6.4 Linear dependence and independence

Why this matters

Suppose a collection of vectors spans a space. If one vector can already be built from the others, it is redundant. Linear dependence and independence are the tools that detect that redundancy.

Definition

Linear independence

Vectors $u_1, u_2, \ldots, u_n$ are linearly independent if

\alpha_1 u_1 + \alpha_2 u_2 + \cdots + \alpha_n u_n = 0

forces $\alpha_1 = \alpha_2 = \cdots = \alpha_n = 0$ .

Definition

Linear dependence

Vectors $u_1, u_2, \ldots, u_n$ are linearly dependent if there is a nontrivial choice of scalars, not all zero, such that

\alpha_1 u_1 + \alpha_2 u_2 + \cdots + \alpha_n u_n = 0.

Use the checker below to test whether a small list of vectors behaves like a redundant family or a genuinely independent one.

Read and try

Test one set for dependence

The live checker compares small vector sets and explains whether a nontrivial linear relation exists.

Verdict

Independent

The only way to solve c1e1 + c2e2 = 0 is c1 = c2 = 0, so this pair is linearly independent.

Key relation

No nontrivial linear relation appears.

A simple reading rule

The notes give two very useful consequences:

a dependent family has at least one vector that is a linear combination of the others;
an independent family has no such vector.

That is the fastest way to read the idea in practice.

Worked example

A dependent family has a relation

The notes give

u_3 = u_1 + u_2.

Rewriting this gives

u_1 + u_2 - u_3 = 0.

That is a nontrivial linear relation, so the family is linearly dependent.

Proof

Why dependence means one vector is redundant

Common mistake

Dependent does not mean useless

A dependent list can still span a space. It only means the list has redundancy. That is why dependence is useful when we want to trim a spanning set down.

Matrix criterion and pivot criterion

Suppose we place $u_1,\dots,u_n$ as columns of a matrix

A = [u_1\ u_2\ \cdots\ u_n].

Then a relation

\alpha_1 u_1 + \cdots + \alpha_n u_n = 0

is exactly the same as solving

A\alpha = 0,\quad \alpha = (\alpha_1,\dots,\alpha_n)^T.

So linear independence means the homogeneous system has only the trivial solution.

Theorem

Equivalent matrix test for independence

The vectors $u_1,\dots,u_n$ are linearly independent if and only if the matrix $A=[u_1\ \cdots\ u_n]$ has a pivot in every column (equivalently, no free variable in $A\alpha=0$ ).

Worked example

Use row reduction to test independence

Let

u_1=\begin{bmatrix}1\\2\\1\end{bmatrix},\quad u_2=\begin{bmatrix}2\\4\\3\end{bmatrix},\quad u_3=\begin{bmatrix}1\\1\\0\end{bmatrix}.

Set $A=[u_1\ u_2\ u_3]$ . Row reduce:

\begin{bmatrix} 1 & 2 & 1\\ 2 & 4 & 1\\ 1 & 3 & 0 \end{bmatrix} \xrightarrow{R_2-2R_1,\ R_3-R_1} \begin{bmatrix} 1 & 2 & 1\\ 0 & 0 & -1\\ 0 & 1 & -1 \end{bmatrix} \xrightarrow{R_2\leftrightarrow R_3} \begin{bmatrix} 1 & 2 & 1\\ 0 & 1 & -1\\ 0 & 0 & -1 \end{bmatrix}.

There is a pivot in each column, so the three vectors are linearly independent.

Quick check

Are $e_1, e_2, e_3$ in $R^3$ linearly independent?

Try the vector equation $\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3 = 0$ .

Solution

Answer

Quick check

Can a set that contains the zero vector be linearly independent?

Test the definition directly.

Solution

Answer

Quick check

If a set has 5 vectors in $R^3$ , can it be linearly independent?

Use pivots and the number of rows.

Solution

Answer

Exercises

Quick check

Determine whether `{(1,0,1),(2,1,3),(0,1,1)}` is independent.

Set them as columns, reduce, and read pivot columns.

Solution

Guided solution

Quick check

Show that `{(1,2,3),(2,4,6),(1,0,1)}` is dependent by writing one vector from others.

Look for a direct scalar-multiple relation first.

Solution

Guided solution

Dependence means redundancy

Theorem

Equivalent redundancy test

A list of vectors is linearly dependent if and only if one vector in the list can be written as a linear combination of the others.

