Suppose $v$ and $w$ are linearly independent. Show that the cross product of $v$ and $w$ is linearly independent. Suppose $v$ and $w$ are linearly independent. Show that the cross product of $v$ and $w$ is linearly independent. 
I understand how to calculate the cross product of two linearly independent vectors and have found that the result is linearly independent. I don't understand how to construct a abstract proof to represent all cases and answer the question.
Cheers 
 A: Show that the cross product $v\times w$ is orthogonal to $v$ and $w$. Since $v$ and $w$ are linearly independent, this implies that the three vectors are linearly independent. Indeed suppose that
$$
c_1v+c_2w+c_3(v\times w)=0\tag{0}
$$
for some $c_i$. Then take the inner product with respect to $v\times w$ of both sides to yield that
$$
c_3|v\times w|^2=0
$$
so $c_3=0$ (since $v\neq 0$ and $w\neq 0$). Hence $(0)$ becomes
$$
c_1v+c_2w=0.
$$
But $v$ and $w$ are linearly independent so $c_1=c_2=c_3=0$ as desired.
A: Suppose $ w, v, w \times v$ were linearly dependent.  Then there exists scalars a and b such that $$ aw + bv = w \times v $$ 
“Dot producting” both sides of this equation with $w \times v$ gives $0$ on one side and a positive number on the other. 
A: One can define the cross product of $\vec a$ and $\vec b$ to be a vector perpendicular to both $\vec a$ and $\vec b$, with the right hand rule used to determine the direction,  and magnitude equal to the area of the parallelogram spanned by $\vec a$ and $\vec b$, that is, $\mid \vec a\mid\mid \vec b\mid\sin\theta$, where $\theta$ is the angle between them. 
If we use this definition then we are done right away.
According to Wikipedia,  the equivalency with the determinant definition is not that easy. 
A: When vector a and b are orthogonal then
$a.b = 0$, and a and b are linearly independent, i.e. $a+b \ne 0$.
Now v and w are linearly independent so $k_1v+k_2w \ne 0$, we take the dot product of this non zero vector with the cross product of v,w say u.
$u.(k_1v+k_2w) = k_1u.v+k_2u.w$, now as u is orthogonal to v and w we have
$u.(k_1v+k_2w) = k_1u.v+k_2u.w = 0 + 0 = 0$
So $u+(k_1v+k_2w)\ne 0$, which is equivalent to $k_3u+k_1v+k_2w\ne 0$ 
(because if $k_3u+k_1v+k_2w= 0$ then assuming $k_3$ is non zero, $u+(k_1/k_3)v+(k_2/k_3)w= 0$ )
that indicates $u,v,w$ are linearly independent.
