A basis of a vector space V is a set of vectors that is both linearly independent and spans V (i.e., every vector in V is a linear combination of them). Intuitively, it is a minimal toolkit that covers the whole space. Bases are not unique: for R^2, each of {(1,0),(0,1)}, {(1,1),(1,-1)} is a basis. The number of vectors in a basis is the dimension of the space.

Linear Independence, Linear Dependence, and Bases

Q: What is linear independence?

A set of vectors {v1,...,vk} is linearly independent if c1*v1 + ... + ck*vk = 0 holds only when all ci = 0. Intuitively, no vector in the set can be expressed as a linear combination of the others.

Q: How do you determine whether vectors are linearly independent?

Arrange the vectors as columns of a matrix A. The vectors are linearly independent if and only if rank(A) equals the number of vectors. For square matrices, this is equivalent to det(A) != 0.

Q: What is the Wronskian?

The Wronskian is a determinant used to test the linear independence of functions f1, ..., fn that are n-1 times differentiable. If the determinant of the matrix formed by the functions and their derivatives is nonzero at some point, then the functions are linearly independent. The converse does not hold in general, but is an equivalence when the functions are analytic.

Q: Do three linearly independent vectors have to be orthogonal?

No. Mutually orthogonal nonzero vectors are linearly independent, but the converse is false. For example, (1,0) and (1,1) are linearly independent but not orthogonal. Linear independence only requires that none of the vectors is a combination of the others; orthogonality is a stronger condition.

Goals

Understand linear independence and linear dependence from three perspectives: definition, geometric meaning, and determination methods. Common stumbling blocks for beginners are addressed with particular care.

1. Intuitive Understanding

1.1 Are there "redundant" vectors?

Given a collection of vectors, if any one of them can be built from combinations of the others, that vector is "redundant".

Example: Consider $\boldsymbol{v}_1 = (1, 0)$, $\boldsymbol{v}_2 = (0, 1)$, $\boldsymbol{v}_3 = (2, 3)$.

Since $\boldsymbol{v}_3 = 2\boldsymbol{v}_1 + 3\boldsymbol{v}_2$, the vector $\boldsymbol{v}_3$ is "redundant".

Figure 1: v₃ can be built from v₁, v₂ — from the origin, go 2v₁ along x and 3v₂ along y to land exactly on v₃

1.2 Meaning of Linear Independence

Linearly independent: No vector can be expressed as a combination of the others.

Linearly dependent: At least one vector can be expressed as a combination of the others.

2. Formal Definition

2.1 Linear Combination

A linear combination of vectors $\boldsymbol{v}_1, \ldots, \boldsymbol{v}_k$ is:

$$c_1\boldsymbol{v}_1 + c_2\boldsymbol{v}_2 + \cdots + c_k\boldsymbol{v}_k$$

The scalars $c_1, \ldots, c_k$ are called coefficients.

2.2 Definition of Linear Independence

Definition: Vectors $\boldsymbol{v}_1, \ldots, \boldsymbol{v}_k$ are linearly independent if

$$c_1\boldsymbol{v}_1 + c_2\boldsymbol{v}_2 + \cdots + c_k\boldsymbol{v}_k = \boldsymbol{0}$$

holds only when $c_1 = c_2 = \cdots = c_k = 0$.

2.3 Definition of Linear Dependence

Definition: Vectors $\boldsymbol{v}_1, \ldots, \boldsymbol{v}_k$ are linearly dependent if there exist

\begin{equation}c_1\boldsymbol{v}_1 + c_2\boldsymbol{v}_2 + \cdots + c_k\boldsymbol{v}_k = \boldsymbol{0} \label{eq:linearly-dependent}\end{equation}

non-trivial coefficients (not all zero).

2.4 Equivalent Formulation

Linearly dependent ⇔ at least one vector can be written as a linear combination of the others.

This can be shown by rearranging the defining equation $\eqref{eq:linearly-dependent}$. Assume $c_1 \neq 0$ among the non-trivial coefficients (by relabelling, we can always put a non-zero one first):

Move $c_2\boldsymbol{v}_2 + \cdots + c_k\boldsymbol{v}_k$ in $\eqref{eq:linearly-dependent}$ to the right-hand side: $$c_1\boldsymbol{v}_1 = -c_2\boldsymbol{v}_2 - \cdots - c_k\boldsymbol{v}_k$$
Divide both sides by $c_1 \neq 0$: $$\boldsymbol{v}_1 = -\dfrac{c_2}{c_1}\boldsymbol{v}_2 - \cdots - \dfrac{c_k}{c_1}\boldsymbol{v}_k$$

Thus $\boldsymbol{v}_1$ is expressed as a linear combination of the others. The same argument works for any non-zero $c_i$, so "a non-trivial linear combination equals zero" and "at least one vector is a linear combination of the others" are equivalent.

