Linear-1 7

Linear-1 7

Interchangeability of LID #lemma

\begin{matrix} Let vector space V over F \\ Let A \subseteq V - linear independence, B \subseteq V, s p (B) = V \\ Then \\ \forall \vec{a} \in A \exists \vec{b} \in B : \vec{b} \notin A ∖ {\vec{a}} \land (A ∖ {\vec{a}}) \cup {\vec{b}} is a linear independence \\ Proof: \\ Let \exists \vec{a} \in A : \forall \vec{b} \in B : \vec{b} \in A ∖ {\vec{a}} \lor (A ∖ {\vec{a}}) \cup {\vec{b}} is a linear dependence \\ If \vec{b} \in A ∖ {\vec{a}} ⟹ \vec{b} \in s p ({\vec{a}})) \\ A is a linear independence ⟹ A ∖ {\vec{a}} is a linear independence \\ If \vec{b} \notin A ∖ {\vec{a}} ⟹ [\vec{b} \notin s p (A ∖ {\vec{a}}) ⟺ (A ∖ {\vec{a}}) \cup {\vec{b}} is a linear independence] \\ ⟹ \vec{b} \in s p (A ∖ {\vec{a}}) \\ \forall \vec{b} \in B : \vec{b} \in s p (A ∖ {\vec{a}}) ⟹ B \subseteq s p (A ∖ {\vec{a}}) ⟹ s p (B) \subseteq s p (A ∖ {\vec{a}}) \\ ⟹ V \subseteq s p (A ∖ {\vec{a}}) \land A \subseteq V ⟹ V = s p (A ∖ {\vec{a}}) \\ A is a linear independence ⟺ s p (A ∖ {\vec{a}}) \subset s p (A) \subseteq V \\ ⟹ s p (A ∖ {\vec{a}}) \subset V - Contradiction! \end{matrix}

LID is not larger than the spanning set #lemma

\begin{matrix} Let vector space V over F \\ V is finitely generated: \exists X : | X | = n \land s p (X) = V \\ Let A \subseteq V - linear independence, B \subseteq V, s p (B) = V \\ Then | A | \leq | B | \\ Proof: \\ Let | B | = n \\ B = {b_{1}, b_{2}, \dots b_{n}} \\ Let | A | > | B | \\ A = {a_{1}, a_{2}, \dots, a_{n + 1}, \dots} \\ By the lemma above: \\ \exists b_{1} \in B : b \notin A ∖ {a_{1}} \land (A ∖ {a_{1}}) \cup {b_{1}} is a linear independence \\ By induction: (A ∖ {a_{1}, a_{2}, \dots, a_{n}}) \cup B is a linear independence \\ Let C = (A ∖ {a_{1}, a_{2}, \dots, a_{n}}) \cup B, B \subseteq C \\ By the lemma above: \exists b_{x} \in B : b_{x} \notin C ∖ {a_{n + 1}} - Contradiction! \\ ⟹ | A | \leq | B | \\ Let B be an infinite set \\ \exists C \subseteq V : | C | = n \land s p (C) = V \\ ⟹ | A | \leq | C | \leq | B | ⟹ | A | \leq | B | \end{matrix}

Basis of the vector space #definition

\begin{matrix} Let B \subseteq V over F \\ s p (B) = V \land B is a linear independence \end{matrix}

Dimension of the vector space #definition

\begin{matrix} B is the basis of V over F \\ Dimension of V over F is \\ d i m_{F} V = | B | \end{matrix}

Dimensions of bases #lemma

\begin{matrix} Let B, C be two bases of V \\ Then | B | = | C | \\ Proof: \\ C is a linear independence and s p (B) = V \\ ⟹ | C | \leq | B | \\ B is a linear independence and s p (C) = V \\ ⟹ | B | \leq | C | \\ ⟹ | B | = | C | \end{matrix}

Example

\begin{matrix} V = R^{2} over R \\ A = {(\begin{matrix} 0 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix})} is a basis of V \\ More than that, it is called the standard basis of V \\ d i m_{R} R^{2} = 2 \end{matrix}

Standard basis #definition

\begin{matrix} Let V = F^{n} over F \\ {e_{1}, \dots, e_{n}} is called standard basis of V \\ Where \forall k \in [1, n] : (e_{i})_{k} = {\begin{cases} 1 & k = i \\ 0 & otherwise \end{cases} \end{matrix}

Basis properties #lemma

\begin{matrix} Let V be a finitely generated vector space over F \\ Then \\ 1. & \exists A : A is basis of V \\ 2. & B is a basis of V ⟺ B is a maximal linear independence in V \\ 3. & B is a basis of V ⟺ B is a minimal set generating V \end{matrix}

"Of the three" #lemma

\begin{matrix} B \subseteq V \\ 1. s p (B) = V \\ 2. B is a linear independence \\ 3. | B | = d i m V \\ If two out of three are true, then all are true \\ Proof: \\ \underset{―}{(1) and (2)} ⟹ (3) by definition of basis \\ \underset{―}{Let (2) and (3)} \\ B is a linear independence, | B | = d i m V \\ B \subseteq V \land B is a linear independence ⟹ | B | = n \\ B = {b_{1}, b_{2}, \dots, b_{n}} \\ Let s p (B) \neq V ⟺ \exists v \in V : v \notin s p (B) \\ ⟹ B \cup {v} is a linear independence \\ Let C be a basis of V \\ ⟹ | B | = d i m V = | C | \\ s p (C) = V and | C | < | B \cup {v} | - Contradiction! \\ ⟹ s p (B) = V \\ \underset{―}{Let (1) and (3)} \\ s p (B) = V and | B | = d i m V \\ Let C be a basis of V \\ | C | = d i m V = | B | \\ Let B be a linear dependence \\ ⟺ \exists v \in V : s p (B ∖ {v}) = s p (B) = V \\ ⟹ s p (B ∖ {v}) = V ⟹ | C | > | B ∖ {v} | - Contradiction! \\ ⟹ B is a linear independence \end{matrix}

\begin{matrix} d i m V = n ⟹ {\begin{cases} A is a linear independence ⟹ | A | \leq n \\ s p (A) = V ⟹ | A | \geq n \end{cases} \end{matrix}

"Useful" #lemma

\begin{matrix} Let V be a finitely generated vector space over F \\ U, W \subseteq V - vector subspaces of V \\ Then \\ U \subseteq W ⟹ d i m U \leq d i m W \\ U = W ⟺ (U \subseteq W \land d i m U = d i m W) \\ Proof: \\ It is obvious that U = W ⟹ (U \subseteq W \land d i m U = d i m W) \\ Now let U \subseteq W \land d i m U = d i m W = n \\ \exists B = {u_{1}, u_{2}, \dots, u_{n}} which is a basis of U \\ B \subseteq U \subseteq W \\ B is a linear independence \land d i m W = | B | ⟹ U = s p (B) = W ⟹ U = W \end{matrix}

Dimension theorem

\begin{matrix} Let V be a finitely generated vector space over F \\ U, W \subseteq V - vector subspaces of V \\ Then d i m (U + W) = d i m (U) + d i m (W) - d i m (U \cap W) \end{matrix}