Proof

Why dependence and redundancy are the same

Worked example

A redundant vector can be removed without changing the span

Let

u_1 = \begin{bmatrix}1\\2\\1\end{bmatrix},\qquad u_2 = \begin{bmatrix}0\\1\\1\end{bmatrix},\qquad u_3 = \begin{bmatrix}1\\3\\2\end{bmatrix}.

Here $u_3 = u_1 + u_2$ . So every vector of the form

a u_1 + b u_2 + c u_3

can be rewritten as

(a+c)u_1 + (b+c)u_2.

Therefore

\operatorname{Span}\{u_1,u_2,u_3\} = \operatorname{Span}\{u_1,u_2\}.

The third vector changes the description, but not the span itself.

Theorem

A redundant vector can be removed without changing the span

If one vector in a list is a linear combination of the others, then deleting it does not change the span of the list.

Proof

Why the span stays the same

Common mistake

Dependent does not mean the span gets smaller

A dependent list can still span a whole space. Dependence only says that at least one vector is unnecessary for generating the span.

Column-matrix criterion and null-space viewpoint

Put the vectors into a matrix

A = [u_1\ u_2\ \cdots\ u_n].

Then the relation

\alpha_1u_1+\cdots+\alpha_nu_n=0

is exactly the homogeneous system $A\alpha=0$ , where $\alpha=(\alpha_1,\ldots,\alpha_n)^T$ .

Theorem

Matrix test for dependence

The vectors $u_1,\dots,u_n$ are linearly independent if and only if the homogeneous system $A\alpha=0$ has only the trivial solution. Equivalently, the null space N(A) contains only 0.

Proof

Why the homogeneous system controls dependence

Worked example

Read a dependence relation from row reduction

Take

A= \begin{bmatrix} 1 & 0 & 1\\ 2 & 1 & 3\\ 1 & 1 & 2 \end{bmatrix},

whose columns are $u_1=(1,2,1)^T$ , $u_2=(0,1,1)^T$ , and $u_3=(1,3,2)^T$ . Row reduction gives

\begin{bmatrix} 1 & 0 & 1\\ 2 & 1 & 3\\ 1 & 1 & 2 \end{bmatrix} \xrightarrow{R_2-2R_1,\ R_3-R_1} \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 1\\ 0 & 1 & 1 \end{bmatrix} \xrightarrow{R_3-R_2} \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 1\\ 0 & 0 & 0 \end{bmatrix}.

The third column is not a pivot column, so $\alpha_3$ is free in $A\alpha=0$ . Set $\alpha_3=1$ . Then the reduced system gives $\alpha_1=-1$ and $\alpha_2=-1$ , so

-u_1-u_2+u_3=0,

or equivalently $u_3=u_1+u_2$ .

Theorem

Pivot criterion

After row reducing $A$ , the list $u_1,\dots,u_n$ is linearly independent if and only if every column is a pivot column. If one column is not a pivot column, then there is a free variable in $A\alpha=0$ , so a nontrivial relation exists.

Common mistake

Row reduction is allowed because it preserves the homogeneous solution set

When you row reduce $A$ for independence testing, you are not changing the columns themselves into new vectors you want to study. You are simplifying the equation $A\alpha=0$ . Row-equivalent matrices have the same homogeneous solutions, so they have the same dependence relations.

Fast low-dimensional tests

Theorem

Two nonzero vectors are independent exactly when neither is a scalar multiple of the other

For a pair {u,v} with both vectors nonzero, the set is linearly independent if and only if one vector is not a scalar multiple of the other.

Proof

Why two vectors reduce to a scalar-multiple test

Worked example

Two vectors in $R^2$

Let

u = \begin{bmatrix}1\\2\end{bmatrix}, \qquad v = \begin{bmatrix}2\\4\end{bmatrix}.

Since $v=2u$ , the pair is dependent. Geometrically, both vectors point in the same direction.

Worked example

Three vectors in $R^2$ must be dependent

If $u_1,u_2,u_3 \in R^2$ , then the matrix $A=[u_1\ u_2\ u_3]$ has size $2\times 3$ . After row reduction it can have at most two pivots, so one column is not a pivot column. Therefore $A\alpha=0$ has a nontrivial solution and the three vectors are linearly dependent.

Worked example

Three vectors in $R^3$ that lie in one plane

If three vectors in $R^3$ all lie in the plane $z=0$ , then they all belong to the span of $e_1$ and $e_2$ . But that plane is already generated by two independent directions, so a third vector cannot add a new independent direction. The list is dependent.

Theorem

Too many vectors in $R^m$ force dependence

Any list of more than m vectors in $R^m$ is linearly dependent.

