Basis & Dimension of Vector Spaces

🎯 Learning Objectives

• Define a basis of a vector space
• Prove that all bases of a finite-dimensional space have the same size (dimension)
• Determine whether a given set is a basis for a vector space
• Find a basis for a subspace given in parametric or implicit form
• Compute the dimension of common vector spaces
• Use dimension to reason about subspaces

📋 Prerequisites

• Linear Independence and Span (Week 3) — a basis is an independent spanning set
• Gaussian Elimination (Week 2) — used to find bases computationally
• A basis is the "minimal" set of vectors you need to describe the whole space

1. Intuition: The Skeleton of a Space

• "3 blocks east, 2 blocks north"
• That's a coordinate relative to two reference directions (east, north). A basis is a set of reference directions that lets you describe every point in a vector space uniquely and efficiently. (Diagram) Key properties of a basis:

1. Spanning: Every vector in the space can be written as a combination of basis vectors
2. Independence: The representation is unique (no redundancy)

2. Definition of a Basis

Definition (Basis). A set $B = \{\mathbf{b}_1, \mathbf{b}_2, \dots, \mathbf{b}_n\}$ of vectors in a vector space $V$ is a basis of $V$ if:

1. $B$ is linearly independent
2. $B$ spans $V$ (i.e., $\text{span}(B) = V$) Example 1: Standard basis of $\mathbb{R}^n$

The standard basis of $\mathbb{R}^3$ is:

Independence: $c_1(1,0,0) + c_2(0,1,0) + c_3(0,0,1) = (c_1, c_2, c_3) = (0,0,0) \Rightarrow c_1=c_2=c_3=0$ ✓

Spanning: Any $(x,y,z) = x\mathbf{e}_1 + y\mathbf{e}_2 + z\mathbf{e}_3$ ✓ Example 2: A non-standard basis of $\mathbb{R}^2$

$B = \{(1,2), (3,-1)\}$

Independence: Check $c_1(1,2) + c_2(3,-1) = (0,0)$:

Independent ✓

Spanning: For any $(x,y)$, solve $(x,y) = c_1(1,2) + c_2(3,-1)$: $c_1 = \frac{x+3y}{7}$, $c_2 = \frac{2x-y}{7}$ — solution exists for all $(x,y)$ ✓

So $B$ is a basis. Coordinates of $(x,y)$ in this basis are $\left(\frac{x+3y}{7}, \frac{2x-y}{7}\right)$.

2.1 Uniqueness of Representation

Theorem. If $B = \{\mathbf{b}_1, \dots, \mathbf{b}_n\}$ is a basis of $V$, then every $\mathbf{v} \in V$ has a unique representation:

3. Dimension

3.1 All Bases Have the Same Size

Theorem (Invariance of Dimension). If $V$ has a basis with $n$ vectors, then every basis of $V$ has exactly $n$ vectors. Proof Sketch

Let $B_1$ have $n$ vectors and $B_2$ have $m$ vectors, both bases of $V$.

• Since $B_1$ spans $V$ and $B_2$ is independent, the Exchange Theorem (Steinitz) gives $m \leq n$.
• Since $B_2$ spans $V$ and $B_1$ is independent, $n \leq m$.
• Therefore $m = n$. This number is called the dimension of $V$, denoted $\dim(V)$. Definition (Dimension). The dimension of a vector space $V$ is the number of vectors in any basis of $V$. If $V = \{\mathbf{0}\}$, $\dim(V) = 0$.

3.2 Dimensions of Common Spaces

4. Finding Bases

4.1 From a Spanning Set to a Basis

1. Write the vectors as rows of a matrix
2. Row reduce to RREF
3. The non-zero rows form a basis for the span Algorithm (Column method):
4. Write the vectors as columns of a matrix
5. Row reduce to RREF
6. The original columns corresponding to pivot columns form a basis

Example 3: Finding a basis from a spanning set

Find a basis for $\text{span}\{(1,2,1), (2,3,4), (3,5,5), (1,1,3)\}$.

Row method:

Non-zero rows: $\{(1,0,5), (0,1,-2)\}$ form a basis. Dimension = 2.

Column method:

Pivots in columns 1 and 2. So original vectors $\{(1,2,1), (2,3,4)\}$ form a basis.

4.2 Finding a Basis for a Subspace Defined by Equations

1. Solve the system using Gaussian elimination
2. Express the solution in parametric form
3. The vectors multiplying the free parameters form a basis for the null space

Example 4: Basis for a plane subspace

Find a basis for $W = \{(x,y,z) \in \mathbb{R}^3 \mid x + 2y - z = 0\}$.

$x = -2y + z$, $y, z$ free.

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

4.3 Finding a Basis for the Column Space

Example 5: Basis for column space

Pivots in columns 1 and 2. Basis for $C(A)$: columns 1 and 2 of original $A$:

5. Dimension and Subspaces

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

• $\dim(W) \leq \dim(V)$
• $\dim(W) = \dim(V)$ if and only if $W = V$ Example 6: Dimensions of subspaces of $\mathbb{R}^3$

6. Edge Cases & Gotchas

7. Common Pitfalls

❌ Pitfall 1: Confusing dimension of a vector space with number of entries

❌ Pitfall 2: Taking RREF columns as basis vectors

❌ Pitfall 3: Thinking a set must be orthogonal to be a basis

❌ Pitfall 4: A spanning set with too few vectors

8. Formula Summary Table

9. 📝 Practice Questions

Q1: Is this a basis?

