Linear Independence

Quick answer

Vectors v₁ through v_k are linearly independent when c₁v₁ + ⋯ + c_kv_k = 0 forces every coefficient to be zero. If some combination with a nonzero coefficient gives zero, the set is dependent, and one of the vectors is a combination of the others, so it adds nothing to the span. The test is a homogeneous system: the set is independent exactly when the matrix with those vectors as columns has a pivot in every column. A set with more vectors than components is always dependent.

What you'll learn

State what linear independence means
Test a set of vectors by row reduction
Connect independence to pivot columns and free variables
Recognize sets that are dependent at a glance

Redundant directions

Two vectors along the same line carry one direction between them. The span of $\{(1, 3), (2, 6)\}$ is the same line as the span of $\{(1, 3)\}$ alone. The second vector is redundant. Linear independence is the precise version of “no redundancy”.

The vectors $\mathbf{v}_1, \ldots, \mathbf{v}_k$ are linearly independent when $c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k = \mathbf{0}$ only for $c_1 = c_2 = \cdots = c_k = 0$ . Otherwise they are linearly dependent.

Every set admits the all-zero coefficients, called the trivial combination. The question is whether any other combination also lands on $\mathbf{0}$ .

Independent and dependent pairs

Why the test is a homogeneous system

The equation $c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k = \mathbf{0}$ is a vector equation in the unknown coefficients. Matching components turns it into a homogeneous linear system whose coefficient matrix has the vectors as its columns. That system always has the trivial solution, so the only question is whether it has others, and free variables decide that. The vectors are independent exactly when the matrix holding them as columns has a pivot in every column. A column without a pivot is a free variable, and a free variable produces a nonzero solution.

Two consequences come for free:

A set containing $\mathbf{0}$ is dependent: take coefficient $1$ on the zero vector and $0$ elsewhere.
More vectors than components means dependence. A matrix with $n$ rows has at most $n$ pivots, so $k > n$ columns cannot all hold one.

Worked examples

Common mistakes

Practice problems

Are $(2, 4)$ and $(3, 6)$ independent?

Answer

No

Full solution

$(3, 6) = 1.5(2, 4)$ , so $3(2, 4) - 2(3, 6) = (0, 0)$ with nonzero coefficients.
Are $(1, 0)$ and $(0, 1)$ independent?

Answer

Yes

Full solution

$c_1(1, 0) + c_2(0, 1) = (c_1, c_2)$ , which is $(0, 0)$ only when both coefficients are zero.
Are $(1, 2, 3)$ , $(2, 4, 6)$ and $(3, 6, 9)$ independent?

Answer

No

Full solution

All three lie on one line through the origin: the second is twice the first and the third is three times it. The matrix reduces to a single pivot.
Are $(2, 1, 0)$ , $(1, 2, 1)$ and $(0, 1, 2)$ independent?

Answer

Yes

Full solution

Row reducing gives three pivots. The determinant of the matrix is $4$ , and a nonzero determinant is another way to see that no column is redundant.
Can five vectors in ℝ⁴ be independent?

Answer

No

Full solution

Four rows allow at most four pivots, so the fifth column has none and gives a free variable.
Is $\{(1, 2), (0, 0)\}$ independent?

Answer

No

Full solution

$0(1, 2) + 1(0, 0) = (0, 0)$ uses a nonzero coefficient. Any set containing the zero vector is dependent.
Write a dependence relation among $(1, 2)$ , $(3, 1)$ and $(5, 5)$ .

Answer

$2(1, 2) + (3, 1) - (5, 5) = (0, 0)$

Full solution

Solving $c_1(1, 2) + c_2(3, 1) = (5, 5)$ gives $c_1 = 2$ and $c_2 = 1$ , and moving $(5, 5)$ to the other side gives the relation.
A $4 \times 6$ matrix has how many pivot columns at most? Are its columns independent?

Answer

At most four; the columns are dependent.

Full solution

Each pivot needs its own row, and there are four rows. With six columns, at least two lack a pivot.
The columns of a $3 \times 3$ matrix are independent. How many solutions does $A\mathbf{x} = \mathbf{0}$ have?

Answer

One: the trivial solution

Full solution

Independence means a pivot in every column, so there are no free variables and $\mathbf{x} = \mathbf{0}$ is the only solution.
A student says $(1, 0, 1)$ , $(0, 1, 1)$ and $(1, 1, 2)$ are independent, since no one of them is a multiple of another. What went wrong?

Hint

Add the first two.

Answer

The third is the sum of the first two, so the set is dependent.

Full solution

For two vectors, dependence does mean one is a multiple of the other. For three or more, a vector can be a combination of several others without being a multiple of any single one: $(1, 1, 2) = (1, 0, 1) + (0, 1, 1)$ .

Frequently asked questions

What does linearly independent mean?

The only linear combination of the vectors that equals the zero vector is the one with every coefficient zero.

How do you test for independence?

Put the vectors in the columns of a matrix and row reduce. The set is independent exactly when every column contains a pivot.

What does dependence say about the vectors?

At least one of them is a linear combination of the others, so removing it leaves the span unchanged.

Are two vectors independent when neither is a multiple of the other?

Yes. For two vectors, dependence means exactly that one is a scalar multiple of the other, including the case of a zero vector.

Can four vectors in ℝ³ be independent?

No. A matrix with three rows has at most three pivots, so a fourth column has none and the set is dependent.

What to learn next

Matrix Multiplication and Ax as a Combination of ColumnsRead Ax as a linear combination of the columns of A, multiply matrices of any compatible size, and see the product AB as doing B first and then A.

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