\times v) \cdot u =0 $$ but,$(u\times v) \cdot u = 0$ by the construction of the cross product. Hence, $$ c_1u \cdot u +c_2v \cdot u =0 $$ Likewise, take dot product with $v$ to uncover: $$ c_1u \cdot v +c_2v \cdot v =0 $$ We can solve these equations to show $c_1=c_2=0$. Notice $u \cdot u$ and $v \cdot v$ are nonzero as if they were zero it would force $u$ or $v$ to be zero which contradicts the supposed linear independence of $\{u,v \}$. I leave the details to the reader, but, it should be relatively easy algebra.
Next, take dot-product with $u \times v$, $$ c_1u \cdot (u \times v)+c_2v\cdot (u \times v)+c_3(u\times v)\cdot (u \times v) =0 $$ observe the first two terms are trivial whereas the third returns: $$ c_3 ||u \times v||^2 = 0.$$ But, $||u \times v|| = ||u||||v|| \sin \theta$ and we already know $\theta \neq 0, \pi$ as $u,v$ cannot be colinear hence $||u \times v||^2 \neq 0$ and we conclude $c_3=0$. There is probably a slicker proof. This is merely the first thing that comes to mind.