# Proof that two vectors, not a scalar multiple of one another, are linearly independent

I have the following statement:

If $\ \vec{x}, \vec{y} \in \mathbb R^n$ such that neither $\,\vec x$ nor $\,\vec y$ is a scalar multiple of the other, then $\{\vec x, \vec y\}$ is linearly independent.

How would I go about proving this? It seems really hard to prove something that is obviously true.

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Suppose not. Then you can find scalars $c_1$ and $c_2$ with $c_1^2+c_2^2\neq 0$ s.t. $c_1x+c_2y=0$. WLOG assume $c_1\neq 0$. This tells you $x=-\frac{c_2y}{c_1}$ a contradiction since neither vector is a scalar multiple of the other.
$c_1^2+c_2^2\neq 0$ Where did that come from? I don't remember learning that. AFAIK, the only condition is that $c_1 and$c_2 not equal to zero. – No_name Jan 16 '13 at 4:49
When $c_1, c_2$ are real numbers, the condition that $c_1$ and $c_2$ are not both $0$ is equivalent to $c_1^2 + c_2^2 \neq 0$. However, this is a bit too clever, because the two conditions are no longer equivalent over other scalar fields such as the complex numbers (which are usually introduced at the end of a one-semester linear algebra course when dealing with eigenvector and eigenvalue problems). – Michael Joyce Jan 16 '13 at 4:53
are not both zero right? a short-hand of saying $(a,b)\neq (0,0)$ is $a^2+b^2\neq 0$. You can verify it for yourself, its pretty entertaining. And as @MichaelJoyce points out, please avoid this notation unless you are certain that you are working with real numbers. – mathemagician Jan 16 '13 at 4:53