Linear Independence

No Redundant Vectors

A set of vectors is linearly independent if none of them can be written as a linear combination of the others — no vector is redundant. The precise definition uses the zero vector: $v_1, \dots, v_k$ are independent when the only way to have $$c_1 v_1 + c_2 v_2 + \dots + c_k v_k = \mathbf{0}$$ is with all coefficients zero ($c_1 = c_2 = \dots = c_k = 0$). If some nontrivial combination (not all zero) gives $\mathbf{0}$, the vectors are linearly dependent — and then one of them can be solved for in terms of the others, exposing the redundancy.

Some quick tests and facts:

• Two vectors are dependent exactly when one is a scalar multiple of the other (parallel).
• Any set containing the zero vector is automatically dependent (take its coefficient to be 1, the rest 0).
• In $\mathbb{R}^n$, any set of more than $n$ vectors is automatically dependent — you cannot have more than $n$ independent directions in $n$-dimensional space.

Checking independence is, once again, a linear system: $c_1 v_1 + \dots + c_k v_k = \mathbf{0}$ is a homogeneous system in the $c_i$. The vectors are independent exactly when this system has only the trivial (all-zero) solution — which happens precisely when the matrix with those vectors as columns has full column rank (a pivot in every column).

Independence is the key structural property: it means "no wasted vectors," which combined with spanning gives the notion of a basis in the next lesson.

Common pitfall: reading the definition as "no vector is a multiple of another." That is the test only for a set of two vectors. With three or more, a set can be dependent even though no single vector is a scalar multiple of another — e.g. a third vector might be the sum of the first two. The correct definition is that the only combination equal to $\mathbf{0}$ is the all-zero one.

Three vectors from the origin where the accent third vector is drawn as exactly the sum of the first two (dependent), beside three vectors pointing in genuinely different directions (independent).

$$c_1 v_1 + \dots + c_k v_k = \mathbf{0} \Rightarrow \text{all } c_i = 0$$