LA-2.4

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Basis & Dimension

A basis is a minimal spanning set and a maximal independent set. The number of elements in any basis is the dimension—the most important invariant of a vector space.

Learning Objectives

Define basis as a linearly independent spanning set
Prove that all bases of a space have the same cardinality (dimension theorem)
Compute coordinates with respect to a given basis
Understand the coordinate isomorphism between V and F^n
Distinguish finite and infinite dimensional spaces
Find bases for common vector spaces (R^n, polynomials, matrices)
Apply the dimension formula for subspaces
Perform change of basis calculations
Use dimension arguments to solve problems
Connect dimension to rank of matrices

Prerequisites

Linear independence and dependence (LA-2.3)
Spanning sets and linear span (LA-2.2)
Vector space definition and axioms (LA-2.1)
Matrix row reduction and rank (LA-1.4)
Basic set theory notation

Historical Context

The concept of dimension evolved gradually through the 19th century. While geometric intuition suggested that lines are 1-dimensional, planes are 2-dimensional, and space is 3-dimensional, the abstract notion of dimension for arbitrary vector spaces was formalized by Hermann Grassmann (1844) and Giuseppe Peano (1888). The proof that dimension is well-defined—that all bases have the same size—relies on the Steinitz Exchange Lemma (1913), which Ernst Steinitz proved in his foundational work on field theory. This result is sometimes called the "Replacement Theorem" and is one of the most important theorems in linear algebra.

1. Basis

A basis is the "perfect" set of vectors for a vector space: it has exactly the right number of vectors to span the entire space without any redundancy. Think of it as a minimal set of building blocks from which every vector in the space can be uniquely constructed.

Definition 1.1: Basis

A basis of a vector space $V$ over a field $F$ is a set $\mathcal{B} \subseteq V$ that satisfies:

Linear independence: No vector in $\mathcal{B}$ is a linear combination of the others
Spanning: $\text{span}(\mathcal{B}) = V$

Remark 1.1: Intuition

A basis gives you:

Completeness: Every vector can be expressed using basis vectors
Non-redundancy: No basis vector is "extra"—removing any would lose spanning
Uniqueness: Each vector has exactly one representation in terms of the basis

Theorem 1.1: Basis Equivalences (TFAE)

For a finite set $\mathcal{B} = \{v_1, \ldots, v_n\}$ in a vector space $V$ , the following are equivalent:

$\mathcal{B}$ is a basis for $V$
Every $v \in V$ can be written uniquely as $v = \alpha_1 v_1 + \cdots + \alpha_n v_n$
$\mathcal{B}$ is a maximal linearly independent set
$\mathcal{B}$ is a minimal spanning set

Proof of Theorem 1.1:

(1) ⇒ (2): Since $\mathcal{B}$ spans, every $v$ can be written as $v = \sum \alpha_i v_i$ . If also $v = \sum \beta_i v_i$ , then $\sum (\alpha_i - \beta_i) v_i = 0$ . By independence, $\alpha_i = \beta_i$ for all $i$ .

(2) ⇒ (3): Unique representation implies independence (the zero vector has only the trivial representation). Adding any $w \notin \mathcal{B}$ creates dependence since $w$ already has a representation.

(3) ⇒ (4): If $\mathcal{B}$ is maximal independent but doesn't span, there exists $w \notin \text{span}(\mathcal{B})$ . Then $\mathcal{B} \cup \{w\}$ is independent, contradicting maximality.

(4) ⇒ (1): Minimal spanning implies independence (if dependent, remove a redundant vector while maintaining spanning).

∎

Example 1.1: Standard Basis of ℝⁿ

The standard basis of $\mathbb{R}^n$ is:

\mathcal{E} = \{e_1, e_2, \ldots, e_n\}

where $e_i = (0, \ldots, 0, 1, 0, \ldots, 0)$ has 1 in position $i$ and 0 elsewhere.

For example, in $\mathbb{R}^3$ :

e_1 = (1, 0, 0), \quad e_2 = (0, 1, 0), \quad e_3 = (0, 0, 1)

Verification:

Independent: $\sum \alpha_i e_i = 0$ implies $(\alpha_1, \ldots, \alpha_n) = (0, \ldots, 0)$
Spanning: Any $(x_1, \ldots, x_n) = x_1 e_1 + \cdots + x_n e_n$

Example 1.2: Standard Basis of Polynomials

For $P_n(F)$ = polynomials of degree ≤ $n$ , the standard basis is:

\{1, x, x^2, \ldots, x^n\}

This has $n + 1$ elements.

Example: $3x^2 - 2x + 5 = 5 \cdot 1 + (-2) \cdot x + 3 \cdot x^2$

Example 1.3: Standard Basis of Matrices

For $M_{m \times n}(F)$ , the standard basis consists of matrices $E_{ij}$ with 1 in position $(i, j)$ and 0 elsewhere.

For $M_{2 \times 2}(\mathbb{R})$ :

E_{11} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, E_{12} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, E_{21} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, E_{22} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}

Any matrix $A = (a_{ij})$ can be written as $A = \sum_{i,j} a_{ij} E_{ij}$ .

