help with a vector proof 
Let $\textbf{u}$ and $\textbf{v}$ be non-zero vectors. Show that any two-dimensional vector can be expressed in the form
$$s \textbf{u} + t \textbf{v},$$where $s$ and $t$ are real numbers, if and only if of the vectors $\textbf{u}$ and $\textbf{v}$, one vector is not a scalar multiple of the other vector.

Below is my proof, but I'm not sure if my proof by contradiction makes sense and also how to take care of the "if and only if" part in the problem statement.

We know that a pair of vectors that can generate the whole plane is linearly independent, so let's try to show that $s \textbf{u} + t \textbf{v}$ is linearly independent. For this to be linearly independent, then $s=t=0$. If this were linearly dependent, at least one of $s$ and $t$ must not equal $0$.
Suppose for the sake of contradiction that $\textbf{v} = a\textbf{u}$, where $a$ is a real scalar multiple. Also, let $\textbf{u} = \dbinom{\textbf{u_1}}{\textbf{u_2}}$ and $\textbf{v} = \dbinom{\textbf{v_1}}{\textbf{v_2}}$. Then $\textbf{v_1} = a\textbf{u_1}$ and $\textbf{v_2} = a\textbf{u_u}$. Then $$\dbinom{s\textbf{u_1} + at\textbf{u_1}}{s\textbf{u_2} + at\textbf{u_2}} =\dbinom{0}{0} .$$ So $(s+at)\textbf{u_1} = 0$ and $(s+at)\textbf{u_2} = 0$. Since $\textbf{u}$ is a nonzero vactor, this means $s+at =0$. However, because $s,a,t$ are all real numbers, this does not necessarily say whether $s$ and $t$ must be nonzero. Therefore, the assumption that $\textbf{v}$ was a scalar multiple of $\textbf{u}$ (and thus the vectors were linearly dependent) must be wrong, and $\textbf{u}$ and $\textbf{v}$ are linearly independent.

 A: You seem to have a fairly deep misunderstanding of what the phrase "linearly independent" means. So, here's the definition:

Let $V$ be a vector space over $\mathbb R$, and let $S\subseteq V$. The set $S$ is linearly depenedent if there exists a finite nonempty set $\{v_1,v_2,\dots v_n\}\subseteq S$ and a set of real numbers $\alpha_1,\alpha_2.\dots,\alpha_n$ such that not all $\alpha_i$ are equal to $0$ and that $$\alpha_1v_1 + \alpha_2v_2+\cdots + \alpha_n v_n=0.$$
The set $S$ is linearly independent if it is not linearly dependent.


So, to clarify:
Linear dependence is the property of a set of vectors.

Knowing this definition, it should now be clear that the sentence you wrote:

so let's try to show that $s \textbf{u} + t \textbf{v}$ is linearly independent.

Is nonsensical. For every $s$ and $t\in\mathbb R$, $s\textbf u + \textbf v$ is simply a vector, not a set of vectors, and therefore you cannot speak about it being "linearly independent". Furthermore, the following sentence also makes very little sense:

For this to be linearly independent, then $s=t=0$ -

what? If $s=t=0$, then the vector $s\textbf u + \textbf v$ is equal to $0$, why woult that be "linearly independent"?

I suggest you start the proof over, and ditch the whole concept of "linear dependency" you are obviously not familiar with. Remember, the statement you want to proove is an "if and only if statement", so you need to prove two statements:

*

*If neither vector is a scalar multiple of the other, then any $2$-dimensional vector can be expressed in the form of $s\textbf u + \textbf v$.

*If any $2$-dimnensional vector can be expressed in the form of $s\textbf u + \textbf v$, then neither vector is a scalar multiple of the other.

I suggest you first prove point (1) with a fairly brute-force approach. That is, you set $v=\begin{bmatrix}{v_1\\v_2}\end{bmatrix}$ and $u=\begin{bmatrix}{u_1\\u_2}\end{bmatrix}$ and set $c=\begin{bmatrix}{c_1\\c_2}\end{bmatrix}$ and try to find the $s$ and $t$ such that $su+tv=c$.
A: Like 5xum wrote, you have misunderstanding about what linearly independent vectors are.
We need to show that $\textbf{u}$ is linearly independent to $\textbf{v}$:

Given that $\textbf{u}\ne a\textbf{v}\iff\textbf{u}-a\textbf{v}\ne0\iff\textbf{u}+b\textbf{v}\ne0$. Hence they are linearly independent


You can do this work without showing linearly independent:

Assuming that the span of $\textbf v, \textbf u$ is $\Bbb R$, i.e. any two-dimensional vector can be expressed using those $2$.
Let's $\textbf u=a\textbf v$,
So $c\textbf u+b\textbf v=c\textbf u+ab\textbf u=\begin{bmatrix}abcu_1\\abcu_2\end{bmatrix}\ne\begin{bmatrix}u_1+1\\u_2\end{bmatrix}$(why is that?)
Hence $\textbf u\ne a\textbf v$
Assuming $\textbf u\ne a\textbf v$, 
If so I have $a\textbf u+b\textbf v=\hat e_x$ and $c\textbf u+d\textbf v=\hat e_y$, $\hat e_{y,x}$ are the standard basis. From those to vector I can have any vector so $p\hat e_x+q\hat e_y=p(a\textbf u+b\textbf v)+q(c\textbf u+d\textbf v)=(pa+qc)\textbf u+(pb+qd)\textbf v=\textbf x$ for any arbitrary vector $\textbf x$.
And done
