How the formal definition works... I'm not sure how a linear combination constructed in a certain way can tell you the variables are independent or not.

Perhaps a more intutive definition of linear dependence is as follows:

A set of vectors is linearly dependent if one of its vectors is a linear combination of the other vectors.

Loosely speaking, this means that some / all of the vectors lie in a plane / line.

Why set the linear combination equations to $\mathbf{0}$. I don't see how setting to zero helps determine independence or not.

The definition above can be deduced from the formal definition:

If $\{ \mathbf{v_1}, \dots, \mathbf{v_n} \}$ is linearly dependent, then there exists some non-zero $\alpha_1, \dots, \alpha_n$ such that

$$ \alpha_1 \mathbf{v_1} + \alpha_2 \mathbf{v_2} + \dots + \alpha_n \mathbf{v_n} = \mathbf{0} $$

Without loss of generality, suppose $\alpha_1 \neq 0$ then we have

$$ \mathbf{v_1} = - \frac{\alpha_2}{\alpha_1} \mathbf{v_2} - \dots - \frac{\alpha_n}{\alpha_1} \mathbf{v_n} $$

So $\mathbf{v_1}$ is a linear combination of the other vectors.

Why choose $\alpha_k$ to be non-zero in one case. It seems arbitrary.

In the proof above, we need not work with $\alpha_1$. We could have chosen any non-zero $\alpha_k$ to show that $\mathbf{v_k}$ is a linear combination of the other vectors.

How the formal definition works... I'm not sure how a linear combination constructed in a certain way can tell you the variables are independent or not.

Perhaps a more intutive definition of linear dependence is as follows:

A set of vectors is linearly dependent if one of its vectors is a linear combination of the other vectors.

In other words, one of its vectors "depends linearly" on the other vectors. This definition can be deduced from the formal definition:

If $\{ \mathbf{v}_1, \dots, \mathbf{v}_n \}$ is linearly dependent, then there exists some non-zero $\alpha_1, \dots, \alpha_n$ such that

$$ \alpha_1 \mathbf{v}_1 + \alpha_2 \mathbf{v}_2 + \dots + \alpha_n \mathbf{v}_n = \mathbf{0} $$

Without loss of generality, suppose $\alpha_1 \neq 0$ then we have

$$ \mathbf{v}_1 = - \frac{\alpha_2}{\alpha_1} \mathbf{v}_2 - \dots - \frac{\alpha_n}{\alpha_1} \mathbf{v}_n $$

So $\mathbf{v}_1$ is a linear combination of the other vectors.

Why set the linear combination equations to $\mathbf{0}$. I don't see how setting to zero helps determine independence or not.

In the proof above, we did not set the right-hand side to zero. It is zero only because everything is moved to the left-hand side.

Why choose $\alpha_k$ to be non-zero in one case. It seems arbitrary.

In the proof above, we need not work with $\alpha_1$. We could have chosen any non-zero $\alpha_k$ to show that $\mathbf{v}_k$ is a linear combination of the other vectors.

How the formal definition works... I'm not sure how a linear combination constructed in a certain way can tell you the variables are independent or not.

Perhaps a more intutive definition of linear dependence is as follows:

A set of vectors is linearly dependent if one of its vectors is a linear combination of the other vectors.

Why set the linear combination equations to $\mathbf{0}$. I don't see how setting to zero helps determine independence or not.

The definition above can be deduced from the formal definition:

If $\{ \mathbf{v_1}, \dots, \mathbf{v_n} \}$ is linearly dependent, then there exists some non-zero $\alpha_1, \dots, \alpha_n$ such that

$$ \alpha_1 \mathbf{v_1} + \alpha_2 \mathbf{v_2} + \dots + \alpha_n \mathbf{v_n} = \mathbf{0} $$

Without loss of generality, suppose $\alpha_1 \neq 0$ then we have

$$ \mathbf{v_1} = - \frac{\alpha_2}{\alpha_1} \mathbf{v_2} - \dots - \frac{\alpha_n}{\alpha_1} \mathbf{v_n} $$

So $\mathbf{v_1}$ is a linear combination of the other vectors.

Why choose $\alpha_k$ to be non-zero in one case. It seems arbitrary.

Referring to the proof above, if $\alpha_1 = 0$ then we can find a non-zero $\alpha_k$ and repeat the proof to show that $\mathbf{v_k}$ is a linear combination of the other vectors.

How the formal definition works... I'm not sure how a linear combination constructed in a certain way can tell you the variables are independent or not.

Perhaps a more intutive definition of linear dependence is as follows:

A set of vectors is linearly dependent if one of its vectors is a linear combination of the other vectors.

Loosely speaking, this means that some / all of the vectors lie in a plane / line.

Why set the linear combination equations to $\mathbf{0}$. I don't see how setting to zero helps determine independence or not.

The definition above can be deduced from the formal definition:

If $\{ \mathbf{v_1}, \dots, \mathbf{v_n} \}$ is linearly dependent, then there exists some non-zero $\alpha_1, \dots, \alpha_n$ such that

$$ \alpha_1 \mathbf{v_1} + \alpha_2 \mathbf{v_2} + \dots + \alpha_n \mathbf{v_n} = \mathbf{0} $$

Without loss of generality, suppose $\alpha_1 \neq 0$ then we have

$$ \mathbf{v_1} = - \frac{\alpha_2}{\alpha_1} \mathbf{v_2} - \dots - \frac{\alpha_n}{\alpha_1} \mathbf{v_n} $$

So $\mathbf{v_1}$ is a linear combination of the other vectors.

Why choose $\alpha_k$ to be non-zero in one case. It seems arbitrary.

In the proof above, we need not work with $\alpha_1$. We could have chosen any non-zero $\alpha_k$ to show that $\mathbf{v_k}$ is a linear combination of the other vectors.