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Show $\{v_1, v_2 \}$ is Linearly Independent iff neither $v_1$ or $v_2$ is a multiple of each other. (where $v_1, v_2$ are vectors).

Going forward. Assume {$v_1, v_2$} is LI. Thus, if $c_1v_1 + c_2v_2 = 0$ $c_1 = c_2 = 0$. Assume for sake of contradiction that $v_1 = dv_2, d \in \mathbb{R}$. Thus $v_2 (c_2 + dc_1) = 0$, since $v_2 \ne 0$, it must be $c_2 + dc_1 = 0$ thus $d = c_2/-c_1$, but since $c_1 = c_2 = 0$, we get $d = 0/0$ undefined, contradiction.

The other direction, prove the contrapositive.

{$v_1, v_2$} is LD thus case where $c_1 = 0$ only, $c_2 = 0$ only, and both $c_1, c_2$ not zero.

Proving first two cases is trivial as we get $c_2v_2 = 0 = v_1$ and otherwise $c_1v_2 = 0 = v_2$.

Assume $c_1, c_2$ non zero, then we have $v_1 = -c_2/c_1v_2$, which is a multiple of $v_2$.

Is the proof complete?

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I don't like the forward direction. Showing that some set of calculations requires you to divide zero by zero is not a contradiction. Instead, observe that if $\nu_1 = d\nu_2$ then $\nu_1-d\nu_2=0$ is a linear combination with coefficients not all zero that equals zero, violating LI.

The other direction looks correct.

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$\{v_1, v_2 \}$ is linearly independent $\Leftrightarrow $ neither $v_1$ or $v_2$ is a multiple of each other.

Proof.

$\Rightarrow)$ (By contradiction). Without loss of generality, if $v_2=\lambda v_1$ then $(-\lambda)v_1+\underbrace{1}_{\ne 0}v_2=0$ and $\{v_1, v_2 \}$ is linearly dependent (contradiction).

$\Leftarrow)$ (By contradiction). If $\{v_1, v_2 \}$ is linearly dependent then, $\lambda_1v_1+\lambda_2v_2=0$ with $\lambda_1\ne 0$ or $\lambda_2\ne 0.$ Suppose without loss of generality that $\lambda_1\ne 0.$ Then, $$\lambda_1v_1+\lambda_2v_2=0\Rightarrow \lambda_1^{-1}\left(\lambda_1v_1+\lambda_2v_2\right)=\lambda_1^{-1}0$$ $$\Rightarrow \lambda_1^{-1}\lambda_1v_1+\lambda_1^{-1}\lambda_2v_2=0\Rightarrow v_1=\left(-\lambda_1^{-1}\lambda_2\right)v_2\text{ (contradiction).}$$

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