Example 1.4: Non-Standard Bases

In $\mathbb{R}^2$ , besides the standard basis $\{(1,0), (0,1)\}$ , other bases include:

$\{(1, 1), (1, -1)\}$ — diagonal directions
$\{(1, 0), (1, 1)\}$ — one standard, one diagonal
$\{(2, 3), (1, 2)\}$ — any two non-parallel vectors

Key insight: Any two linearly independent vectors in $\mathbb{R}^2$ form a basis.

Theorem 1.2: Extension Theorem

Every linearly independent set in a finite-dimensional vector space $V$ can beextended to a basis of $V$ .

Proof of Theorem 1.2:

Let $S$ be linearly independent in $V$ . If $\text{span}(S) = V$ , then $S$ is already a basis.

Otherwise, there exists $v \notin \text{span}(S)$ . Then $S \cup \{v\}$ is still independent (by the extension criterion).

Repeat this process. Since $V$ is finite-dimensional, there's an upper bound on independent set sizes (any spanning set size). So the process terminates with a basis.

∎

Theorem 1.3: Reduction Theorem

Every spanning set of a finite-dimensional vector space $V$ contains a basis.

Proof of Theorem 1.3:

Let $S$ span $V$ . If $S$ is independent, it's a basis.

Otherwise, some $v \in S$ is a linear combination of others: $v = \sum_{w \neq v} \alpha_w w$ .

Then $S \setminus \{v\}$ still spans $V$ . Repeat until independent.

∎

Corollary 1.1: Existence of Basis

Every finite-dimensional vector space has a basis.

Remark 1.2: The Empty Basis

The zero space $\{0\}$ has the empty set as its (unique) basis. This is consistent: the empty set is vacuously independent, and its span is $\{0\}$ (by convention, the empty sum equals the zero vector).

Example 1.5: Basis for Upper Triangular Matrices

For 2×2 upper triangular matrices over $\mathbb{R}$ :

U_2 = \left\{ \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} : a, b, c \in \mathbb{R} \right\}

Standard basis:

\left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\}

$\dim(U_2) = 3$ .

Example 1.6: Basis for Trace-Zero Matrices

For 2×2 matrices with trace 0:

\text{tr}(A) = a + d = 0 \implies d = -a

General element: $\begin{pmatrix} a & b \\ c & -a \end{pmatrix} = a\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + c\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$

$\dim(\text{sl}_2) = 3$ (This is the Lie algebra $\mathfrak{sl}_2$ ).

Theorem 1.4: Basis from Row Reduction

If $v_1, \ldots, v_m$ are vectors in $F^n$ , form the matrix with these as rows and row reduce. The original vectors corresponding to pivot rows form a basis for $\text{span}\{v_1, \ldots, v_m\}$ .

Example 1.7: Basis via Row Reduction

Problem: Find a basis for $\text{span}\{(1, 2, 1), (2, 4, 3), (1, 2, 2)\}$ .

Solution:

\begin{pmatrix} 1 & 2 & 1 \\ 2 & 4 & 3 \\ 1 & 2 & 2 \end{pmatrix} \to \begin{pmatrix} 1 & 2 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}

Pivots in rows 1 and 2. Basis: $\{(1, 2, 1), (2, 4, 3)\}$ . Dimension = 2.

2. Dimension

The dimension of a vector space is the number of vectors in any basis. The remarkable fact is that this number is the same for all bases—a consequence of the Steinitz Exchange Lemma. Dimension is the most important invariant of a vector space.

Theorem 2.1: Dimension is Well-Defined

Any two bases of a finite-dimensional vector space have the same number of elements.

Proof of Theorem 2.1:

Let $\mathcal{B}_1$ and $\mathcal{B}_2$ be two bases with $m$ and $n$ elements respectively.

$\mathcal{B}_1$ is independent and $\mathcal{B}_2$ spans, so by the Fundamental Inequality: $m \leq n$ .

$\mathcal{B}_2$ is independent and $\mathcal{B}_1$ spans, so: $n \leq m$ .

Therefore $m = n$ .

∎

Definition 2.1: Dimension

The dimension of a finite-dimensional vector space $V$ , denoted $\dim(V)$ or $\dim_F(V)$ , is the number of elements in any basis of $V$ .

By convention: $\dim(\{0\}) = 0$ .

Example 2.1: Standard Dimensions

Vector Space	Standard Basis	Dimension
$\mathbb{R}^n$	$\{e_1, \ldots, e_n\}$	$n$
$P_n(F)$	$\{1, x, \ldots, x^n\}$	$n + 1$
$M_{m \times n}(F)$	$\{E_{ij}\}$	$mn$
$\mathbb{C}$ over $\mathbb{R}$	$\{1, i\}$	2
$\{0\}$	$\emptyset$	0

Theorem 2.2: Dimension Criteria

Let $\dim(V) = n$ . Then:

Any $n$ linearly independent vectors form a basis
Any $n$ spanning vectors form a basis
Any set of more than $n$ vectors is linearly dependent
Any set of fewer than $n$ vectors cannot span $V$

Proof of Theorem 2.2:

(1): If $S$ has $n$ independent vectors but doesn't span, extend to a basis of size $>$ n. Contradiction.

