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# 19a Proof of a characterization of linear independence.
Let us prove this characterization:
> **A characterization of linear independence.**
> A list of $k\ge 2$ vectors $v_{1},v_{2},\ldots,v_{k}$ is linearly independent if and only if no one vector is a linear combination of the others.
We will use just the definition of these words. It is in this sense that a linear independent set has no redundant vectors -- that no one vector can be made out of some linear combination of the others. Since this is a characterization (if and only if), we need to show both directions. This time, I will demonstrate them with "proof by contradiction".
$\blacktriangleright$ Proof.
$(\implies)$ Suppose $v_{1},v_{2},\ldots,v_{k}$ is a list of $k\ge 2$ linearly independent vectors. We wish to show no one vector is a linear combination of the others.
**Suppose to the contrary** that one of the vectors is a linear combination of the others, let us say, without loss $v_{1}$ is a linear combination of $v_{2},\ldots,v_{k}$. This means we have $$
v_{1}=c_{2}v_{2}+\cdots + c_{k}v_{k}
$$But we see that this means $$
1v_{1}-c_{2}v_{2}-\cdots -c_{k}v_{k}=\mathbf{0},
$$which is a nontrivial linear combination for the zero vector, as the coefficient in front of $v_{1}$ is $1\neq 0$. So this implies the list is in fact linearly dependent, a contradiction!
Hence no one vector is a linear combination of the others.
$(\impliedby)$ Suppose in the list of $k\ge 2$ vectors $v_{1},\ldots,v_{k}$, no one vector is a linear combination of the others. We wish to show that this list is linearly independent.
**Suppose to the contrary** that this list is in fact linearly dependent. That means there is a nontrivial linear combination $$
c_{1}v_{1}+c_{2}v_{2} + \cdots +c_{k}v_{k}=\mathbf{0}
$$where some $c_{j}\neq 0$. Then we can solve for $v_j$ and get $$
v_{j}= -\frac{1}{c_{j}}\underbrace{(c_{1}v_{1}+\cdots +c_{k}v_{k})}_{\text{without \(v_j\)}},
$$which shows $v_{j}$ is a linear combination of the others, a contradiction!
Hence this list of vectors $v_{1},\ldots,v_{k}$, is linearly independent.
We have as claimed. $\blacksquare$
**Sort of important but not really but not not really remark.** Here the $k \ge 2$ condition is just so the term "others" make sense -- that having at least two vectors allows you to talk about the other vectors from one of them. However we can also relax the statement to $k \ge 1$ vectors. In this case, there are no "other vectors" from one of the vectors. If we take the linear combination of no vectors as the zero vector (which we do by defining span of nothing to be the appropriate zero vector), the argument above still hold. And if we go further, even the statement $k\ge 0$ make sense. When there is no vector, an empty list or set is automatically defined as linearly independent, and the statement "no one vector is a linear combination of the other" is vacuously true for an empty set. Anyway, edge case consideration is important, but it can be a nuance.