---
title: "Vectors, spans and dependence"
description: "Decide whether vectors are linearly independent by checking if only zero coefficients give the zero vector, with parallel, zero and three-vector cases."
canonical_url: "https://fanout.sh/ml-math/lessons/spaces-linear-dependence-and-independence-t-0caozq9"
md_url: "https://fanout.sh/ml-math/lessons/spaces-linear-dependence-and-independence-t-0caozq9.md"
access: "public"
---

# Vectors, spans and dependence

Decide whether vectors are linearly independent by checking if only zero coefficients give the zero vector, with parallel, zero and three-vector cases.

Use vector coordinates to describe everything a collection can generate, decide when a vector is redundant, and distinguish two different kinds of multiplication.

## Can one vector be made from the others?

A linear combination scales vectors by real coefficients and adds them. Their span is the set of all such combinations. If one listed vector is already a combination of the others, including it does not enlarge the span. We call this redundancy linear dependence. Work with column vectors in $\mathbb R^3$, the space of three real coordinates; tuples below are shorthand for columns.

$$u=(1,0,1),\qquad v=(0,1,1),\qquad w=(1,2,3)=u+2v$$

Every combination of u, v and w can be rewritten using just u and v: $au+bv+cw=(a+c)u+(b+2c)v$. The vector w therefore adds no new reachable points. Move the terms of $w=u+2v$ to one side: $u+2v-w=0$. Here 0 means the zero vector $(0,0,0)$. The coefficients $(1,2,-1)$ are not all zero.

$$c_1v_1+\cdots+c_kv_k=0$$

For k vectors of the same length, ask which real coefficients make their combination the zero vector.

The vectors are linearly independent if the only solution is $c_1=\cdots=c_k=0$, called the trivial solution. They are linearly dependent if there is any solution with at least one nonzero coefficient. That does not require every coefficient to be nonzero. To see why the test detects redundancy, choose a nonzero coefficient in a zero relation, move the other terms across, and divide by that coefficient: its vector becomes a combination of the others.

### Solve the whole zero-combination equation

1. For our u, v and w, $au+bv+cw=0$ gives $a+c=0$, $b+2c=0$, and $a+b+3c=0$.
2. The first two equations give $a=-c$ and $b=-2c$. Substitution makes the third $-c-2c+3c=0$, so it places no further restriction.
3. Let $c=t$ be any real number. All coefficient solutions are $(a,b,c)=(-t,-2t,t)$. Taking $t=1$ gives $-u-2v+w=0$, a nontrivial relation proving dependence.

The pair u, v is independent: $au+bv=(a,b,a+b)=0$ forces $a=0$ and $b=0$. Three vectors can also be independent. With $e_3=(0,0,1)$, the equation $au+bv+ce_3=0$ forces $a=0$, $b=0$, then $c=0$. Trying one proposed relation and finding that it fails would not prove independence; we must rule out every nontrivial solution.

The pair shortcut has a limit

Two vectors are dependent exactly when one is a scalar multiple of the other. For example, $2u$ and u satisfy $1(2u)-2u=0$. For three or more vectors, pairwise checks are insufficient: none of u, v, w is a scalar multiple of another, yet $u+2v-w=0$. A collection containing a zero vector is always dependent: put coefficient 1 on that zero vector and 0 on every other vector.

The same test works in any number of coordinates. Put the vectors in the columns of A and solve the homogeneous system $Ac=0$, where c holds the coefficients. A pivot in every column leaves only the trivial solution and proves independence. A free variable permits a nontrivial solution and proves dependence. This system is always consistent because the all-zero coefficient choice always works.

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