(2): If $S$ spans with $n$ vectors but is dependent, reduce to a basis of size $<$ n. Contradiction.

(3): By Fundamental Inequality: |independent| ≤ |spanning| = n.

(4): By Fundamental Inequality: |spanning| ≥ |independent| = n.

∎

Corollary 2.1: Subspace Dimension

If $W$ is a subspace of a finite-dimensional space $V$ , then:

$W$ is finite-dimensional
$\dim(W) \leq \dim(V)$
$\dim(W) = \dim(V) \iff W = V$

Proof of Corollary 2.1:

(1), (2): Any independent set in $W$ is independent in $V$ , so has size ≤ dim(V). Thus $W$ has a basis of finite size.

(3): If dim(W) = dim(V) = n, a basis of $W$ has $n$ independent vectors in $V$ , hence is a basis of $V$ . So span = V.

∎

Theorem 2.3: Dimension Formula for Sums

For subspaces $U$ and $W$ of $V$ :

\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W)

Proof of Theorem 2.3:

Let $\{u_1, \ldots, u_k\}$ be a basis of $U \cap W$ .

Extend to bases: $\{u_1, \ldots, u_k, v_1, \ldots, v_m\}$ of $U$ and $\{u_1, \ldots, u_k, w_1, \ldots, w_n\}$ of $W$ .

One can show $\{u_1, \ldots, u_k, v_1, \ldots, v_m, w_1, \ldots, w_n\}$ is a basis for $U + W$ .

Count: $k + m + n = (k + m) + (k + n) - k = \dim(U) + \dim(W) - \dim(U \cap W)$ .

∎

Example 2.2: Dimension of Sum

In $\mathbb{R}^3$ , let $U = \text{span}\{(1,0,0), (0,1,0)\}$ (xy-plane) and $W = \text{span}\{(0,1,0), (0,0,1)\}$ (yz-plane).

$\dim(U) = 2$ , $\dim(W) = 2$
$U \cap W = \text{span}\{(0,1,0)\}$ (y-axis), so $\dim(U \cap W) = 1$
$\dim(U + W) = 2 + 2 - 1 = 3$

Indeed, $U + W = \mathbb{R}^3$ .

Remark 2.1: Finite vs Infinite Dimensional

A vector space is:

Finite-dimensional: has a finite spanning set (equivalently, a finite basis)
Infinite-dimensional: not finite-dimensional

Examples of infinite-dimensional spaces:

$F[x]$ — all polynomials (basis: $1, x, x^2, x^3, \ldots$ )
$C[0,1]$ — continuous functions on [0,1]
$\ell^2$ — square-summable sequences

Example 2.3: Dimension of Kernel

For a linear map $T: V \to W$ , the nullity is $\dim(\ker T)$ .

If $T$ is represented by matrix $A$ , then:

\text{nullity}(A) = n - \text{rank}(A)

This is the rank-nullity theorem (covered in Part III).

Example 2.4: Dimension of Image

For $T: V \to W$ , the rank is $\dim(\text{im } T)$ .

Key relationship:

\dim(V) = \dim(\ker T) + \dim(\text{im } T)

Theorem 2.4: Dimension and Isomorphism

Two finite-dimensional vector spaces over the same field are isomorphic if and only if they have the same dimension.

Proof of Theorem 2.4:

(⇒) If $T: V \to W$ is an isomorphism and $\mathcal{B}$ is a basis of $V$ , then $T(\mathcal{B})$ is a basis of $W$ (bijection preserves size).

(⇐) If $\dim(V) = \dim(W) = n$ , choose bases $\mathcal{B}_V$ and $\mathcal{B}_W$ . Define $T(v_i) = w_i$ and extend linearly. This is an isomorphism.

∎

Corollary 2.2: Classification

Every $n$ -dimensional vector space over $F$ is isomorphic to $F^n$ . Thus, up to isomorphism, there is exactly one $n$ -dimensional space over $F$ for each $n \geq 0$ .

Remark 2.2: Dimension as Complete Invariant

For finite-dimensional spaces, dimension completely characterizes the isomorphism class. Two spaces are "the same" (structurally) iff they have the same dimension over the same field.

3. Coordinates

Once we fix a basis, every vector gets a unique set of coordinates—the coefficients in its representation as a linear combination of basis vectors. This gives us a powerful correspondence between abstract vectors and concrete column vectors in $F^n$ .

Definition 3.1: Ordered Basis

An ordered basis is a basis together with a specific ordering of its elements:

\mathcal{B} = (v_1, v_2, \ldots, v_n)

The ordering matters for defining coordinates.

Definition 3.2: Coordinate Vector

Let $\mathcal{B} = (v_1, \ldots, v_n)$ be an ordered basis of $V$ . For any $v \in V$ , write:

v = \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n

The coordinate vector of $v$ with respect to $\mathcal{B}$ is:

[v]_{\mathcal{B}} = \begin{pmatrix} \alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n \end{pmatrix} \in F^n

Remark 3.1: Uniqueness of Coordinates

Since $\mathcal{B}$ is a basis, the representation $v = \sum \alpha_i v_i$ isunique. So the coordinate vector $[v]_{\mathcal{B}}$ is well-defined.