Is $B = \{(1,1), (1,-1)\}$ a basis for $\mathbb{R}^2$?

Solution: Check independence: $c_1(1,1)+c_2(1,-1) = (c_1+c_2, c_1-c_2) = (0,0)$ ⇒ $c_1+c_2=0$, $c_1-c_2=0$ ⇒ $c_1=c_2=0$. Independent ✓

Check spanning: For any $(x,y)$, solve $c_1+c_2 = x$, $c_1-c_2 = y$ ⇒ $c_1 = \frac{x+y}{2}$, $c_2 = \frac{x-y}{2}$. Exists for all $(x,y)$ ✓

Yes, $B$ is a basis. Q2: Not a basis — dependent

Is $B = \{(1,2,3), (2,4,6), (1,1,1)\}$ a basis for $\mathbb{R}^3$?

Solution: $(2,4,6) = 2(1,2,3)$, so the set is dependent. Not a basis.

Dimension of span is at most 2 (actually $\{(1,2,3), (1,1,1)\}$ spans a plane). Q3: Not a basis — doesn't span

Is $B = \{(1,0,0), (0,1,0)\}$ a basis for $\mathbb{R}^3$?

Solution: It spans the $xy$-plane only, not all of $\mathbb{R}^3$ (e.g., $(0,0,1)$ cannot be formed). Not a basis. Q4: Dimension of a subspace

Find a basis and the dimension of $W = \{(x,y,z,w) \mid x - y + 2z = 0, y + z - w = 0\}$.

Solution: Solve:

$y, z$ free.

Basis: $\{(1,1,0,1), (-2,0,1,1)\}$. Dimension = 2. Q5: Basis for polynomial subspace

Find a basis for $W = \{p(x) \in P_2 \mid p(1) = 0\}$.

Solution: $p(x) = a + bx + cx^2$, $p(1) = a + b + c = 0$, so $a = -b - c$.

$p(x) = (-b-c) + bx + cx^2 = b(x-1) + c(x^2-1)$.

Basis: $\{x-1, x^2-1\}$. Dimension = 2. Q6: Extending to a basis

Extend $\{(1,2,1)\}$ to a basis of $\mathbb{R}^3$.

Solution: Add standard basis vectors and eliminate dependent ones.

Pivots in rows 1-3, so $\{(1,2,1), (1,0,0), (0,1,0)\}$ is a basis (we added $\mathbf{e}_1$ and $\mathbf{e}_2$). Q7: Dimension of sum and intersection

In $\mathbb{R}^3$, $W_1 = \text{span}\{(1,0,0), (0,1,0)\}$ ($xy$-plane), $W_2 = \text{span}\{(0,1,0), (0,0,1)\}$ ($yz$-plane). Find $\dim(W_1 \cap W_2)$ and $\dim(W_1 + W_2)$.

Solution: $W_1 \cap W_2 = \text{span}\{(0,1,0)\}$ ($y$-axis), dimension 1.

$W_1 + W_2 = \mathbb{R}^3$, dimension 3.

Check: $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2) = 2 + 2 - 1 = 3$ ✓ Q8: Basis for matrix space

Find a basis for the subspace of $2 \times 2$ symmetric matrices.

Solution: A symmetric $2 \times 2$ matrix is

.

Basis:

. Dimension = 3. Q9: Proving dimension property

Prove: If $W$ is a subspace of $\mathbb{R}^n$ and $\dim(W) = n$, then $W = \mathbb{R}^n$.

Proof: $\dim(W) = n = \dim(\mathbb{R}^n)$. Since $W \subseteq \mathbb{R}^n$ and both have the same finite dimension, they must be equal. Q10: Coordinate vector

Find $[\mathbf{v}]_B$ where $\mathbf{v} = (5,3)$ and $B = \{(1,2), (2,1)\}$.

Solution: Solve $(5,3) = c_1(1,2) + c_2(2,1)$:

From first: $c_1 = 5-2c_2$. Substitute: $2(5-2c_2) + c_2 = 10 - 4c_2 + c_2 = 10 - 3c_2 = 3 \Rightarrow c_2 = \frac{7}{3}$. Then $c_1 = 5 - 2(\frac{7}{3}) = 5 - \frac{14}{3} = \frac{1}{3}$.

.

🔗 Cross-References

• Next topic: Rank & Dimension via Gaussian Elimination
• Week 5 (Rank-Nullity): Links dimensions of null space and column space
• Week 6 (Matrix Representation): The matrix of a linear transformation depends on the choice of basis
• Week 8 (Gram-Schmidt): Builds an orthonormal basis from any basis
• BSCS2004 (ML Foundations): Dimensionality reduction, feature selection Join Discord Previous3.3 Linear IndependenceNext4.2 Rank & Dimension