Linear-1 5

1

\begin{matrix} U = {(\begin{matrix} b \\ - c \\ c \end{matrix}) | b, c \in R}, W = {(\begin{matrix} a \\ 2 a \\ 0 \end{matrix}) | a \in R} \end{matrix}

1a

\begin{matrix} Prove: U, W are vector spaces over R \\ Proof: \\ 1. Let b = 0, c = 0 ⟹ (\begin{matrix} b \\ - c \\ c \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) ⟹ 0 \in U \\ Let a = 0 ⟹ (\begin{matrix} a \\ 2 a \\ 0 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) ⟹ 0 \in W \\ 2. Let u_{1}, u_{2} \in U, w_{1}, w_{2} \in W, α, β \in R \\ u_{1} + α u_{2} = (\begin{matrix} b_{1} + α b_{2} \\ - c_{1} - α c_{2} \\ c_{1} + α c_{2} \end{matrix}) \overset{- c_{1} - α c_{2} = - (c_{1} + α c_{2})}{⟹} u_{1} + α u_{2} \in U \\ ⟹ U is a vector space \\ w_{1} + β w_{2} = (\begin{matrix} a_{1} + β a_{2} \\ 2 a_{1} + β 2 a_{2} \\ 0 + β \cdot 0 \end{matrix}) \overset{2 a_{1} + β 2 a_{2} = 2 (a_{1} + β a_{2})}{⟹} w_{1} + β w_{2} \in W \\ ⟹ W is a vector space \end{matrix}

1b

\begin{matrix} Prove: U \oplus W = R^{3} \\ Proof: \\ U + W = {(\begin{matrix} b \\ - c \\ c \end{matrix}) + (\begin{matrix} a \\ 2 a \\ 0 \end{matrix}) | a, b, c \in R} = {(\begin{matrix} b + a \\ - c + 2 a \\ c \end{matrix}) | a, b, c \in R} \\ Let a, b, c \in R, x, y, z \in R \\ (\begin{matrix} b + a \\ - c + 2 a \\ c \end{matrix}) = (\begin{matrix} x \\ y \\ z \end{matrix}) ⟺ {\begin{matrix} a + b + 0 c = x \\ 2 a + 0 b - c = y \\ 0 a + 0 b + c = z \end{matrix} \\ (\begin{array}{cccc} 1 & 1 & 0 & x \\ 2 & 0 & - 1 & y \\ 0 & 0 & 1 & z \end{array}) \to (\begin{array}{cccc} 1 & 1 & 0 & x \\ 0 & - 2 & - 1 & y - 2 x \\ 0 & 0 & 1 & z \end{array}) \to (\begin{array}{cccc} 1 & 1 & 0 & x \\ 0 & - 2 & 0 & y - 2 x + z \\ 0 & 0 & 1 & z \end{array}) \\ \to (\begin{array}{cccc} 1 & 0 & 0 & \frac{y + z}{2} \\ 0 & 1 & 0 & \frac{2 x - y - z}{2} \\ 0 & 0 & 1 & z \end{array}) ⟹ {\begin{matrix} a = \frac{y + z}{2} \\ b = \frac{2 x - y - z}{2} \\ c = z \end{matrix} \\ ⟹ \forall v \in R^{3} : \exists u \in U, w \in W : u + w = v ⟹ U + W = R^{3} (1) \\ (\begin{matrix} b \\ - c \\ c \end{matrix}) = (\begin{matrix} a \\ 2 a \\ 0 \end{matrix}) ⟺ {\begin{matrix} c = 0 \\ - c = 2 a \\ b = a \end{matrix} ⟺ {\begin{matrix} a = 0 \\ b = 0 \\ c = 0 \end{matrix} ⟺ U \cap W = {0} (2) \\ (1) and (2) ⟹ U \oplus W = R^{3} \end{matrix}