Example 3.1: Standard Coordinates

In $\mathbb{R}^3$ with the standard basis $\mathcal{E} = (e_1, e_2, e_3)$ :

[(2, -3, 5)]_{\mathcal{E}} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix}

For the standard basis, coordinates equal components.

Example 3.2: Non-Standard Coordinates

In $\mathbb{R}^2$ , let $\mathcal{B} = ((1, 1), (1, -1))$ .

Find $[(3, 1)]_{\mathcal{B}}$ .

Solution: Solve $(3, 1) = \alpha(1, 1) + \beta(1, -1)$ :

\begin{cases} \alpha + \beta = 3 \\ \alpha - \beta = 1 \end{cases} \implies \alpha = 2, \beta = 1

So $[(3, 1)]_{\mathcal{B}} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$ .

Theorem 3.1: Coordinate Isomorphism

Let $\mathcal{B}$ be an ordered basis of an $n$ -dimensional space $V$ . The map:

\phi_{\mathcal{B}}: V \to F^n, \quad v \mapsto [v]_{\mathcal{B}}

is a linear isomorphism.

Proof of Theorem 3.1:

Linear: If $v = \sum \alpha_i v_i$ and $w = \sum \beta_i v_i$ , then:

v + w = \sum (\alpha_i + \beta_i) v_i \implies [v + w]_{\mathcal{B}} = [v]_{\mathcal{B}} + [w]_{\mathcal{B}}

Similarly for scalar multiplication.

Bijective:

Injective: $[v]_{\mathcal{B}} = 0 \implies v = 0$
Surjective: For any $(\alpha_1, \ldots, \alpha_n) \in F^n$ , the vector $v = \sum \alpha_i v_i$ maps to it

∎

Corollary 3.1: Finite-Dimensional Spaces are Isomorphic to Fⁿ

Every $n$ -dimensional vector space over $F$ is isomorphic to $F^n$ .

Remark 3.2: Why Coordinates Matter

The coordinate isomorphism lets us:

Reduce abstract problems to concrete matrix calculations
Represent linear maps as matrices
Use computers to solve linear algebra problems
Transfer results from $F^n$ to any $n$ -dimensional space

Example 3.3: Polynomial Coordinates

In $P_2(\mathbb{R})$ with basis $\mathcal{B} = (1, x, x^2)$ :

[3x^2 - 2x + 5]_{\mathcal{B}} = \begin{pmatrix} 5 \\ -2 \\ 3 \end{pmatrix}

The coefficients of the polynomial become coordinates.

Example 3.4: Matrix Coordinates

In $M_{2 \times 2}(\mathbb{R})$ with standard basis $(E_{11}, E_{12}, E_{21}, E_{22})$ :

\left[ \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \right]_{\mathcal{E}} = \begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \end{pmatrix}

Theorem 3.2: Change of Basis

Let $\mathcal{B}$ and $\mathcal{B}'$ be two ordered bases of $V$ . There exists an invertible matrix $P$ (the change of basis matrix) such that:

[v]_{\mathcal{B}'} = P^{-1} [v]_{\mathcal{B}}

The columns of $P$ are $[v_1']_{\mathcal{B}}, \ldots, [v_n']_{\mathcal{B}}$ .

Example 3.5: Change of Basis Calculation

Let $\mathcal{B} = ((1, 0), (0, 1))$ and $\mathcal{B}' = ((1, 1), (1, -1))$ in $\mathbb{R}^2$ .

Change of basis matrix:

P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \quad P^{-1} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}

Example: For $v = (3, 1)$ :

[v]_{\mathcal{B}} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}, \quad [v]_{\mathcal{B}'} = P^{-1} [v]_{\mathcal{B}} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}

Theorem 3.3: Coordinates and Linear Operations

For any ordered basis $\mathcal{B}$ and vectors $u, v \in V$ , scalar $\alpha \in F$ :

$[u + v]_{\mathcal{B}} = [u]_{\mathcal{B}} + [v]_{\mathcal{B}}$
$[\alpha v]_{\mathcal{B}} = \alpha [v]_{\mathcal{B}}$
$[0]_{\mathcal{B}} = 0$ (zero vector maps to zero vector)

This is why coordinates give an isomorphism—they preserve all vector space operations.

Example 3.6: Operations via Coordinates

In $\mathbb{R}^2$ with $\mathcal{B} = ((1, 1), (1, -1))$ :

Let $u = (4, 2)$ and $v = (1, 3)$ .

Find coordinates:

$[u]_{\mathcal{B}} = (3, 1)^T$ (since $u = 3(1,1) + 1(1,-1)$ )
$[v]_{\mathcal{B}} = (2, -1)^T$ (since $v = 2(1,1) - 1(1,-1)$ )

Add in coordinates: $[u]_{\mathcal{B}} + [v]_{\mathcal{B}} = (5, 0)^T$ .

