Showing a vector belongs in the span Without using dimension, I have to show
From: question
$\overrightarrow{x} \in \mathbb{R}^3$ is also in span $\{v_1, v_2, n\}$ where $n = v_1 \times v_2$. Where $v_1, v_ 2\in \mathbb{R}^3$ as well.
We can let $x = \langle x_1, x_2, x_3 \rangle$, but that isn't very helpful.
I know the set is linearly independent, could that be used? Should I write their sum in the form of components?
 A: I don't know why you wouldn't want to use any dimension arguments. Let me outline the most common approach.
General fact: Let $V$ be a finite-dimensional vector space, say $\dim(V)=n$. Suppose that $\left\{v_1,v_2, \dots , v_n\right\}$ is a linearly independent subset, then this set is generating.
This statement is covered in any standard textbook (and course) on linear algebra. The proof of this fact is easy once you know what the dimension of a vector space is. But to properly define the dimension of a vector space, you first need to know that given two different bases of the same vector space, they have the same amount of vectors. The proof of the latter fact essentially boils down to the lemma of Steinitz.
Having said that, let's return to the problem at hand. You have three vectors in $\mathbb{R}^3$ which are linearly independent. Since $\dim(\mathbb{R}^3)=3$, the above discussions yields that thee linearly independent vectors automatically form a basis.
Without some explicit information on $v_1,v_2$, it's going to be difficult to show that the set is generating directly. I'm not sure whether you even can give a "direct" argument. Showing that a set is generating directly is often more difficult than showing linear independence, this technique of adding the dimension into the discussion is often useful to avoid having to do that explicitly (which might not be possible).
A: An approach: if $\{\mathbf{u},\mathbf{v},\mathbf{u\times v}\}$ is linearly independent then, for all $\mathbf{x}\in\mathbb{R}^3$, the system $$\mathbf{x}=\lambda_1\mathbf{u}+\lambda_2\mathbf{v}+\lambda_3(\mathbf{u\times v})$$ has solution (besides, unique solution) because $$\text{rank }[\mathbf{u},\mathbf{v},\mathbf{u\times v}]=\text{rank }[\mathbf{u},\mathbf{v},\mathbf{u\times v},\mathbf{x}]=3$$ that is, $\mathbf{x}\in \text{Span }\{\mathbf{u},\mathbf{v},\mathbf{u\times v}\}.$
A: I'll use the notation from the previous answer, since I don't like subscripts. We are given two vectors $\mathbf{u}$ and $\mathbf{v}$, and we want to show that any given vector $\mathbf{x}$ is in the span of $\{\mathbf{u}, \mathbf{v}, \mathbf{n}\}$, where $\mathbf{n} = \mathbf{u} \times \mathbf{v}$.
First let's focus on the point $\mathbf{w} = p \mathbf{u} + q \mathbf{v}$ that is the foot of the perpendicular from $\mathbf{x}$ to the plane spanned by $\mathbf{u}$ and $\mathbf{v}$. The scalars $p$ and $q$ are unknown, as yet -- we have to find them. To get perpendicularity, we need $p$ and $q$ to satisfy the equations
$$(\mathbf{x} - \mathbf{w}) \cdot \mathbf{u} = (\mathbf{x} - p \mathbf{u} - q \mathbf{v}) \cdot \mathbf{u} = 0
$$
$$
(\mathbf{x} - \mathbf{w}) \cdot \mathbf{v} = (\mathbf{x} - p \mathbf{u} - q \mathbf{v}) \cdot \mathbf{v} = 0
$$
In other words, $p$ and $q$ that are solutions of the linear system
$$
p (\mathbf{u} \cdot \mathbf{u}) + q (\mathbf{u} \cdot \mathbf{v}) = \mathbf{x} \cdot \mathbf{u}
$$
$$
p (\mathbf{u} \cdot \mathbf{v}) + q (\mathbf{v} \cdot \mathbf{v}) = \mathbf{x} \cdot \mathbf{v}
$$
A solution exists because the determinant of this system is 
$(\mathbf{u} \cdot \mathbf{v})(\mathbf{v} \cdot \mathbf{v}) - (\mathbf{u} \cdot \mathbf{v})^2$, which is positive unless $\mathbf{u}$ and $\mathbf{v}$ are linearly independent. Or, you can just see that the system has a solution by calculating it via Cramer's rule.
