Linear-1 7

\begin{matrix} Let V be a vector space \\ S, T \subseteq V - vector subspaces of V \\ S \subseteq s p (T) \land T \subseteq s p (S) ⟺ s p (S) = s p (T) \\ Proof: \\ Let S \subseteq s p (T) \land T \subseteq s p (S) \\ S \subseteq s p (T) ⟹ s p (S) \subseteq s p (s p (T)) = s p (T) \\ T \subseteq s p (S) ⟹ s p (T) \subseteq s p (s p (S)) = s p (S) \\ s p (S) \subseteq s p (T) \land s p (T) \subseteq s p (S) ⟹ s p (S) = s p (T) \\ Let s p (S) = s p (T) \\ T \subseteq s p (T) = s p (S) \\ S \subseteq s p (S) = s p (S) \end{matrix}

\begin{matrix} S = {α x - 1, 4 x^{2} - x - 2, 15 x^{2}} \\ α (α x - 1) + β (4 x^{2} - x - 2) + γ (15 x^{2}) = 0 \\ (α - 2 β) 1 + (2 α - β) x + (4 β + 15 γ) x^{2} = 0 \\ (\begin{array}{cccc} 1 & - 2 & 0 & 0 \\ 2 & - 1 & 0 & 0 \\ 0 & 4 & 15 & 0 \end{array}) \to (\begin{array}{cccc} 1 & - 2 & 0 & 0 \\ 0 & - 5 & 0 & 0 \\ 0 & 4 & 15 & 0 \end{array}) \to (\begin{array}{cccc} 1 & - 2 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 15 & 0 \end{array}) \\ ⟹ S is a linear independence \end{matrix}

\begin{matrix} S = {v_{1}, \dots, v_{n}} \subseteq V \\ T = {v_{1} + v_{1}, \dots, v_{n} + v_{1}} \subseteq V \\ S is a linear independence ⟺ T is a linear independence \\ Proof: \\ Let S be a linear independence \\ \sum_{i = 1}^{n} α_{i} (v_{i} + v_{1}) = \sum_{i = 1}^{n} α_{i} v_{i} + \sum_{i = 1}^{n} α_{i} v_{1} = (α_{1} + \sum_{i = 1}^{n} α_{i}) v_{1} + \sum_{i = 2}^{n} α_{i} v_{i} = 0 \\ This is a linear combination of S \\ ⟹ {\begin{matrix} α_{1} + \sum_{i = 1}^{n} α_{i} = 0 \\ α_{2} = 0 \\ \dots \\ α_{n} = 0 \end{matrix} ⟹ 2 α_{1} = 0 ⟹ α_{1} = 0 (over R) \\ ⟹ T is a linear independence \\ Let T be a linear independence \\ \sum_{i = 1}^{n} α_{i} v_{i} = \sum_{i = 1}^{n} α_{i} (v_{i} + v_{1} - v_{1}) = \sum_{i = 1}^{n} α_{i} (v_{i} + v_{1}) + \sum_{i = 1}^{n} - α_{i} v_{1} = \\ = \frac{(α_{1} - \sum_{i = 2}^{n} α_{i})}{2} (v_{1} + v_{1}) + \sum_{i = 2}^{n} α_{i} (v_{i} + v_{1}) = 0 \\ ⟹ {\begin{matrix} \frac{(α_{1} - \sum_{i = 2}^{n} α_{i})}{2} = 0 \\ α_{2} = 0 \\ \dots \\ α_{n} = 0 \end{matrix} ⟹ \frac{α_{1}}{2} = 0 ⟹ α_{1} = 0 \\ ⟹ S is a linear independence \end{matrix}

\begin{matrix} V = R^{n x n} \\ A \in V \\ 1. \exists A^{- 1} ⟹ ⧸ {A, A^{2}} is a linear independence \\ A = I, \exists A^{- 1} = I, A^{2} = I ⟹ {A, A^{2}} is a linear dependence \\ 2. \exists A^{- 1} ⟸ ⧸ {A, A^{2}} is a linear independence \\ A = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}), A^{2} = (\begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) ⟹ {A, A^{2}} is a linear dependence, ∄ A^{- 1} \end{matrix}

\begin{matrix} V = R_{2} [x] \\ S = {1 + x^{2}, x, x + x^{2}} \\ S^{'} = {(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix})} \\ (\begin{matrix} x \\ y \\ z \end{matrix}) = α s_{1}^{'} + β s_{2}^{'} + γ s_{3}^{'} \\ (\begin{array}{cccc} 1 & 0 & 0 & x \\ 0 & 1 & 1 & y \\ 1 & 0 & 1 & z \end{array}) \to (\begin{array}{cccc} 1 & 0 & 0 & x \\ 0 & 1 & 1 & y \\ 0 & 0 & 1 & z - x \end{array}) \to (\begin{array}{cccc} 1 & 0 & 0 & x \\ 0 & 1 & 0 & x + y - z \\ 0 & 0 & 1 & z - x \end{array}) \\ ⟹ {\begin{matrix} s p (S) = R^{3} \\ S^{'} is a linear independence \end{matrix} ⟹ S^{'} is a basis of R^{3} ⟹ S is a basis of R_{2} [x] \end{matrix}

\begin{matrix} V = R^{n \times n} \\ B = (\begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix}) \\ W = {A \in V | A B = B A} \\ W is vector subspace of V ? \\ (\begin{matrix} a & b \\ c & d \end{matrix}) \cdot (\begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix}) = (\begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix}) \cdot (\begin{matrix} a & b \\ c & d \end{matrix}) \\ (\begin{matrix} a & 2 a \\ c & 2 c \end{matrix}) = (\begin{matrix} a + 2 c & b + 2 d \\ 0 & 0 \end{matrix}) \\ {\begin{matrix} c = 0 \\ 2 a - b - 2 d = 0 \end{matrix} \\ (\begin{array}{ccccc} 1 & - \frac{1}{2} & 0 & - 1 \\ 0 & 0 & 1 & 0 \end{array}) ⟹ {(\begin{matrix} \frac{b}{2} + d & b \\ 0 & d \end{matrix}) | \forall b, d \in R} = W \\ ⟹ W = s p ({(\begin{matrix} \frac{1}{2} & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})}) \end{matrix}