Verify: $u + v = (5, 5) = 5(1,1) + 0(1,-1)$ . ✓

Remark 3.3: Matrix of Linear Map

If $T: V \to W$ is linear and we fix bases $\mathcal{B}_V$ , $\mathcal{B}_W$ , then $T$ can be represented by a matrix $A$ such that:

[T(v)]_{\mathcal{B}_W} = A [v]_{\mathcal{B}_V}

This is covered in detail in Part III: Linear Mappings.

Example 3.7: Coordinates in Function Spaces

In $P_2(\mathbb{R})$ with Lagrange interpolation basis at points $0, 1, 2$ :

L_0(x) = \frac{(x-1)(x-2)}{2}, \quad L_1(x) = -x(x-2), \quad L_2(x) = \frac{x(x-1)}{2}

For any polynomial $p(x)$ of degree ≤ 2:

[p]_{\mathcal{L}} = (p(0), p(1), p(2))^T

The coordinates are simply the function values! This is a key idea in interpolation.

4. Worked Examples

Example 4.1: Finding a Basis for a Subspace

Problem: Find a basis for $W = \{(x, y, z) \in \mathbb{R}^3 : x + y - z = 0\}$ .

Solution: Solve $z = x + y$ , so:

W = \{(x, y, x + y) : x, y \in \mathbb{R}\} = \{x(1, 0, 1) + y(0, 1, 1)\}

Basis: $\{(1, 0, 1), (0, 1, 1)\}$ .

Dimension: $\dim(W) = 2$ .

Example 4.2: Extending to a Basis

Problem: Extend $\{(1, 1, 0)\}$ to a basis of $\mathbb{R}^3$ .

Solution: Need 2 more independent vectors.

Try $e_1 = (1, 0, 0)$ : Independent from $(1, 1, 0)$ ? Yes (not scalar multiple).

Try $e_3 = (0, 0, 1)$ : Independent from both? Yes (third component non-zero).

Basis: $\{(1, 1, 0), (1, 0, 0), (0, 0, 1)\}$ .

Example 4.3: Dimension of Solution Space

Problem: Find dim(ker A) for:

A = \begin{pmatrix} 1 & 2 & 1 & 0 \\ 2 & 4 & 3 & 1 \\ 1 & 2 & 2 & 1 \end{pmatrix}

Solution: Row reduce:

\begin{pmatrix} 1 & 2 & 1 & 0 \\ 2 & 4 & 3 & 1 \\ 1 & 2 & 2 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 2 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}

Rank = 2 (two pivots). By rank-nullity:

\dim(\ker A) = 4 - 2 = 2

Example 4.4: Dimension Argument

Problem: If $\{v_1, v_2, v_3\}$ spans $\mathbb{R}^3$ , prove it's a basis.

Solution: We know $\dim(\mathbb{R}^3) = 3$ .

Any 3 spanning vectors in a 3-dimensional space must be a basis (by dimension criteria).

Alternatively: if dependent, could reduce to a smaller spanning set, contradicting that minimum is 3.

Example 4.5: Intersection Dimension

Problem: In $\mathbb{R}^4$ , let $U$ and $W$ be 3-dimensional subspaces. What are the possible values of $\dim(U \cap W)$ ?

Solution: By dimension formula:

\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W) = 6 - \dim(U \cap W)

Since $U + W \subseteq \mathbb{R}^4$ , we have $\dim(U + W) \leq 4$ .

So $6 - \dim(U \cap W) \leq 4$ , giving $\dim(U \cap W) \geq 2$ .

Also $\dim(U \cap W) \leq 3$ (subspace of $U$ ).

Answer: $\dim(U \cap W) \in \{2, 3\}$ .

Example 4.6: Basis for Symmetric Matrices

Problem: Find a basis for symmetric 2×2 matrices.

Solution: Symmetric matrices have form $\begin{pmatrix} a & b \\ b & c \end{pmatrix}$ .

Basis:

\left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\}

Dimension: 3.

Example 4.7: Coordinates in Polynomial Space

Problem: Find $[p(x)]_{\mathcal{B}}$ where $p(x) = x^2 + 2x + 3$ and $\mathcal{B} = (1, 1+x, 1+x+x^2)$ .

Solution: Find $a, b, c$ such that:

x^2 + 2x + 3 = a \cdot 1 + b(1+x) + c(1+x+x^2)

Comparing coefficients:

Constant: $a + b + c = 3$
$x$ : $b + c = 2$
$x^2$ : $c = 1$

So $c = 1, b = 1, a = 1$ .

Answer: $[p(x)]_{\mathcal{B}} = (1, 1, 1)^T$ .

Example 4.8: Basis for Row Space

Problem: Find a basis for the row space of:

A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 4 & 0 \\ 1 & 2 & -1 \end{pmatrix}

Solution: Row reduce:

\begin{pmatrix} 1 & 2 & 1 \\ 2 & 4 & 0 \\ 1 & 2 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & 2 & 1 \\ 0 & 0 & -2 \\ 0 & 0 & -2 \end{pmatrix} \to \begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}

Basis: $\{(1, 2, 0), (0, 0, 1)\}$ . Dimension = 2.

