Linear Independence, Bases, and Dimension

Every basis of a finite-dimensional vector space has the same number of elements — this fact, proved via the Steinitz exchange lemma, makes dimension well-defined. We build to it through linear combinations, span, and the precise criterion for linear independence.

Folio Official

March 1, 2026

1. Linear Combinations and Span

Definition 1 (Linear combination).

Let

v_{1}, \dots, v_{n}

be vectors in a vector space

V

and let

a_{1}, \dots, a_{n} \in K

be scalars. The expression

a_{1} v_{1} + a_{2} v_{2} + \dots + a_{n} v_{n}

is called a linear combination of

v_{1}, \dots, v_{n}

Definition 2 (Span).

Let

S = {v_{1}, \dots, v_{n}} \subseteq V

. The set of all linear combinations of elements of

S

span (S) = {i = 1 \sum n a_{i} v_{i} a_{i} \in K},

is called the span of

S

. When

span (S) = V

, we say that

S

spans

V

(or that

S

is a spanning set for

V

Example 3.

R^{3}

, we have

span {(1, 0, 0), (0, 1, 0)} = {(x, y, 0)}

, the

x y

-plane. Adding the third standard vector yields

span {(1, 0, 0), (0, 1, 0), (0, 0, 1)} = R^{3}

2. Linear Independence and Dependence

Definition 4 (Linear independence).

Vectors

v_{1}, \dots, v_{n} \in V

are linearly independent if

a_{1} v_{1} + a_{2} v_{2} + \dots + a_{n} v_{n} = 0 ⟹ a_{1} = a_{2} = \dots = a_{n} = 0.

If the vectors are not linearly independent, they are called linearly dependent.

Theorem 5.

The vectors

v_{1}, \dots, v_{n}

are linearly dependent if and only if some

v_{k}

can be expressed as a linear combination of the remaining vectors.

Proof.

(\Rightarrow)

By linear dependence, there exist scalars

a_{1}, \dots, a_{n}

, not all zero, such that

\sum a_{i} v_{i} = 0

. Choose

k

with

a_{k} \neq = 0

. Then

v_{k} = - a_{k}^{- 1} \sum_{i \neq = k} a_{i} v_{i}

(\Leftarrow)

v_{k} = \sum_{i \neq = k} b_{i} v_{i}

, then

\sum_{i \neq = k} b_{i} v_{i} - v_{k} = 0

is a nontrivial linear relation. □

Example 6.

The vectors

v_{1} = (1, 0, 0)

v_{2} = (0, 1, 0)

v_{3} = (1, 1, 0)

are linearly dependent, since

v_{3} = v_{1} + v_{2}

The vectors

v_{1} = (1, 0, 0)

v_{2} = (0, 1, 0)

v_{3} = (0, 0, 1)

are linearly independent. Indeed,

a_{1} (1, 0, 0) + a_{2} (0, 1, 0) + a_{3} (0, 0, 1) = (0, 0, 0)

immediately gives

a_{1} = a_{2} = a_{3} = 0

3. Bases

Definition 7 (Basis).

A set

{v_{1}, \dots, v_{n}}

of vectors in a vector space

V

is a basis for

V

if it satisfies the following two conditions simultaneously:

${v_{1}, \dots, v_{n}}$ spans $V$ .
${v_{1}, \dots, v_{n}}$ is linearly independent.

Theorem 8 (Unique representation with respect to a basis).

{v_{1}, \dots, v_{n}}

is a basis for

V

, then every vector

v \in V

can be written uniquely as

v = a_{1} v_{1} + a_{2} v_{2} + \dots + a_{n} v_{n} .

Proof.

Existence follows from the fact that the basis spans

V

. For uniqueness, suppose

v = \sum a_{i} v_{i} = \sum b_{i} v_{i}

. Then

\sum (a_{i} - b_{i}) v_{i} = 0

, and linear independence forces

a_{i} - b_{i} = 0

for each

i

, so

a_{i} = b_{i}

. □

Definition 9 (Coordinates).

