1. The Central Question
When can one vector be expressed through the others?
Imagine you have a family of vectors $\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}$. A natural question arises: is any one of these vectors redundant — that is, can it be written as a combination of the remaining ones?
For example, if $\mathbf{v}_3 = 2\mathbf{v}_1 - \mathbf{v}_2$, then $\mathbf{v}_3$ is "dependent" on $\mathbf{v}_1$ and $\mathbf{v}_2$. But this formulation privileges $\mathbf{v}_3$ — it singles out one vector as the "dependent" one. We can rewrite the same relationship symmetrically:
$$ 2\mathbf{v}_1 - \mathbf{v}_2 - \mathbf{v}_3 = \mathbf{0} $$Now no vector is singled out. This leads to the symmetric definition that treats all vectors equally.
📌 Core Definitions
Linear Dependence
A family of vectors $\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}$ is linearly dependent if there exist scalars $c_1, c_2, \ldots, c_k$, not all zero, such that:
$$ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0} $$Linear Independence
A family of vectors $\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}$ is linearly independent if the equation
$$ c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0} $$has only the trivial solution $c_1 = c_2 = \cdots = c_k = 0$.
- Dependence = there exists at least one choice of coefficients, not all zero, that makes the combination vanish.
- Independence = the only choice of coefficients that makes the combination vanish is the all-zero choice.
The trivial combination (all $c_i = 0$) always gives $\mathbf{0}$ for any set of vectors. That is not evidence of dependence! Dependence requires a nontrivial combination that yields $\mathbf{0}$.