Since $\mathbf{x} - \mathbf{w}$ is orthogonal to both $\mathbf{u}$ and $\mathbf{v}$, it must be parallel to $\mathbf{n} = \mathbf{u} \times \mathbf{v}$. So, there is a scalar $k$ such that 
$\mathbf{x} - \mathbf{w} = k \mathbf{n}$. But then 
$$
\mathbf{x} = \mathbf{w} + k \mathbf{n} = p \mathbf{u} + q \mathbf{v} + k \mathbf{n}
$$
which means that $\mathbf{x}$ lies in the span of $\{\mathbf{u}, \mathbf{v}, \mathbf{n}\}$.
A: Another approach ...
We can actually show something more general: if the three vectors $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ are not coplanar, then they span $\mathbb{R}^3$. Here's how:
Given a vector $\mathbf{r}$,  we need to show that it can be written as a linear combination of $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$. In other words, we have to find scalars $x$, $y$, $z$ such that
$$
x\mathbf{u} + y\mathbf{v} + z\mathbf{w} = \mathbf{r}
$$
This is a linear system of equations, which can be written
$$
\left[\begin{matrix}
\leftarrow & \mathbf{u} & \rightarrow \\
\leftarrow & \mathbf{v} & \rightarrow \\
\leftarrow & \mathbf{w} & \rightarrow \\
\end{matrix}
\right]
\left[
\begin{matrix}
x  \\
y  \\
z  \\
\end{matrix}\right] = \mathbf{r}
$$
The system has a solution provided the determinant of its matrix is non-zero. But this determinant is just $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$. This triple product measures the volume of the parallelipiped having $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ as edges, so it will be non-zero if $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ are not coplanar.
If you don't know anything about matrices and determinants, you can just apply Cramer's rule to solve the linear system. Again, you will see that the system has a unique solution provided $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \ne 0$.
A: And yet another approach ...
Again, we're going to show that if the three vectors $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ are not coplanar, then they span $\mathbb{R}^3$. 
If $\mathbf{u} \cdot (\mathbf{v} × \mathbf{w}) \ne 0$, we can define three magic new vectors
$\tilde{\mathbf{u}}$, $\tilde{\mathbf{v}}$, $\tilde{\mathbf{w}}$ by the equations
$$
\tilde{\mathbf{u}} = \frac{\mathbf{v} × \mathbf{w}}{\mathbf{u} \cdot (\mathbf{v} × \mathbf{w})} \quad ; \quad
\tilde{\mathbf{v}} = \frac{\mathbf{w} × \mathbf{u}}{\mathbf{u} \cdot (\mathbf{v} × \mathbf{w})} \quad ; \quad
\tilde{\mathbf{w}} = \frac{\mathbf{u} × \mathbf{v}}{\mathbf{u} \cdot (\mathbf{v} × \mathbf{w})}
$$
These three vectors are called the dual vectors of $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$, or the reciprocal vectors.
If $\mathbf{x}$ is any given vector, it is straightforward to verify that
$$
\mathbf{x} = 
(\mathbf{x} \cdot \tilde{\mathbf{u}})\mathbf{u} + 
(\mathbf{x} \cdot \tilde{\mathbf{v}})\mathbf{v} + 
(\mathbf{x} \cdot \tilde{\mathbf{w}})\mathbf{w}
$$
This shows that $\mathbf{x}$ lies in the span of $\{\mathbf{u}, \mathbf{v}, \mathbf{w} \}$, and even gives you explicit formulae for the coefficients. This is cute, but it doesn't give you much insight into the geometry of the situation.
The last equation giving $\mathbf{x}$ is just a disguised statement of Cramer's rule, so this approach is strongly related to the previous one.