Example 4.9: Dimension via Parametrization

Problem: Find dim of $W = \{(x, y, z, w) : x - y + z = 0, y - w = 0\}$ .

Solution: From equations: $y = w$ , $x = y - z = w - z$ .

Parametrize with free variables $z, w$ :

(x, y, z, w) = (w - z, w, z, w) = z(-1, 0, 1, 0) + w(1, 1, 0, 1)

Basis: $\{(-1, 0, 1, 0), (1, 1, 0, 1)\}$ . dim(W) = 2.

Example 4.10: Proving Equality of Subspaces

Problem: Let $U = \text{span}\{(1, 1, 0), (0, 1, 1)\}$ and $W = \{(x, y, z) : x - y + z = 0\}$ . Show $U = W$ .

Solution:

Step 1: Check $U \subseteq W$ :

$(1, 1, 0)$ : $1 - 1 + 0 = 0$ ✓
$(0, 1, 1)$ : $0 - 1 + 1 = 0$ ✓

Step 2: Dimensions:

$\dim(U) = 2$ (two independent vectors)
$\dim(W) = 3 - 1 = 2$ (one constraint in $\mathbb{R}^3$ )

Since $U \subseteq W$ and $\dim(U) = \dim(W)$ , we have $U = W$ .

Example 4.11: Direct Sum Dimension

Problem: If $V = U \oplus W$ (direct sum), prove $\dim(V) = \dim(U) + \dim(W)$ .

Solution: Direct sum means $U \cap W = \{0\}$ and $U + W = V$ .

By dimension formula:

\dim(V) = \dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W) = \dim(U) + \dim(W) - 0

Example 4.12: Finding Change of Basis Matrix

Problem: Find the change of basis matrix from $\mathcal{B} = ((1, 0), (0, 1))$ to $\mathcal{B}' = ((1, 2), (3, 4))$ .

Solution: Express $\mathcal{B}'$ vectors in terms of $\mathcal{B}$ :

$(1, 2) = 1 \cdot (1, 0) + 2 \cdot (0, 1)$ , so $[(1, 2)]_{\mathcal{B}} = (1, 2)^T$
$(3, 4) = 3 \cdot (1, 0) + 4 \cdot (0, 1)$ , so $[(3, 4)]_{\mathcal{B}} = (3, 4)^T$

P = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}

To convert coordinates: $[v]_{\mathcal{B}'} = P^{-1} [v]_{\mathcal{B}}$ .

5. Common Mistakes

Mistake 1: Confusing spanning set with basis

A spanning set may be larger than a basis. Example: $\{e_1, e_2, e_1 + e_2\}$ spans $\mathbb{R}^2$ but is NOT a basis (dependent, has 3 vectors in a 2D space).

Mistake 2: Forgetting dimension of polynomial spaces

$P_n(F)$ has dimension $n + 1$ , not $n$ ! The degree ≤ n polynomials include constants, so the basis $\{1, x, \ldots, x^n\}$ has $n + 1$ elements.

Mistake 3: Assuming coordinates are unique without a basis

Coordinates are only unique when the representing set is a basis. For a dependent spanning set, the same vector can have multiple representations.

Mistake 4: Ignoring the ordering of basis vectors

Coordinates depend on the ORDER of basis vectors. If $\mathcal{B} = (v_1, v_2)$ , then $[v_1 + v_2]_{\mathcal{B}} = (1, 1)^T$ . Swap the order and coordinates change!

Mistake 5: dim(U ∩ W) = dim(U) ∩ dim(W)?

NO! $\dim(U \cap W)$ is NOT $\min(\dim U, \dim W)$ . Use the dimension formula: $\dim(U \cap W) = \dim U + \dim W - \dim(U + W)$ .

Mistake 6: Thinking the zero vector can be a basis element

A basis must be linearly independent. Any set containing the zero vector is dependent, so $0$ can never be part of a basis.

Mistake 7: Wrong dimension after change of basis

Change of basis does NOT change dimension! The same space can be described with different bases, but the number of basis vectors is always the same (the dimension).

Mistake 8: Forgetting dimension depends on the field

$\mathbb{C}$ over $\mathbb{C}$ is 1-dimensional, but $\mathbb{C}$ over $\mathbb{R}$ is 2-dimensional. Always specify the field!

✓ Basis Verification Checklist

To verify $\{v_1, \ldots, v_n\}$ is a basis for $V$ :

Check independence: Solve $\sum \alpha_i v_i = 0$ . Only trivial solution?
Check spanning: Can every $v \in V$ be written as $\sum \alpha_i v_i$ ?
OR: Check count: If |S| = dim(V), need only one of above!

Shortcut for n vectors in n-dim space:

If independent → automatically spans → basis
If spans → automatically independent → basis

6. Key Takeaways

Basis = Independent + Spanning

A basis is the "perfect size": big enough to span, small enough to be independent.

