Let $\vec{x}$ and $\vec{y}$ be non-zero vectors. Show that any two-dimensional vector can be expressed in the form $s \vec{x} + t \vec{y}$, where $a$ and $b$ are real numbers, if and only if of the vectors $\vec{x}$ and $\vec{y}$, one vector is not a scalar multiple of the other vector.
I am not quite sure how to approach this problem. Must the vectors $\vec{x}$ and $\vec{y}$ parallel for this to be true? I just have a suspicion, but I am actually struggling to find a good place to begin.
Suppose $\vec{x} = k \vec{y}$. Then, $s\vec{x} + t\vec{y} = (s+kt)\vec{x}$. The locus of such points as $s$ and $t$ range over the reals is a line. So, not every two-dimensional vector can be expressed as $$s\vec{x} + t\vec{y}.$$For the other direction, let $\vec{x} = \begin{pmatrix} a \\ b \end{pmatrix}$ and $\vec{y} = \begin{pmatrix} c \\ d \end{pmatrix}$. Let $\begin{pmatrix} w \\ z \end{pmatrix}$ be an arbitrary two-dimensional vector.
We want to find real numbers $s$ and $t$ such that $$s \begin{pmatrix} a \\ b \end{pmatrix} + t \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} w \\ z \end{pmatrix}.$$This gives us the system of equations \begin{align*} as + ct &= w, \\ bs + dt &= z. \end{align*}Solving this system in $s$ and $t$, we find $$s = \frac{dw - cz}{ad - bc} \quad \text{and} \quad t = \frac{az - bw}{ad - bc}.$$ Both of these quantities are well-defined as long as $ad \neq bc$. Thus, we can conclude if we show that $ad = bc$ implies that of the two non-zero vectors $\vec{x} = \begin{pmatrix} a \\ b \end{pmatrix}$ and $\vec{y} = \begin{pmatrix} c \\ d \end{pmatrix}$, one vector is a scalar multiple of the other vector.
So, suppose $ad = bc$, If $a \ne 0$, then $d = \frac{c}{a} \cdot b$, so $$\frac{c}{a} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} c \\d \end{pmatrix}.$$ The case where $c \ne 0$ is similar. If $a= c =0$, then both vectors are of the form $\begin{pmatrix} 0 \\ z \end{pmatrix}$, so one must be a multiple of the other.
The statement is as it says, true if and only if the two vectors $\vec x$ and $\vec y$ are not parallel. The reason for this is that a vector can be treated as a translation of the coordinate plane.The vector $\vec x$ is a translation of the plane, with a magnitude equal to $|x|$ and along its own direction. Multiplying a scalar with the vector simply scales the translation (hence the name) Thus a translation $\vec v$ can be treated as the sum of an infinite number of arbitrary translations of the coordinate plane, but once you restrict the translations to two (non-parallel) directions there is only one possible combination, $a \vec x + b \vec y$. A way of quantifying it is to use projections. Projecting in any two basis directions (You know there are two of them as we are dealing with 2-d space) $\vec v = v_1 \vec i +v_2 \vec j$ and similarly for $\vec x$ and $\vec y$ you get a system of linear equations $v_1 = ax_1 + by_1$ and $v_2=ax_2+by_2$ which always has a unique solution in $a$ and $b$, unless $\frac{x_1}{x_2}=\frac{y_1}{y_2}$ which would imply $\vec x$ and $\vec y $ to be parallel.