[Axler] Span and Linear Independence

Linear Algebra: Span and Linear Independence

Based on Linear Algebra Done Right by Sheldon Axler (Chapter 1)

This post covers the foundational concepts of Linear Combinations, Span, and the crucial relationship between Linear Independence and Spanning lists.

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0. Notation Reminder

• $\mathbf{F}$: Denotes either the field $\mathbf{R}$ (real numbers) or $\mathbf{C}$ (complex numbers).
• $V$: A vector space over $\mathbf{F}$.
• List: An ordered collection of elements (e.g., $(6, -4, 7)$ in $\mathbf{F}^3$).

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1. Linear Combinations

Given vectors $v_1, \dots, v_m$ in $V$, a linear combination is an expression of the form:

$$ a_1 v_1 + \dots + a_m v_m $$

where $a_1, \dots, a_m \in \mathbf{F}$.

Example in $\mathbf{R}^3$:
The vector $(13, -1, 7)$ is a linear combination of $(2, 1, -1)$ and $(1, -2, 4)$ because: $$ 5(2, 1, -1) + 3(1, -2, 4) = (13, -1, 7) $$ However, $(13, -1, 6)$ is not a linear combination of these vectors (no solution exists).

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2. Span

Definition

The span of $v_1, \dots, v_m$ is the set of all linear combinations of these vectors:

$$ \span(v_1, \dots, v_m) = \{ a_1 v_1 + \dots + a_m v_m : a_j \in \mathbf{F} \} $$

Key Result: The span is the smallest subspace of $V$ containing the list $v_1, \dots, v_m$.

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3. Finite-Dimensional Vector Spaces

$V$ is called finite-dimensional if there exists a finite list of vectors that spans $V$. Otherwise, it is infinite-dimensional.

• Finite: $\mathbf{F}^3$ (spanned by standard basis).
• Infinite: $\mathbf{F}^\infty$ (space of all infinite sequences).

Note: Most of linear algebra focuses on finite-dimensional spaces.

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4. Linear Independence

A list $v_1, \dots, v_m$ is linearly independent if the only way to make the zero vector is with all zero scalars:

$$ a_1 v_1 + \dots + a_m v_m = 0 \implies a_1 = \dots = a_m = 0 $$

Examples

• A single vector $(v)$ is independent $\iff v \ne 0$.
• The list $(1, x, x^2)$ in polynomial space is independent.
• Any list containing the vector $\mathbf{0}$ is dependent.

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5. Linear Dependence Lemma

If a list $v_1, \dots, v_m$ is linearly dependent, there exists an index $j$ such that:

1. $v_j$ is a linear combination of the preceding vectors $(v_1, \dots, v_{j-1})$.
2. Removing $v_j$ does not change the span: $$ \span(v_1, \dots, \hat{v}_j, \dots, v_m) = \span(v_1, \dots, v_m) $$

Example: In the list $((2, 1, -1), (1, -2, 4), (13, -1, 7))$, the third vector is a combo of the first two. Removing it preserves the span.

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6. Theorem: Independent vs. Spanning

Theorem

In a finite-dimensional vector space, the length of any linearly independent list is less than or equal to the length of any spanning list.

$$ \text{Length(Independent)} \le \text{Length(Spanning)} $$

Applications

• In $\mathbf{R}^3$, no list of 4 vectors can be linearly independent.
  (Reason: Standard basis spans $\mathbf{R}^3$ with length 3. Since $4 \not\le 3$, length 4 cannot be independent.)
• In $\mathbf{R}^4$, 3 vectors cannot span the space.
  (Reason: Standard basis is independent with length 4. Since $4 \not\le 3$, length 3 cannot span.)