2

\begin{matrix} V is a vector space over F \\ U, W - vector subspaces of V \\ U \oplus W = V \\ Prove or disprove: U \cup W = V \\ Disproof: \\ Let V = R^{2} \\ U = {(\begin{matrix} a \\ 0 \end{matrix}) | a \in R}, W = {(\begin{matrix} 0 \\ b \end{matrix}) | b \in R} \\ U \oplus W = V \\ U \cup W = {(\begin{matrix} a \\ 0 \end{matrix}), (\begin{matrix} 0 \\ b \end{matrix}) | a, b \in R} \\ (\begin{matrix} 1 \\ 1 \end{matrix}) \notin U \cup W ⟹ U \cup W \neq V \end{matrix}

3

\begin{matrix} V is a vector space over F \\ U, W - vector subspaces of V \\ Prove or disprove: \forall V : \exists U, W : U \oplus W = V \\ Proof: \\ Let V be a vector space over F \\ V is also a vector subspace of V \\ {0} is also a vector subspace of V \\ Then \forall V : \exists U = V, W = {0} : U \oplus W = V \end{matrix}

4a

\begin{matrix} B_{1} = {x^{2} + x + 1, x^{2} - 2 x + 1} \\ x \overset{?}{\in} s p (B_{1}) \\ 1 \cdot (x^{2} + x + 1) + (- 1) \cdot (x^{2} - 2 x + 1) = 3 x \\ α (x^{2} + x + 1) + β (x^{2} - 2 x + 1) = x ⟹ {\begin{matrix} α = \frac{1}{3} \\ β = - \frac{1}{3} \end{matrix} \\ ⟹ x \in s p (B_{1}) \end{matrix}

4b

\begin{matrix} B_{2} = {(\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 2 \\ 1 \end{matrix})} \\ (\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}) \overset{?}{\in} s p (B_{2}) \\ α (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) + β (\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}) = (\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}) ⟺ {\begin{matrix} α + β = 0 \\ α + 2 β = 1 \\ α + β = 2 \end{matrix} ⟺ \emptyset \\ ⟹ (\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}) \notin s p (B_{2}) \end{matrix}

4c

\begin{matrix} B_{3} = {(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 2 \\ 1 \end{matrix})} \\ (\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}) \overset{?}{\in} s p (B_{3}) \\ α (\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}) + β (\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}) = (\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}) ⟺ {\begin{matrix} α + β = - 1 \\ 2 β = 1 \\ α + β = - 1 \end{matrix} ⟹ {\begin{matrix} α = - \frac{3}{2} \\ β = \frac{1}{2} \end{matrix} \\ ⟹ (\begin{matrix} - 1 \\ 1 \\ - 1 \end{matrix}) \in s p (B_{3}) \end{matrix}