Given a basis

{v_{1}, \dots, v_{n}}

, the scalars

(a_{1}, \dots, a_{n})

in the unique representation

v = \sum a_{i} v_{i}

are called the coordinates of

v

with respect to that basis.

4. The Steinitz Exchange Lemma

Theorem 10 (Steinitz exchange lemma).

Let

V

be a vector space. If

{v_{1}, \dots, v_{m}}

spans

V

and

{w_{1}, \dots, w_{n}}

is linearly independent, then

n \leq m .

Moreover,

n

of the vectors

v_{1}, \dots, v_{m}

can be replaced by

w_{1}, \dots, w_{n}

without losing the spanning property.

Proof.

Since

w_{1}

lies in

span {v_{1}, \dots, v_{m}}

, we can write

w_{1} = \sum_{j = 1}^{m} c_{j} v_{j}

. At least one coefficient is nonzero (because

w_{1} \neq = 0

). Re-indexing if necessary, assume

c_{1} \neq = 0

. Then

v_{1} = c_{1}^{- 1} (w_{1} - \sum_{j = 2}^{m} c_{j} v_{j})

, so

{w_{1}, v_{2}, \dots, v_{m}}

still spans

V

Repeat this process for

w_{2}, w_{3}, \dots

in turn. At the

k

-th step, we add

w_{k}

to the spanning set and remove one of the

v

's. The linear independence of

w_{1}, \dots, w_{k}

guarantees that a removable

v

always exists.

n > m

, then after

m

steps all the

v

's would be exhausted and

{w_{1}, \dots, w_{m}}

would span

V

. But then

w_{m + 1}

would be a linear combination of

w_{1}, \dots, w_{m}

, contradicting the linear independence of

{w_{1}, \dots, w_{m + 1}}

. Therefore

n \leq m

. □

5. Dimension

Theorem 11 (All bases have the same size).

If a vector space

V

has a finite basis, then every basis of

V

has the same number of elements.

Proof.

Let

{v_{1}, \dots, v_{m}}

and

{w_{1}, \dots, w_{n}}

be two bases of

V

. The first is a spanning set and the second is linearly independent, so

n \leq m

by the Steinitz exchange lemma. Reversing the roles gives

m \leq n

, and therefore

m = n

. □

Definition 12 (Dimension).

The number of elements in any basis of a vector space

V

is called the dimension of

V

and is denoted

dim V

. If no finite basis exists, we write

dim V = \infty

Theorem 13.

dim V = n

, then:

Every linearly independent set in $V$ contains at most $n$ vectors.
Any $n$ linearly independent vectors in $V$ form a basis.
Any $n$ vectors that span $V$ form a basis.

Proof.

(1) This is an immediate consequence of the Steinitz exchange lemma.

(2) Suppose

{w_{1}, \dots, w_{n}}

is linearly independent but does not span

V

. Then there exists

v \in V

with

v \in / span {w_{1}, \dots, w_{n}}

, and

{w_{1}, \dots, w_{n}, v}

is a linearly independent set of

n + 1

vectors, contradicting (1).

(3) Suppose

{v_{1}, \dots, v_{n}}

spans

V

but is linearly dependent. Then some

v_{k}

is a linear combination of the rest, and removing it leaves

n - 1

vectors that still span

V

. But

V

has a basis of

n

linearly independent vectors, so by Steinitz we would need

n \leq n - 1

, a contradiction. □

Example 14.

$dim R^{n} = n$ : the standard basis is ${e_{1}, \dots, e_{n}}$ .
$dim K [x]_{\leq n} = n + 1$ : a basis is ${1, x, x^{2}, \dots, x^{n}}$ .
$dim M_{m \times n} (K) = mn$ : a basis is ${E_{ij}}$ , where $E_{ij}$ is the matrix with $1$ in position $(i, j)$ and $0$ elsewhere.

Linear Algebra Algebra Textbook Linear Independence Basis Dimension

Folio Official

Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

1 followers·107 articles

Linear Independence, Bases, and Dimension

Folio Official

March 1, 2026

1. Linear Combinations and Span

Definition 1 (Linear combination).

