Linear-1 6

\begin{matrix} s p (A) \subseteq s p (B) \overset{?}{⟹} A \subseteq B \\ No: \\ A = {(\begin{matrix} 1 \\ 0 \end{matrix})}, B = {(\begin{matrix} 2 \\ 0 \end{matrix})} \\ A \subseteq B \overset{?}{⟹} s p (A) \subseteq s p (B) \\ Yes: \\ \dots \end{matrix}

\begin{matrix} S = s p (S) ⟺ S is a vector subspace of V \end{matrix}

\begin{matrix} s p ({(\begin{matrix} 0 \\ 1 \end{matrix})}) + s p ({(\begin{matrix} 0 \\ 1 \end{matrix})}) = s p ({(\begin{matrix} 0 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix})}) \end{matrix}

Sum of spans is a span of union #lemma

\begin{matrix} Prove: s p (A) + s p (B) = s p (A \cup B) \\ Proof: \\ Let v \in s p (A) + s p (B) \\ v = \underset{\in A \subseteq (A \cup B)}{\underset{⏟}{\sum_{i = 1}^{n} α_{i} u_{i}}} + \underset{\in B \subseteq (A \cup B)}{\underset{⏟}{\sum_{i = 1}^{k} β_{i} w_{i}}} ⟹ v \in (A \cup B) \\ ⟹ s p (A) + s p (B) \subseteq s p (A \cup B) \\ Let v \in s p (A \cup B) \\ v = \sum_{i = 1}^{n} α_{i} v_{i} \\ \forall i \in [1, n] : v_{i} \in A \lor v_{i} \in B \\ Let’s reorder sum the following way: \\ v = \underset{v_{i} \in A}{\underset{⏟}{\sum_{i = 1}^{k} α_{i} v_{i}}} + \underset{v_{i} \in B}{\underset{⏟}{\sum_{i = k + 1}^{n} α_{i} v_{i}}} \\ ⟹ v = \underset{\in s p (A)}{\underset{⏟}{u}} + \underset{\in s p (B)}{\underset{⏟}{w}} \\ ⟹ v \in s p (A) + s p (B) ⟹ s p (A \cup B) \subseteq s p (A) + s p (B) \end{matrix}

Linear dependence/independence #definition

\begin{matrix} V is a vector space over F, S \subseteq V \\ S is called linear dependence if exist \\ s_{1}, s_{2}, \dots, s_{n} \in S \\ α_{1}, α_{2}, \dots, α_{n} \in F \\ such that α_{1} s_{1} + \dots α_{n} s_{n} = 0 \\ S is called linear independence if for all \\ s_{1}, s_{2}, \dots, s_{n} \in S \\ α_{1}, α_{2}, \dots, α_{n} \in F \\ such that α_{1} s_{1} + \dots α_{n} s_{n} = 0 \\ implies α_{1} = α_{2} = \dots = α_{n} = 0 \end{matrix}

Note

0 \in S ⟹ S is a linear dependence

Linear dependence properties #theorem

\begin{matrix} V is a vector space over F, A \subseteq V \\ 1. A is a linear dependence \\ 2. \exists v \in A : v \in s p (A ∖ {v}) \\ 3. \exists v \in A : s p (A) = s p (A ∖ {v}) \\ (1) ⟺ (2) ⟺ (3) \\ Proof: \\ 1. Let (1) \\ Let α_{1} \neq 0 by linear dependence \\ α_{1} v_{1} + \dots + α_{n} v_{n} = 0 \\ v_{1} = (- \frac{α_{2}}{α_{1}}) v_{2} + \dots + (- \frac{a_{n}}{a_{1}}) v_{n} \\ \forall i \in [2, n] : v_{i} \in A \land v_{i} \neq v_{1} ⟹ v_{1} \in s p (A ∖ {v_{1}}) (2) \\ ⟹ (1) ⟹ (2) \\ 2. Let (2) \\ A ∖ {v} \subseteq A ⟹ s p (A ∖ {v}) \subseteq s p (A) \\ Let u \in s p (A) \\ u = \sum_{i = 1}^{n} α_{i} {\vec{a}}_{i} \\ If \forall i \in [1, n] : v \neq a_{i} ⟹ u \in s p (A ∖ {v}) \\ If \exists i \in [1, n] : v = a_{i} \\ Let i = 1 by commutativity of addition, v = {\vec{a}}_{1} \\ v \in s p (A ∖ {v}) ⟹ v = \sum_{j = 2}^{n} β_{j} {\tilde{a}}_{j} \\ ⟹ u = α_{1} \sum_{j = 1}^{n} β_{j} {\tilde{a}}_{j} + \sum_{j = 2}^{n} α_{j} {\vec{a}}_{j} \\ \forall i \in [1, n], j \in [2, n] : ({\tilde{a}}_{i} \in A ∖ {v} \land {\vec{a}}_{j} \in A ∖ {v}) ⟹ u \in s p (A ∖ {v}) \\ ⟹ s p (A) \subseteq s p (A ∖ {v}) ⟹ s p (A) = s p (A ∖ {v}) \\ 3. Let (3) \\ v \in A \underset{A \subseteq s p (A)}{⟹} v \in s p (A) ⟹ v \in s p (A) ∖ {v} \\ ⟹ v = α_{1} {\vec{a}}_{1} + \dots α_{n} {\vec{a}}_{n} \\ v + (- α_{1}) {\vec{a}}_{1} + \dots (- α_{n}) {\vec{a}}_{n} = 0 \\ v = 1 \cdot v ⟹ {1, - α_{1}, \dots, - α_{n}} is a non-trivial linear combination \\ ⟹ A is a linear dependence ⟹ (2) ⟹ (3) \\ (1) ⟹ (2) ⟹ (3) ⟹ (1) ⟹ (1) ⟺ (2) ⟺ (3) \end{matrix}

Linear independence properties #theorem

\begin{matrix} V is a vector space over F \\ 1. A \subseteq V is a linear independence, v \in V, v \notin A \\ v \notin s p (A) ⟺ A \cup {v} is a linear independence \\ 2. A \subseteq V is a linear independence, v \in A \\ s p (A ∖ {v}) \neq s p (A) \\ Proof: \\ 1. \\ Let v \notin s p (A) \\ v_{1}, \dots, v_{n} \in A \cup {v} \\ α_{1}, \dots, α_{n} \in F \\ α_{1} v_{1} + \dots + α_{n} v_{n} = 0 \\ If \forall i \in [1, n] : v \neq v_{i} ⟹ α_{1} = \dots = α_{n} = 0 \\ If \exists i \in [1, n] : v = v_{i} \\ Let i = 1 by commutativity of addition, v = v_{1} \\ α_{1} v + α_{2} v_{2} + \dots + α_{n} v_{n} = 0 \\ α_{1} v = - α_{2} v_{2} - \dots - α_{n} v_{n} \\ ⟹ α_{1} v \in s p (A) \\ {\begin{matrix} α_{1} v \in s p (A) \\ v \notin s p (A) \end{matrix} ⟹ α_{1} = 0 ⟹ α_{1} = \dots = α_{n} = 0 \\ ⟹ A \cup {v} is a linear independence \\ 2. \\ v \in A ⟹ v \in s p (A) \\ v \in A ⟹ (A ∖ {v}) \cup {v} = A \\ A is a linear independence ⟹ A ∖ {v} is a linear independence \\ (1) ⟹ v \notin s p (A ∖ {v}) \\ {\begin{matrix} v \in s p (A) \\ v \notin s p (A ∖ {v}) \end{matrix} ⟹ s p (A) \neq s p (A ∖ {v}) \end{matrix}