Linear Dependence and Multiples

To prove that $v_1$ and $v_2$ are linearly dependent if and only if one is a scalar multiple of the other, we'll address both implications of the "if and only if" statement.

Forward Direction (⇒):

Assume $v_1$ and $v_2$ are linearly dependent.

By definition, this means there exist scalars $c_1$ and $c_2$, not both zero, such that:

$$c_1 v_1 + c_2 v_2 = 0.$$

We consider two cases based on which scalar is nonzero.

1. If $c_1 \ne 0$:

   $$c_1 v_1 = -c_2 v_2 \implies v_1 = -\frac{c_2}{c_1} v_2.$$

   Hence, $v_1$ is a scalar multiple of $v_2$.

2. If $c_2 \ne 0$:

   $$c_2 v_2 = -c_1 v_1 \implies v_2 = -\frac{c_1}{c_2} v_1.$$

   Thus, $v_2$ is a scalar multiple of $v_1$.

In either case, one vector is a scalar multiple of the other.

Backward Direction (⇐):

Assume one vector is a scalar multiple of the other.

Without loss of generality, suppose $v_1 = \lambda v_2$ for some scalar $\lambda$. Then:

$$v_1 - \lambda v_2 = 0.$$

This can be rewritten as:

$$1 \cdot v_1 + (-\lambda) v_2 = 0.$$

Since the scalars $1$ and $-\lambda$ are not both zero, this shows that $v_1$ and $v_2$ are linearly dependent.

Similarly, if $v_2 = \mu v_1$, we can show linear dependence in the same way.

Conclusion:

$v_1$ and $v_2$ are linearly dependent if and only if one is a scalar multiple of the other.

Answer: Because one is a scalar multiple of the other exactly when they are linearly dependent.