Let

v_{1}, \dots, v_{n}

be vectors in a vector space

V

and let

a_{1}, \dots, a_{n} \in K

be scalars. The expression

a_{1} v_{1} + a_{2} v_{2} + \dots + a_{n} v_{n}

is called a linear combination of

v_{1}, \dots, v_{n}

Definition 2 (Span).

Let

S = {v_{1}, \dots, v_{n}} \subseteq V

. The set of all linear combinations of elements of

S

span (S) = {i = 1 \sum n a_{i} v_{i} a_{i} \in K},

is called the span of

S

. When

span (S) = V

, we say that

S

spans

V

(or that

S

is a spanning set for

V

Example 3.

R^{3}

, we have

span {(1, 0, 0), (0, 1, 0)} = {(x, y, 0)}

, the

x y

-plane. Adding the third standard vector yields

span {(1, 0, 0), (0, 1, 0), (0, 0, 1)} = R^{3}

2. Linear Independence and Dependence

Definition 4 (Linear independence).

Vectors

v_{1}, \dots, v_{n} \in V

are linearly independent if

a_{1} v_{1} + a_{2} v_{2} + \dots + a_{n} v_{n} = 0 ⟹ a_{1} = a_{2} = \dots = a_{n} = 0.

If the vectors are not linearly independent, they are called linearly dependent.

Theorem 5.

The vectors

v_{1}, \dots, v_{n}

are linearly dependent if and only if some

v_{k}

can be expressed as a linear combination of the remaining vectors.

Proof.

(\Rightarrow)

By linear dependence, there exist scalars

a_{1}, \dots, a_{n}

, not all zero, such that

\sum a_{i} v_{i} = 0

. Choose

k

with

a_{k} \neq = 0

. Then

v_{k} = - a_{k}^{- 1} \sum_{i \neq = k} a_{i} v_{i}

(\Leftarrow)

v_{k} = \sum_{i \neq = k} b_{i} v_{i}

, then

\sum_{i \neq = k} b_{i} v_{i} - v_{k} = 0

is a nontrivial linear relation. □

Example 6.

The vectors

v_{1} = (1, 0, 0)

v_{2} = (0, 1, 0)

v_{3} = (1, 1, 0)

are linearly dependent, since

v_{3} = v_{1} + v_{2}

The vectors

v_{1} = (1, 0, 0)

v_{2} = (0, 1, 0)

v_{3} = (0, 0, 1)

are linearly independent. Indeed,

a_{1} (1, 0, 0) + a_{2} (0, 1, 0) + a_{3} (0, 0, 1) = (0, 0, 0)

immediately gives

a_{1} = a_{2} = a_{3} = 0

3. Bases

Definition 7 (Basis).

A set

{v_{1}, \dots, v_{n}}

of vectors in a vector space

V

is a basis for

V

if it satisfies the following two conditions simultaneously:

${v_{1}, \dots, v_{n}}$ spans $V$ .
${v_{1}, \dots, v_{n}}$ is linearly independent.

Theorem 8 (Unique representation with respect to a basis).

{v_{1}, \dots, v_{n}}

is a basis for

V

, then every vector

v \in V

can be written uniquely as

v = a_{1} v_{1} + a_{2} v_{2} + \dots + a_{n} v_{n} .

Proof.

Existence follows from the fact that the basis spans

V

. For uniqueness, suppose

v = \sum a_{i} v_{i} = \sum b_{i} v_{i}

. Then

\sum (a_{i} - b_{i}) v_{i} = 0

, and linear independence forces

a_{i} - b_{i} = 0

for each

i

, so

a_{i} = b_{i}

. □

Definition 9 (Coordinates).

Given a basis

{v_{1}, \dots, v_{n}}

, the scalars

(a_{1}, \dots, a_{n})

in the unique representation

v = \sum a_{i} v_{i}

are called the coordinates of

v

with respect to that basis.

4. The Steinitz Exchange Lemma

Theorem 10 (Steinitz exchange lemma).