3. Geometric Meaning

3.1 The 2-Dimensional Case

Two vectors $\boldsymbol{v}_1$, $\boldsymbol{v}_2$ are:

Linearly independent: not on a common line through the origin (they span a plane)
Linearly dependent: on a common line through the origin (parallel, or one of them is the zero vector)

3.2 The 3-Dimensional Case

Three vectors $\boldsymbol{v}_1$, $\boldsymbol{v}_2$, $\boldsymbol{v}_3$ are:

Linearly independent: not contained in a common plane through the origin (they span the whole space)
Linearly dependent: all contained in some plane through the origin

Figure 2: The 3D case — Left: three linearly independent vectors span 3-space (a parallelepiped). Right: three linearly dependent vectors are squeezed into the same plane

3.3 Generalization

In $n$-dimensional space:

Any set of $n+1$ or more vectors is necessarily linearly dependent
$n$ linearly independent vectors span the whole space

Recap: "basis" (full treatment in Chapter 1 §5)

A basis of a space $V$ is a set of vectors $\{\boldsymbol{e}_1, \ldots, \boldsymbol{e}_n\}$ satisfying both:

Linearly independent — no $\boldsymbol{e}_i$ can be built from the others (no waste)
Spans $V$ — every vector in $V$ is a linear combination of $\boldsymbol{e}_1, \ldots, \boldsymbol{e}_n$ (reaches everywhere)

Intuitively, "a minimal toolkit that covers the whole space". Too few and it fails to cover; too many and there is redundancy.

Examples:

Standard basis of $\mathbb{R}^2$: $\{(1,0), (0,1)\}$ — 2 vectors span the plane
Standard basis of $\mathbb{R}^3$: $\{(1,0,0), (0,1,0), (0,0,1)\}$ — 3 vectors span 3-space
In $\mathbb{R}^2$, $\{(1,0), (1,1)\}$ is also a basis (non-standard bases are possible)
In $\mathbb{R}^2$, $\{(1,0), (2,0)\}$ is not a basis (linearly dependent; cannot reach the $y$ direction)

The number of vectors in a basis is called the dimension of the space; $\mathbb{R}^n$ has dimension $n$. Non-uniqueness of a basis, plus bases for polynomial / matrix / function spaces are discussed in Chapter 1 §5.

4. Determination Methods

Prerequisites for this section — We use the rank of a matrix and the determinant (det) below. Detailed definitions:

Rank: the maximum number of linearly independent column vectors (equivalently, row vectors). Details in Chapter 11.
Determinant: a scalar defined for square matrices, with $\det(A) \neq 0$ ⇔ columns are linearly independent. Derivation and computation in Chapter 3–Chapter 5; a gentler introduction in the Intro §7.

In this section we introduce them as mechanical tools for deciding linear independence.

4.1 Using the Rank of a Matrix

Arrange the vectors as columns of a matrix $A = (\boldsymbol{v}_1 | \boldsymbol{v}_2 | \cdots | \boldsymbol{v}_k)$.

Test: $\boldsymbol{v}_1, \ldots, \boldsymbol{v}_k$ are linearly independent ⇔ $\mathrm{rank}(A) = k$.

Intuitively, $\mathrm{rank}(A)$ is "the dimension of the space spanned by the columns of $A$". If all $k$ vectors are linearly independent, the rank is $k$; if there is any linear dependence, the rank is less than $k$.

4.2 The Square Matrix Case

When the number of vectors equals the dimension ($n$ vectors in $n$-dimensional space):

Test: $\boldsymbol{v}_1, \ldots, \boldsymbol{v}_n$ are linearly independent ⇔ $\det(A) \neq 0$.

The determinant vanishes exactly when "the space spanned by the columns has dimension less than $n$", which is precisely the linearly dependent case. Chapter 5 revisits this fact from the geometric perspective "$\det(A)$ = signed volume of the parallelepiped".

4.3 Worked Example

Are $\boldsymbol{v}_1 = (1, 2, 3)$, $\boldsymbol{v}_2 = (4, 5, 6)$, $\boldsymbol{v}_3 = (7, 8, 9)$ linearly independent?

$$A = \begin{pmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{pmatrix}$$ $$\det(A) = 1(45-48) - 4(18-24) + 7(12-15) = -3 + 24 - 21 = 0$$

Since $\det(A) = 0$, the vectors are linearly dependent. Indeed, $\boldsymbol{v}_3 = 2\boldsymbol{v}_2 - \boldsymbol{v}_1$.

4.4 Linear Independence of Functions: The Wronskian ★Advanced (optional)

This subsection is advanced material — it goes beyond the "vector = number tuple" framework and treats functions as "vectors" belonging to an infinite-dimensional vector space (a function space). It is an application combining differentiation with determinants; feel free to skip it on first reading (it also serves as a preview of the intermediate topic Function Spaces).