Dimension is Well-Defined

All bases have the same size—this is not obvious but follows from the Steinitz Exchange Lemma.

n Vectors in n-dim Space

In an $n$ -dimensional space: $n$ independent vectors = basis = $n$ spanning vectors.

Coordinates = Unique Representation

With an ordered basis, every vector has unique coordinates—this is the key to computation.

Coordinate Isomorphism

Every $n$ -dimensional space is isomorphic to $F^n$ —abstract spaces become concrete!

Dimension Formula

$\dim(U + W) = \dim U + \dim W - \dim(U \cap W)$

Quick Reference Table

Vector Space	Dimension	Standard Basis
ℝⁿ	n	e₁, ..., eₙ
ℂ over ℝ	2	1, i
Pₙ(F)	n + 1	1, x, x², ..., xⁿ
Mₘₓₙ(F)	mn	Eᵢⱼ
Symmetric n×n	n(n+1)/2	Eᵢᵢ, Eᵢⱼ + Eⱼᵢ
Skew-symmetric n×n	n(n-1)/2	Eᵢⱼ - Eⱼᵢ (i < j)

7. Applications

Computer Graphics

3D coordinates use basis vectors. Change of basis rotates/transforms scenes. Homogeneous coordinates add a dimension for projective transformations.

Data Compression

SVD/PCA find low-dimensional subspaces capturing most variance. Dimension reduction = finding a good low-rank basis.

Differential Equations

Solution spaces of linear ODEs are vector spaces. Dimension = order of equation. Finding basis solutions gives the general solution.

Quantum Mechanics

Quantum states live in Hilbert spaces. Observables have eigenbases. Measurements correspond to basis projections.

Example 7.1: Rank-Nullity in Data Science

For a data matrix $A \in \mathbb{R}^{m \times n}$ (m samples, n features):

$\text{rank}(A)$ = dimension of column space = number of "effective" features
$\dim(\ker A)$ = number of linear dependencies in features
If rank < n, there's redundancy in the features (multicollinearity)

Example 7.2: Network Analysis

In a network with $n$ nodes and $m$ edges:

The cycle space has dimension $m - n + c$ (c = connected components)
This counts independent cycles in the network
Used in electrical circuit analysis and graph theory

Remark 7.1: The Power of Dimension

Dimension is a complete invariant for finite-dimensional vector spaces: two spaces are isomorphic if and only if they have the same dimension. This is why dimension is often the first thing we compute about a vector space.

Example 7.3: Machine Learning: Feature Engineering

In machine learning, consider a dataset with 100 features:

If rank of feature matrix is only 10, the "effective dimension" is 10
90 features are redundant (linearly dependent on others)
PCA finds an optimal 10-dimensional basis

Dimension reduction = finding a low-dimensional basis that captures most information.

Example 7.4: Coding Theory

A linear code $C$ is a subspace of $F_q^n$ :

dim(C) = k means 2ᵏ codewords (for binary codes)
A $[n, k, d]$ code has block length n, dimension k, minimum distance d
Generator matrix has k rows (basis of code)
Parity check matrix has n-k rows (basis of orthogonal complement)

Example 7.5: Structural Engineering

In structural analysis, forces and displacements live in vector spaces:

Degrees of freedom = dimension of displacement space
Constraints reduce dimension (fixing a point removes 3 DOF in 3D)
Stiffness matrix relates force and displacement bases

Why Bases Matter in Practice

• Computation: Algorithms work with coordinates, not abstract vectors
• Compression: Good basis → sparse coordinates → efficient storage
• Analysis: Eigenbasis simplifies linear transformations
• Visualization: Project to 2D/3D basis for plotting high-dim data
• Stability: Orthonormal bases minimize numerical errors

8. Quick Reference

Finding a Basis

Write spanning set
Form matrix with vectors as rows/columns
Row reduce
Pivot positions give basis vectors

Finding Coordinates

Set up $v = \sum \alpha_i b_i$
Create system of equations
Solve for coefficients
Coordinates = coefficient vector

Checking if Basis

Count vectors: should equal dim(V)
Check independence (or spanning)
Either condition + right count = basis

Key Formulas

• dim(Fⁿ) = n
• dim(Pₙ) = n + 1
• dim(Mₘₓₙ) = mn
• dim(U + W) = dim(U) + dim(W) - dim(U ∩ W)

Remark 8.1: Looking Ahead

With basis and dimension established, we can now represent linear maps as matrices. The choice of bases for domain and codomain determines the matrix representation. This connection between abstract linear maps and concrete matrices is one of the most powerful ideas in linear algebra.

