Linear independence

Overview §

A set of vectors in a vector space $\{v_1,\cdots ,v_n\}\subset V$ is linearly independent if any linear relation has all coefficients equal to zero.

Redundant vectors can be deleted from an ordered list to get a linearly independent set with the same span as the original (Lemma 27).

Additionally, given a linearly independent set and a spanning set, the size of the linearly independent set must be less than or equal to the size of the span (Theorem 31).

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Linear relations §

Definition: Linear relation

Let $V$ be a vector space over a field $\mathbb{F}$. Given a set $S=\{v_1,\cdots ,v_n\}\subset V$, a linear relation between them is a sequence $(a_1,\cdots ,a_n)\in \mathbb{F}$ whose linear combination produces the zero vector.

$$a_1{v}_1+\cdots a_n{v}_n=\vec{0}$$

$(0,\cdots ,0)$ is the trivial linear relation, which doesn’t say anything “interesting” about the relation between elements of the vector set.

$$0v_1+\cdots +0v_n=\vec{0}$$

A nontrivial linear relation means there exists an $i$ so that $a_i\ne 0$.

The set of linear relations $(a_1,\cdots ,a_n)\in {\mathbb{F}}^n$ for $\{v_1,\cdots ,v_n\}\subset V$ is a linear subspace of ${\mathbb{F}}^n$, since each condition in the subspace definition is satisfied:

• The zero vector is the trivial relation.
• Addition is closed, since the sum of any two sequences $(a_1+b_1,\cdots ,a_n+b_n)$ is a linear relation producing $\vec{0}+\vec{0}=\vec{0}$.
• Scalar multiplication is closed by a similar argument.

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Linear independence §

Definition: Linear independence

A set $S=\{v_1,\cdots ,v_n\}$ is linearly independent if the only linear relation between each vector is the trivial linear relation.

$$a_1{v}_1+\cdots +a_n{v}_n=\vec{0}⟹a_1=0,\cdots ,a_n=0$$

Alternatively, $S$ is linearly independent if there exists a nontrivial linear relation so that $a_i\ne 0$ for some $1\le i\le n$.

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Redundancy §

Definition: Redundant vector

Given vector space $V$ and an ordered of vectors $(v_1,\cdots ,v_n)$ with all $v_i\in V$ for all $1\le i\le n$, a vector $v_i$ is redundant if it can be written as a linear combination of the previous elements in the list.

$$v_i=a_1{v}_1+\cdots +a_{i-1}{v}_{i-1}$$ $$v_i\in \text{span}(v_1,\cdots ,v_{i-1})$$

• (Lemma 27) Given vector space $V$, if $(v_1,\cdots ,v_n)\in V$ is a list of vectors and $v_i$ is redundant for some $1\le i\le n$, then the span of the list does not change does not change by excluding $v_i$.

$$\text{span}(v_1,\cdots ,v_{i-1},v_{i+1},\cdots ,v_n)=\text{span}(v_1,\cdots v_n)$$

• (Corallary 28) Given a list of vectors in a vector space $(v_1,\cdots ,v_n)\in V$, a smaller list $(v_1,\cdots ,v_k)$ can be made by removing the redundant vectors of the first set so that $\text{span}(v_1,\cdots v_n)=\text{span}(v_1,\cdots v_k)$.
• (Proposition 29) Given a finite list $(v_1,\cdots v_n)\in V$, the list of vectors is linearly dependent if and only if there is some $v_i$ that is redundant for $1\le i\le n$.

$$[(v_1,\cdots v_n)\text{ is linearly dependent}]⟺[v_i\text{ is redundant for some }1\le i\le n]$$

• (Corollary 30) Given any set of vectors in a vector space $S=\{v_1,\cdots ,v_n\}\subset V$, there is a subset $S^{\prime }=\{v_{i_1},\cdots ,v_{i_k}\}$ with $S^{\prime }\subset S$ so that $\text{span}(S^{\prime })=\text{span}(S)$ and $S^{\prime }$ is linearly independent.
• (Theorem) The size of a linearly independent set is less than or equal to the size of its spanning set