Linear Independence Explained: Higher Math Guide

A set of vectors $\mathbf{v}_1, \ldots, \mathbf{v}_k$ is linearly independent when the only way to combine them into the zero vector is to use all zero coefficients: $$c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k = \mathbf{0} \quad\text{only when}\quad c_1 = c_2 = \cdots = c_k = 0.$$ If some other combination also gives zero, the set is linearly dependent. In plain words, independent vectors carry no passengers: none of them can be built from the others, so each one adds a new direction to the span.

Without the fog

Margalit and Rabinoff's open textbook Interactive Linear Algebra starts from a picture. Sometimes the span of a set of vectors is smaller than you expect from the number of vectors. That means at least one vector is redundant: you can remove it and the span stays the same. Linear independence is the precise way of saying that no vector is redundant.

The book gives two equivalent tests that match that picture.

The span test. A set is linearly dependent exactly when one of its vectors is in the span of the others. Any such vector can be removed without changing the span.
The increasing span test. Build the set one vector at a time. If the span gets bigger every time you add a vector, the set is linearly independent.

A nonzero combination that gives zero, such as $2\mathbf{v}_1 - \mathbf{v}_2 = \mathbf{0}$, is called a linear dependence relation. Finding one is how you prove dependence. Showing that none exists is how you prove independence.

The recipe: row reduce and look for pivots

Put the vectors into a matrix $A$ as its columns. The vector equation above is the same as the matrix equation $A\mathbf{c} = \mathbf{0}$. So the columns are independent exactly when $A\mathbf{c} = \mathbf{0}$ has only the trivial solution, which happens exactly when $A$ has a pivot position in every column. A column without a pivot means a free variable, and any nonzero value of that free variable gives a dependence relation.

The row reduction (RREF) intuition explainer and the RREF calculator do the mechanical part. The idea to keep is short: pivot in every column means independent; a free column means dependent.

Worked example 1: two vectors in the plane

Are $\mathbf{v}_1 = (1,2)$ and $\mathbf{v}_2 = (3,1)$ independent? Solve $c_1(1,2) + c_2(3,1) = (0,0)$. That is two equations: $$c_1 + 3c_2 = 0, \qquad 2c_1 + c_2 = 0.$$ The first gives $c_1 = -3c_2$. Substitute into the second: $-6c_2 + c_2 = -5c_2 = 0$, so $c_2 = 0$ and then $c_1 = 0$. Only the trivial solution exists, so the vectors are independent.

The picture agrees. Two vectors are dependent exactly when they are collinear, meaning one is a scalar multiple of the other. Here $(3,1)$ is not a multiple of $(1,2)$, so the two arrows point in different directions and together they span the whole plane. The determinant of the matrix with these columns is $1 \cdot 1 - 3 \cdot 2 = -5$, which is not zero, the same verdict from a different angle.

Worked example 2: three vectors in space

Take $\mathbf{u}_1 = (1,0,1)$, $\mathbf{u}_2 = (0,1,1)$ and $\mathbf{u}_3 = (1,1,2)$. Put them in a matrix as columns and row reduce: $$\begin{pmatrix}1&0&1\\0&1&1\\1&1&2\end{pmatrix} \;\to\; \begin{pmatrix}1&0&1\\0&1&1\\0&1&1\end{pmatrix} \;\to\; \begin{pmatrix}1&0&1\\0&1&1\\0&0&0\end{pmatrix}.$$ The first step subtracts row 1 from row 3, and the second subtracts row 2 from row 3. Columns 1 and 2 have pivots. Column 3 does not, so its variable $c_3$ is free and the set is dependent.

Set $c_3 = 1$. The reduced rows say $c_1 + c_3 = 0$ and $c_2 + c_3 = 0$, so $c_1 = -1$ and $c_2 = -1$. That gives the dependence relation $$-\mathbf{u}_1 - \mathbf{u}_2 + \mathbf{u}_3 = \mathbf{0}, \quad\text{that is,}\quad \mathbf{u}_3 = \mathbf{u}_1 + \mathbf{u}_2.$$ Check it directly: $(1,0,1) + (0,1,1) = (1,1,2)$. All three vectors lie in one plane through the origin, so their span is a plane, not all of space. Interactive Linear Algebra notes that three vectors are dependent exactly when they are coplanar.

Worked example 3: too many vectors

Any four vectors in three dimensional space are dependent, whatever they are. Put them in a matrix with 3 rows and 4 columns. Each row holds at most one pivot, so there are at most 3 pivots and at least one column must be free. In general, a wide matrix (more columns than rows) always has dependent columns. A tall matrix may or may not have independent columns, so you still have to check.