Algorithm: Finding Basis from Spanning Set

Input: Spanning set $S = \{v_1, \ldots, v_m\}$

Form matrix $A$ with $v_i$ as columns
Row reduce $A$ to echelon form
Identify pivot columns (positions 1, 2, ...)
Output: Vectors corresponding to pivot columns form basis

Complexity: O(mn²) for m vectors in n dimensions

Algorithm: Extending to Basis

Input: Independent set $S = \{v_1, \ldots, v_k\}$ in $F^n$

Form matrix with $v_i$ as columns
Append standard basis vectors $e_1, \ldots, e_n$
Row reduce the augmented matrix
Take vectors corresponding to first $n$ pivot columns

Result: Basis containing original vectors plus some standard vectors

Dimension Checklist

✓ To find dimension of a subspace:

Find a spanning set (parametrize or from generators)
Reduce to basis (row reduce)
Count basis vectors = dimension

✓ To show dim(U) = dim(W):

Find bases and count, OR
Show U ⊆ W and exhibit dim(U) independent vectors in U

✓ To show U = W (subspaces):

Show U ⊆ W and W ⊆ U, OR
Show U ⊆ W and dim(U) = dim(W)

9. Chapter Summary

Core Definitions

Basis: Linearly independent spanning set
Dimension: Number of vectors in any basis
Coordinate vector: Coefficients in basis representation

Key Theorems

All bases of a space have the same size (dimension is well-defined)
n independent vectors in n-dim space ⟺ basis ⟺ n spanning vectors
Coordinate map is a linear isomorphism: V ≅ Fⁿ
dim(U + W) = dim(U) + dim(W) - dim(U ∩ W)

Key Skills

Find bases via row reduction
Extend independent sets to bases
Compute coordinates with respect to any basis
Calculate change of basis matrices
Use dimension arguments to prove subspace equality

Standard Dimensions

ℝⁿ: n | Pₙ(F): n+1 | Mₘₓₙ(F): mn | Symmetric n×n: n(n+1)/2

Remark 9.1: What's Next

With basis and dimension established, we're ready for:

Direct sums: Decomposing spaces into "independent" subspaces
Linear maps: Functions that preserve vector space structure
Matrix representation: Representing linear maps using coordinates

The dimension of domain and codomain determines the size of representing matrices!

Basis & Dimension Practice

Questions

Correct

Accuracy

What is

\dim(\mathbb{R}^5)

Easy

Not attempted

What is

\dim(M_{2 \times 3}(\mathbb{R}))

Easy

Not attempted

What is

\dim(\mathbb{R}_3[x])

(polynomials of degree ≤ 3)?

Easy

Not attempted

\dim(V) = n

, how many vectors are in any basis of

V

Medium

Not attempted

\mathbb{R}^3

, if

W

is a plane through the origin, what is

\dim(W)

Medium

Not attempted

Can 3 vectors form a basis for

\mathbb{R}^4

Medium

Not attempted

What is

[v]_B

v = (3, 4)

and

B = \{(1, 0), (0, 1)\}

Medium

Not attempted

\{(1,1), (1,-1)\}

a basis for

\mathbb{R}^2

Hard

Not attempted

W

is a subspace of

V

and

\dim(W) = \dim(V)

, what can you conclude?

Hard

Not attempted

What is the dimension of the solution space of

Ax = 0

where

A

3 \times 5

with rank 2?

Hard

Not attempted

\{v_1, v_2\}

is a basis of

V

and

w = 3v_1 - 2v_2

, what is

[w]_B

Medium

Not attempted

What is

\dim(\mathbb{R}[x])

(all polynomials)?

Hard

Not attempted

Frequently Asked Questions

Why is dimension well-defined?

The Steinitz exchange lemma shows that if you have a spanning set of size n and an independent set of size m, then m ≤ n. Applying this both ways to two bases shows they have the same size.

What's the difference between a basis and a spanning set?

A spanning set may have redundant vectors. A basis is a minimal spanning set—remove any vector and it no longer spans. Equivalently, it's a maximal independent set.

How do I find the dimension of a subspace?

Find a basis (e.g., by row reducing the matrix whose rows generate the subspace) and count the basis vectors. Or use the rank of the associated matrix.

Can a space have multiple different bases?

Yes! There are infinitely many bases for any space of dimension ≥ 1. All bases have the same size but contain different vectors. Coordinates depend on basis choice.

What about infinite-dimensional spaces?

Spaces like F[x] (all polynomials) or C[0,1] (continuous functions) are infinite-dimensional—no finite set can span them. They still have bases (by Zorn's lemma) but of infinite cardinality.

How do coordinates change when I change the basis?

If you switch from basis B to basis B', there's a change of basis matrix P such that [v]_B' = P^{-1}[v]_B. The matrix P has columns that are the B-coordinates of the B' basis vectors.

What is the relationship between dimension and rank?

The rank of a matrix equals the dimension of its column space (or row space). For a linear map T: V → W, rank(T) = dim(im(T)) and the rank-nullity theorem says dim(V) = dim(ker T) + dim(im T).

Why is the zero space 0-dimensional?

The zero space {0} has no basis vectors (the empty set is its basis), so dim({0}) = 0. This is consistent: any n vectors containing 0 are dependent, so you can't have a non-empty basis.

Can I always extend an independent set to a basis?

Yes, in finite-dimensional spaces. Start with an independent set, and if it doesn't span, add vectors one by one that aren't in the current span. This process terminates with a basis.

What's the dimension of a sum of subspaces?

For subspaces U and W: dim(U + W) = dim(U) + dim(W) - dim(U ∩ W). This is the dimension formula for sums, analogous to inclusion-exclusion in counting.