Proof

Why more than `m` vectors cannot be independent in $R^m$

Quick checks

Quick check

Is any list containing the zero vector automatically dependent?

Test the definition directly.

Solution

Answer

Quick check

If $u_3 = u_1 + u_2$ , does $\{u_1,u_2,u_3\}$ remain independent?

Write the relation in the standard form $\alpha_1u_1+\alpha_2u_2+\alpha_3u_3=0$ .

Solution

Answer

Quick check

If a $4\times 4$ matrix has four pivot columns, what does that say about its columns?

Use the matrix test.

Solution

Answer

Guided exercises

Quick check

Decide whether `\{(1,1,0),(0,1,1),(1,2,1)\}` is independent, and if not, write one dependence relation.

Set the vectors as columns and look for a free variable after row reduction.

Solution

Guided solution

Quick check

Explain why any subset of a linearly independent set is linearly independent.

Think about what would happen if the subset had its own relation.

Solution

Guided solution

Quick check

If one vector in a list is redundant, what should you try next?

Use the redundancy viewpoint from this page.

Solution

Guided solution

Read this first

This page builds on 6.3 Linear combinations and span and the row-reduction language from 2.3 Gaussian elimination and RREF.

6.4 Linear dependence and independence

MATH1030: Linear algebra I

Why this matters

Linear independence

Linear dependence

Test one set for dependence

A simple reading rule

A dependent family has a relation

Why dependence means one vector is redundant

Common mistake

Dependent does not mean useless

Matrix criterion and pivot criterion

Equivalent matrix test for independence

Use row reduction to test independence

Quick check

Are e1,e2,e3e_1, e_2, e_3e1​,e2​,e3​ in R3R^3R3 linearly independent?

Answer

Can a set that contains the zero vector be linearly independent?

Answer

If a set has 5 vectors in R3R^3R3, can it be linearly independent?

Answer

Exercises

Determine whether {(1,0,1),(2,1,3),(0,1,1)} is independent.

Guided solution

Show that {(1,2,3),(2,4,6),(1,0,1)} is dependent by writing one vector from others.

Guided solution

Dependence means redundancy

Equivalent redundancy test

Why dependence and redundancy are the same

A redundant vector can be removed without changing the span

A redundant vector can be removed without changing the span

Why the span stays the same

Dependent does not mean the span gets smaller

Column-matrix criterion and null-space viewpoint

Matrix test for dependence

Why the homogeneous system controls dependence

Read a dependence relation from row reduction

Pivot criterion

Row reduction is allowed because it preserves the homogeneous solution set

Fast low-dimensional tests

Two nonzero vectors are independent exactly when neither is a scalar multiple of the other

Why two vectors reduce to a scalar-multiple test

Two vectors in R2R^2R2

Three vectors in R2R^2R2 must be dependent

Three vectors in R3R^3R3 that lie in one plane

Too many vectors in RmR^mRm force dependence

Why more than m vectors cannot be independent in RmR^mRm

Quick checks

Is any list containing the zero vector automatically dependent?

Answer

If u3=u1+u2u_3 = u_1 + u_2u3​=u1​+u2​, does {u1,u2,u3}\{u_1,u_2,u_3\}{u1​,u2​,u3​} remain independent?

Answer

If a 4×44\times 44×4 matrix has four pivot columns, what does that say about its columns?

Answer

Guided exercises

Decide whether \{(1,1,0),(0,1,1),(1,2,1)\} is independent, and if not, write one dependence relation.

Guided solution

Explain why any subset of a linearly independent set is linearly independent.

Guided solution

If one vector in a list is redundant, what should you try next?

Guided solution

Read this first

Section mastery checkpoint

Prerequisites

Key terms in this unit

More notes in this series

Are $e_1, e_2, e_3$ in $R^3$ linearly independent?

If a set has 5 vectors in $R^3$ , can it be linearly independent?

Determine whether `{(1,0,1),(2,1,3),(0,1,1)}` is independent.

Show that `{(1,2,3),(2,4,6),(1,0,1)}` is dependent by writing one vector from others.

Two vectors in $R^2$

Three vectors in $R^2$ must be dependent

Three vectors in $R^3$ that lie in one plane

Too many vectors in $R^m$ force dependence

Why more than `m` vectors cannot be independent in $R^m$

If $u_3 = u_1 + u_2$ , does $\{u_1,u_2,u_3\}$ remain independent?

If a $4\times 4$ matrix has four pivot columns, what does that say about its columns?

Decide whether `\{(1,1,0),(0,1,1),(1,2,1)\}` is independent, and if not, write one dependence relation.