Show that $\vec{v}$ can be any vector in a plane where $\vec{v} = a_1\vec{v}_1+a_2\vec{v}_2$

Question

How can I show that $\vec{v}$ can be any vector in a plane where $\vec{v} = a_1\vec{v}_1+a_2\vec{v}_2$?

All vectors start at the origin, $a_1$ and $a_2$ are scalars, $\vec{v}_1$ and $\vec{v}_2$ are vectors that are not scalar multiples of each other.

I know two vectors that are not scalar multiples of each other are on a plane, but I don't know how to show this. (I think "linearly independent" is the right term?)

(This is from Linear and Geometric Algebra by A. Macdonald — I am self studying between semesters.)

Answer 1

If $v_1 = (x_1,y_1), v_2 = (x_2,y_2)$, and let $v = (a_1,a_2)$. Thus: $av_1+bv_2 = a(x_1,y_1)+b(x_2,y_2) = (ax_1+bx_2,ay_1+by_2) = (a_1,a_2) \Rightarrow \begin{cases} ax_1+bx_2 = a_1 \\ ay_1+by_2 = a_2 \end{cases}$. This system of linear equations in variables $a,b$ has a solution because $\left|\begin{pmatrix} x_1 & x_2 \\ y_1 & y_2 \end{pmatrix}\right| \neq 0$ since $v_1 \neq kv_2$ for all $k$.

Answer 2

Since you are working in $\mathbb{R}^2$, the dimension of your vectorial space is $2$. That means that a $base$ of $\mathbb{R}^2$ consist of two linearly independent vectors. (In general, for an $n$-dimensional vector space, a base, consist of $n$ linear independent vectors.)

Now if $B$ is a base of $\mathbb{R}^2$ then you can write any vector $v \in \mathbb{R}^2$ as a linear combination of the vectors of the base $B$. Now in the plane, if two vectors are not scalar multiple of each other, then they are linearly independent. You can prove this by definition as an exercise. Therefore, since they are linearly independent and you have to of them, you can put them together to create a base $B$ for $\mathbb{R}^2$. As I said before, since $B$ is a base, then given any vector $v \in \mathbb{R}^2$ you can write it as a linear combination of the vectors of your base $B$

Answer 3

Recall that a plane is defined by its normal vector $\hat{n}$. So if you have two arbitrary non-parallel vectors then their cross product defines the normal vector of the plane. From there it is sufficient to show that any $\vec{v} = a_1\hat{v_1} + a_2\hat{v}_2$ is perpendicular to $\hat{v_1} \times \hat{v_2}$. Hope this helps!

Answer 4

Since $v_1, v_2$ are linearly independent the matrix $ {\left[\begin{matrix} v_{11} & v_{12} \\ v_{21} & v_{22}\end{matrix}\right]}$ has nonzero determinant, so the linear equations system $ \left[\begin{matrix} v_{11} & v_{12} \\ v_{21} & v_{22}\end{matrix}\right] \cdot \left[\begin{matrix} a \\ b\end{matrix}\right] = \left[\begin{matrix} v_1 \\ v_2\end{matrix}\right]$ has always exactly one solution in $\mathbb{R}²$.