Linear dependence and linear independence I have perused many texts, but cannot seem to get to the root of the problem:
The concepts of linear dependence and linear independence are really throwing me in a loop. Why are those terms called that way? What do they have to do with the way they are defined?
I would like to know these things in reference to the context of the equation $a_1u_1 + a_2u_2 + ... + a_nu_n = 0$ Thank you.
 A: Well, let's start with the case where you want to show that two vectors are linearly dependent. It means that if you write $\alpha_1 \mathbf{x}_1 + \alpha_2 \mathbf{x}_2 = 0$ then at least $\alpha_1$ or $\alpha_2$ is non-zero. Suppose that $\alpha_1$ is non-zero. Since it's non-zero, you can write $\alpha_1 \mathbf{x}_2 = -\alpha_2 \mathbf{x}_2$ and then divide by $\alpha_1 \neq 0$ to get $\mathbf{x}_1 = - \alpha_2 / \alpha_1 .\mathbf{x}_2$. That means $\mathbf{x}_2$ lies on the same line as $\mathbf{x}_1$.
You can reason the same way with $\alpha_1 \mathbf{x}_1 + \alpha_2 \mathbf{x}_2 + \alpha_3 \mathbf{x}_3 = 0$. This time if they are linearly dependent at least one of the coefficients is non-zero, and that vector will be on the same plane as the other two.
This is where the terminology 'linear dependence' comes from. But linear independence is important from the algebraic viewpoint as well. Linear independence enables you to define a basis together with the concept of a spanning set. A spanning set enables you to write every vector in a space as the linear combination of some of them. Linear independence tells you that if you find such a spanning set where all the vectors in the set are linearly independent from each other, then you can write every vector in the space 'uniquely' as a linear combination of them. So, the linear independence of the vectors in the spanning set guarantees the uniqueness of our representation.
For example if $\{\mathbf{x}_1, \cdots, \mathbf{x}_n\}$ is a spanning set of vectors and $\mathbf{x}$ is a vector in our vector space V then we can write $\mathbf{x}$ as $\mathbf{x} = \alpha_1.\mathbf{x}_1 + \cdots + \alpha_n.\mathbf{x}_n$. If we have another representation like $\mathbf{x} = \beta_1.\mathbf{x}_1 + \cdots + \beta_n.\mathbf{x}_n$ then if $\{\mathbf{x}_1, \cdots, \mathbf{x}_n\}$ is a linearly independent set of vectors then by definition we can reason as follows:
$\mathbf{x} = \alpha_1.\mathbf{x}_1 + \cdots + \alpha_n.\mathbf{x}_n$ and $\mathbf{x} = \beta_1.\mathbf{x}_1 + \cdots + \beta_n.\mathbf{x}_n$ $\implies$
$\mathbf{0} = (\beta_1-\alpha_1).\mathbf{x}_1 + \cdots + (\beta_n-\alpha_n).\mathbf{x}_n \implies \beta_1 - \alpha_1 = 0 , \cdots, \beta_n-\alpha_n=0 \implies$ 
$\alpha_1 = \beta_1,\cdots,\alpha_n=\beta_n$ 
A: If $\mathbf{z} \ne 0$ is linearly dependent on $\mathbf{x}$ and $\mathbf{y}$ this means that there exist real numbers $a,b$ such that
$\mathbf{z} = a\mathbf{x} + b\mathbf{z}. \tag{1}$
If $\mathbf{z} \ne 0$ is linearly independent of $\mathbf{x}$ and $\mathbf{y}$ then no $a,b \in \mathbb{R}$ can be found such that (1) holds. This can be extended to more elements in the following way.
If $\mathbf{z} \ne 0$ is linearly dependent on $\{\mathbf{x}_{1}, \ldots \mathbf{x}_{n}\}$ then there exist real numbers $\{a_{1}, \ldots, a_{n}\}$ such that
$\mathbf{z} = \displaystyle \sum_{i=1}^{n}a_{i}\mathbf{x}_{i}$
and if $\mathbf{z}$ is linearly independent of $\{\mathbf{x}_{1}, \ldots \mathbf{x}_{n}\}$ then no such $\{a_{1}, \ldots, a_{n}\}$ can be found.
