Span of two vectors in $\mathbb{R}^2$

Question

The span of two vectors in $\mathbb{R}^2$ neither of which is zero vector, and which are not parallel, is-

1. a point.
2. line in $\mathbb{R}^2$ not running through origin.
3. line in $\mathbb{R}^2$ running through origin.
4. all of $\mathbb{R}^2$.

Thanks for helping.

Answer 1

Let say the your vectors are- $\begin{bmatrix}\alpha_1 \\ \alpha_2\end{bmatrix} \ \& \ \begin{bmatrix}\beta_1 \\ \beta_2 \end{bmatrix}$.

So, $$x_1\begin{bmatrix}\alpha_1 \\ \alpha_2\end{bmatrix} \ + x_2\ \begin{bmatrix}\beta_1 \\ \beta_2 \end{bmatrix} = \begin{bmatrix}\gamma_1 \\ \gamma_2\end{bmatrix}, \text{ where } \gamma_1, \ \gamma_2 \in \mathbb{R}.$$

$$\implies \begin{bmatrix}\alpha_1 && \beta_1\\ \alpha_2 && \beta_2\end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix}\gamma_1 \\ \gamma_2\end{bmatrix}.$$

Say,

$$A=\begin{bmatrix}\alpha_1 && \beta_1\\ \alpha_2 && \beta_2\end{bmatrix}.$$

From the given information in the question, we can say that $A$ is a full rank matrix, so it is invertible.

$$\begin{bmatrix}x_1 \\ x_2 \end{bmatrix} = A^{-1}\begin{bmatrix}\gamma_1 \\ \gamma_2\end{bmatrix}.$$

For any $\gamma_1,\gamma_2$, we can get a unique solution for $x_1,x_2$. So, the vectors span the whole $\mathbb{R}^2$.

Answer 2

The answer is $4)$. To this end, it amounts to showing the following:

the system of equations: $x(a,b)+y(c,d) = (m,n)$ has a solution for any $m,n$. This means:

$ax+cy = m, bx+dy = n$ has at least one solution,and using Cramer's rule, this occurs when $ad - bc \neq 0$ which means the vectors $(a,b), (c,d)$ are non-parallel. Thus we're done.