Left: $(1,2)$ and $(3,1)$ point in different directions and span the plane. Right: $(2,4)$ is twice $(1,2)$, so both lie on one line and the span is only that line. Grid units, not to scale.

Quick facts worth memorising

Two vectors are dependent exactly when one is a scalar multiple of the other.
Any set containing the zero vector is dependent, because $1 \cdot \mathbf{0} = \mathbf{0}$ is a dependence relation.
If part of a set is dependent, the whole set is dependent. Adding vectors can never repair dependence.
For $n$ vectors in $n$ dimensional space, the square matrix with those columns has independent columns exactly when its determinant is not zero. The determinant explainer shows why: dependent columns flatten the unit square or cube, so the volume scale factor is $0$.

Why it matters: bases and dimension

A basis is a set of vectors that is linearly independent and spans the space. Independence is what makes coordinates unique. If $\mathbf{b}_1, \ldots, \mathbf{b}_n$ are independent and $\mathbf{x} = a_1\mathbf{b}_1 + \cdots + a_n\mathbf{b}_n = a_1'\mathbf{b}_1 + \cdots + a_n'\mathbf{b}_n$, subtracting gives a combination equal to zero, so every $a_i - a_i'$ is zero. Without independence, the same vector would have more than one set of coordinates.

Interactive Linear Algebra also shows which vectors to keep. Row reduce the matrix of columns, then delete the columns of the original matrix that have no pivot. The pivot columns that remain are independent and have the same span. Their number is the dimension of the span: one pivot column means a line, two mean a plane, three mean a three dimensional space. Take the pivot columns from the original matrix, not the row reduced one, because row operations change the column span.

The same idea runs through the rest of the course. Eigenvalues are useful for diagonalising a matrix only when there are enough independent eigenvectors. The Gram Schmidt calculator turns an independent list into an orthogonal one. For functions, the Wronskian calculator gives a test: if the Wronskian of a list of differentiable functions is nonzero at some point of an interval, the functions are linearly independent on that interval.

Checkpoints

Are $(2,6)$ and $(3,9)$ independent? If not, write a dependence relation.
Are $(1,0,0)$, $(1,1,0)$ and $(1,1,1)$ independent? Use the increasing span test or row reduce.
Without computing, explain why five vectors in four dimensional space must be dependent.
A $3 \times 3$ matrix has determinant $0$. What does that tell you about its columns?

Common mistakes

Calling one vector dependent. Interactive Linear Algebra stresses that independence and dependence describe a collection of vectors. It does not make sense to say one vector is dependent, or that a matrix is independent.
Assuming every vector in a dependent set is redundant. In a dependent set at least one vector is in the span of the others, but not necessarily every one. In the set $(1,0)$, $(2,0)$, $(0,1)$, the vector $(0,1)$ is not a combination of the other two.
Stopping at "the vectors look different". Three vectors can be pairwise non collinear and still dependent, as in worked example 2. Check the whole set, not pairs.
Using the reduced matrix's columns as the basis. Row reduction tells you which positions are pivots. Take those columns from the original matrix.
Reading a tall matrix as automatically independent. Only the wide case is automatic. A tall matrix still needs a pivot in every column.

Related tools

The RREF calculator finds the pivot columns for the recipe above. The 2 by 2 matrix calculator and the 3 by 3 matrix calculator give the determinant check for square cases, and the Wronskian calculator handles lists of functions. Predict the answer first, then check.

Sources

Last reviewed: 11 November 2026.

Common questions

What does linearly independent mean in simple terms?

A set of vectors is linearly independent when none of them can be built from the others. Formally, the only combination of the vectors that equals the zero vector is the one with every coefficient equal to zero. Each vector then adds a new direction to the span.

How do you check whether vectors are linearly independent?

Put the vectors into a matrix as columns and row reduce. If every column has a pivot, the vectors are independent. If some column has no pivot, its free variable gives a nonzero combination equal to zero, so the vectors are dependent.

How is linear independence related to the determinant?

For n vectors in n dimensional space, the square matrix with those vectors as columns has independent columns exactly when its determinant is not zero. A zero determinant means the columns are dependent and the map flattens space into a lower dimension.

What is the difference between linear independence and a basis?

A basis must do two things: be linearly independent and span the whole space. An independent set that does not span is only part of a basis. For example, (1, 0, 0) and (0, 1, 0) are independent in three dimensional space but span only a plane.

Linear independence explained