Linear Independence, Bases, and Dimension

Beginning with linear combinations and span, we define linear independence, bases, and dimension. The Steinitz exchange lemma establishes that every basis of a finite-dimensional space has the same number of elements, placing dimension on a firm footing.

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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

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

1 followers·105 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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