Let

V

be a vector space. If

{v_{1}, \dots, v_{m}}

spans

V

and

{w_{1}, \dots, w_{n}}

is linearly independent, then

n \leq m .

Moreover,

n

of the vectors

v_{1}, \dots, v_{m}

can be replaced by

w_{1}, \dots, w_{n}

without losing the spanning property.

Proof.

Since

w_{1}

lies in

span {v_{1}, \dots, v_{m}}

, we can write

w_{1} = \sum_{j = 1}^{m} c_{j} v_{j}

. At least one coefficient is nonzero (because

w_{1} \neq = 0

). Re-indexing if necessary, assume

c_{1} \neq = 0

. Then

v_{1} = c_{1}^{- 1} (w_{1} - \sum_{j = 2}^{m} c_{j} v_{j})

, so

{w_{1}, v_{2}, \dots, v_{m}}

still spans

V

Repeat this process for

w_{2}, w_{3}, \dots

in turn. At the

k

-th step, we add

w_{k}

to the spanning set and remove one of the

v

's. The linear independence of

w_{1}, \dots, w_{k}

guarantees that a removable

v

always exists.

n > m

, then after

m

steps all the

v

's would be exhausted and

{w_{1}, \dots, w_{m}}

would span

V

. But then

w_{m + 1}

would be a linear combination of

w_{1}, \dots, w_{m}

, contradicting the linear independence of

{w_{1}, \dots, w_{m + 1}}

. Therefore

n \leq m

. □

5. Dimension

Theorem 11 (All bases have the same size).

If a vector space

V

has a finite basis, then every basis of

V

has the same number of elements.

Proof.

Let

{v_{1}, \dots, v_{m}}

and

{w_{1}, \dots, w_{n}}

be two bases of

V

. The first is a spanning set and the second is linearly independent, so

n \leq m

by the Steinitz exchange lemma. Reversing the roles gives

m \leq n

, and therefore

m = n

. □

Definition 12 (Dimension).

The number of elements in any basis of a vector space

V

is called the dimension of

V

and is denoted

dim V

. If no finite basis exists, we write

dim V = \infty

Theorem 13.

dim V = n

, then:

Every linearly independent set in $V$ contains at most $n$ vectors.
Any $n$ linearly independent vectors in $V$ form a basis.
Any $n$ vectors that span $V$ form a basis.

Proof.

(1) This is an immediate consequence of the Steinitz exchange lemma.

(2) Suppose

{w_{1}, \dots, w_{n}}

is linearly independent but does not span

V

. Then there exists

v \in V

with

v \in / span {w_{1}, \dots, w_{n}}

, and

{w_{1}, \dots, w_{n}, v}

is a linearly independent set of

n + 1

vectors, contradicting (1).

(3) Suppose

{v_{1}, \dots, v_{n}}

spans

V

but is linearly dependent. Then some

v_{k}

is a linear combination of the rest, and removing it leaves

n - 1

vectors that still span

V

. But

V

has a basis of

n

linearly independent vectors, so by Steinitz we would need

n \leq n - 1

, a contradiction. □

Example 14.

$dim R^{n} = n$ : the standard basis is ${e_{1}, \dots, e_{n}}$ .
$dim K [x]_{\leq n} = n + 1$ : a basis is ${1, x, x^{2}, \dots, x^{n}}$ .
$dim M_{m \times n} (K) = mn$ : a basis is ${E_{ij}}$ , where $E_{ij}$ is the matrix with $1$ in position $(i, j)$ and $0$ elsewhere.

Linear Algebra Algebra Textbook Linear Independence Basis Dimension

Folio Official

Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

1 followers·107 articles

Linear Independence, Bases, and Dimension

1. Linear Combinations and Span

2. Linear Independence and Dependence

3. Bases

4. The Steinitz Exchange Lemma

5. Dimension

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Linear Independence, Bases, and Dimension

1. Linear Combinations and Span

2. Linear Independence and Dependence

3. Bases

4. The Steinitz Exchange Lemma

5. Dimension

Share your expertise with the world

More from Folio Official

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