Proving linear dependence iff $x \neq 0, a = xb$

Question

I need to prove linear dependence iff $x \neq 0, a = xb$.

What I have for the forward arrow:

Assume $a, b$ are linearly dependent. Then there exists $x_1, x_2 \in \mathbb{R}$ that are nonzero such that $x_1u + x_2v = 0$. Assume $x_1 \neq 0$. Then:

$$ x_1a + x_2b = 0 $$

$$x_1a = -x_2b $$

$$a = (-x_2/x_1)b$$

Let $x = (-x_2/x_1)$

$$ a = xb $$

First off, how is the proof for the first part of the proof? For the second part, I am unsure how to go about and prove it. Any help would be appreciated.

Edit(after comments):

Assume $a = xb$, for some $x \in \mathbb{R}$. Then $1a + -xb = 0$. In other words, $a$ and $b$ are linearly dependent.

Is this what it is supposed to look like?

Answer 1

Correct statement:

If $a$ and $b$ are vectors in a vector space $V$, then $\{a,b\}$ is linearly dependent if and only if $a=\alpha b$, in which $\alpha \in \mathbb{R}$.

Proof. If the vectors are LD, then there are scalars $\beta$ and $\gamma$ (not both zero) such that $$ \beta a + \gamma b =0_v. $$

Without losing generality, assume that $\beta \ne 0$. Then $$ a = -(\gamma/\beta) b. $$ Select $\alpha = -(\gamma/\beta)$.

Conversely , if $a=\alpha b$, then $1a + (-\beta)b =0$ is a linear dependence relation.