For functions $f_1, \ldots, f_n$ that are $n-1$ times differentiable, the Wronskian is defined as

$$W(f_1, \ldots, f_n) = \det\begin{pmatrix} f_1 & \cdots & f_n \\ f_1' & \cdots & f_n' \\ \vdots & & \vdots \\ f_1^{(n-1)} & \cdots & f_n^{(n-1)} \end{pmatrix}$$

If $W \neq 0$ at some point, the functions are linearly independent.

Caveat: the converse is not true in general. There exist examples where $W \equiv 0$ (identically zero) yet the functions are not linearly dependent (see Peano's counterexample). However, for analytic functions, $W \equiv 0$ ⇔ linearly dependent does hold.

Example: To see that $1, x, x^2$ are linearly independent in $C(\mathbb{R})$, note that $c_1 + c_2 x + c_3 x^2 = 0$ for all $x$ forces $c_1 = c_2 = c_3 = 0$. The Wronskian confirms this:

$$W(1, x, x^2) = \det\begin{pmatrix} 1 & x & x^2 \\ 0 & 1 & 2x \\ 0 & 0 & 2 \end{pmatrix} = 2 \neq 0$$

5. Common Misconceptions

5.1 "Orthogonal" is Different from "Linearly Independent"

Mutually orthogonal nonzero vectors ⇒ linearly independent (true)

Note: the zero vector is a counterexample if it is allowed. $(1,0)$ and $(0,0)$ have inner product 0, yet $0\cdot(1,0) + 1\cdot(0,0) = \boldsymbol{0}$, so they are linearly dependent.

Linearly independent ⇒ orthogonal (false)

Counterexample: $(1, 0)$ and $(1, 1)$ are linearly independent but not orthogonal.

5.2 A Set Containing the Zero Vector is Linearly Dependent

Any set that contains $\boldsymbol{0}$ is automatically linearly dependent.

This is because $1 \cdot \boldsymbol{0} + 0 \cdot \boldsymbol{v}_1 + \cdots = \boldsymbol{0}$ is a non-trivial combination giving $\boldsymbol{0}$.

5.3 Linear Independence Even Applies to a Single Vector

If $\boldsymbol{v} \neq \boldsymbol{0}$, then $\{\boldsymbol{v}\}$ is linearly independent.

If $\boldsymbol{v} = \boldsymbol{0}$, then $\{\boldsymbol{v}\}$ is linearly dependent.

6. Summary

Key Takeaways

Linearly independent: $\displaystyle\sum c_i\boldsymbol{v}_i = \boldsymbol{0}$ forces all $c_i = 0$
Linearly dependent: a non-trivial linear combination gives the zero vector
Geometric meaning: in 2D, not on a common line through the origin; in 3D, not contained in a common plane through the origin
Test: matrix rank, or determinant
Any set of $n+1$ or more vectors in $n$-dimensional space is linearly dependent

Related Pages:

Frequently Asked Questions

Q1. What is linear independence?

A set of vectors $\{\boldsymbol{v}_1, \ldots, \boldsymbol{v}_k\}$ is linearly independent if $c_1\boldsymbol{v}_1 + \cdots + c_k\boldsymbol{v}_k = \boldsymbol{0}$ holds only when every $c_i = 0$. Intuitively, no vector in the set can be written as a linear combination of the others.

Q2. How do you determine whether vectors are linearly independent?

Arrange the vectors as the columns of a matrix $A$. They are linearly independent if and only if $\operatorname{rank}(A)$ equals the number of vectors. For a square matrix this is equivalent to $\det(A) \neq 0$.

Q3. What is the Wronskian?

It is a determinant used to test the linear independence of functions $f_1, \ldots, f_n$ that are $n-1$ times differentiable. If the determinant of the matrix formed by the functions and their derivatives is nonzero at some point, the functions are linearly independent. The converse does not hold in general, but it becomes an equivalence for analytic functions.

Q4. What is a basis?

A basis of a vector space $V$ is a set of vectors that is both linearly independent and spans $V$ (every vector of $V$ is a linear combination of them). Intuitively it is the minimal toolkit that covers the whole space. Bases are not unique: for $\mathbb{R}^2$, both $\{(1,0),(0,1)\}$ and $\{(1,1),(1,-1)\}$ are bases. The number of vectors in a basis is the dimension of the space.

Q5. Do three linearly independent vectors have to be orthogonal?

No. Mutually orthogonal nonzero vectors are linearly independent, but the converse is false. For example $(1,0)$ and $(1,1)$ are linearly independent but not orthogonal. Linear independence only requires that none of the vectors is a combination of the others; orthogonality is a stronger condition.