4d

\begin{matrix} B_{4} = {(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}), (\begin{matrix} 2 & 3 \\ 4 & 1 \end{matrix}), (\begin{matrix} 3 & 4 \\ 1 & 2 \end{matrix}), (\begin{matrix} 4 & 1 \\ 2 & 3 \end{matrix})} \\ (\begin{matrix} 30 & 24 \\ 22 & 24 \end{matrix}) \overset{?}{\in} s p (B_{4}) \\ α (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) + β (\begin{matrix} 2 & 3 \\ 4 & 1 \end{matrix}) + γ (\begin{matrix} 3 & 4 \\ 1 & 2 \end{matrix}) + δ (\begin{matrix} 4 & 1 \\ 2 & 3 \end{matrix}) = (\begin{matrix} 30 & 24 \\ 22 & 24 \end{matrix}) \\ ⟺ {\begin{matrix} α + 2 β + 3 γ + 4 δ = 30 \\ 2 α + 3 β + 4 γ + δ = 24 \\ 3 α + 4 β + γ + 2 δ = 22 \\ 4 α + β + 2 γ + 3 δ = 24 \end{matrix} \\ (\begin{array}{ccccc} 1 & 2 & 3 & 4 & 30 \\ 2 & 3 & 4 & 1 & 24 \\ 3 & 4 & 1 & 2 & 22 \\ 4 & 1 & 2 & 3 & 24 \end{array}) \to (\begin{array}{ccccc} 1 & 2 & 3 & 4 & 30 \\ 0 & 1 & 2 & 7 & 36 \\ 0 & 2 & 8 & 10 & 68 \\ 0 & 7 & 10 & 13 & 96 \end{array}) \to (\begin{array}{ccccc} 1 & 2 & 3 & 4 & 30 \\ 0 & 1 & 2 & 7 & 36 \\ 0 & 0 & 4 & - 4 & - 4 \\ 0 & 0 & 4 & 36 & 156 \end{array}) \\ \to (\begin{array}{ccccc} 1 & 2 & 3 & 4 & 30 \\ 0 & 1 & 2 & 7 & 36 \\ 0 & 0 & 1 & - 1 & - 1 \\ 0 & 0 & 0 & 40 & 160 \end{array}) \to (\begin{array}{ccccc} 1 & 2 & 3 & 4 & 30 \\ 0 & 1 & 2 & 7 & 36 \\ 0 & 0 & 1 & - 1 & - 1 \\ 0 & 0 & 0 & 1 & 4 \end{array}) \to (\begin{array}{ccccc} 1 & 2 & 3 & 0 & 14 \\ 0 & 1 & 2 & 0 & 8 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 4 \end{array}) \\ \to (\begin{array}{ccccc} 1 & 2 & 0 & 0 & 5 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 4 \end{array}) \to (\begin{array}{ccccc} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 4 \end{array}) \\ ⟹ {\begin{matrix} α = 1 \\ β = 2 \\ γ = 3 \\ δ = 4 \end{matrix} ⟹ (\begin{matrix} 30 & 24 \\ 22 & 24 \end{matrix}) \in s p (B_{4}) \end{matrix}

5a

\begin{matrix} B_{2} = {(\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}), (\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}), (\begin{matrix} 2 \\ 1 \\ 1 \end{matrix})} \\ (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) \overset{?}{\in} s p (B_{2}) \\ α (\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}) + β (\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}) + γ (\begin{matrix} 2 \\ 1 \\ 1 \end{matrix}) = (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) \\ (\begin{array}{cccc} 1 & 1 & 2 & 1 \\ 1 & 2 & 1 & 1 \\ 2 & 1 & 1 & 1 \end{array}) \to (\begin{array}{cccc} 1 & 1 & 2 & 1 \\ 0 & 1 & - 1 & 0 \\ 0 & 1 & 3 & 1 \end{array}) \to (\begin{array}{cccc} 1 & 1 & 2 & 1 \\ 0 & 1 & - 1 & 0 \\ 0 & 0 & 4 & 1 \end{array}) \\ ⟹ System has exactly one solution \\ (\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}) + (\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}) + (\begin{matrix} 2 \\ 1 \\ 1 \end{matrix}) = (\begin{matrix} 4 \\ 4 \\ 4 \end{matrix}) \\ ⟹ \frac{1}{4} \cdot ((\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}) + (\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}) + (\begin{matrix} 2 \\ 1 \\ 1 \end{matrix})) = (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) \\ ⟹ {\begin{matrix} α = \frac{1}{4} \\ β = \frac{1}{4} \\ γ = \frac{1}{4} \end{matrix} ⟹ (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) \in s p (B_{2}) \end{matrix}

5b

\begin{matrix} {\begin{matrix} x + y + 2 z = 1 \\ x + 2 y + z = 1 \\ 2 x + y + z = 1 \end{matrix} ⟺ x (\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}) + y (\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}) + z (\begin{matrix} 2 \\ 1 \\ 1 \end{matrix}) = (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) \\ ⟹ {\begin{matrix} x = \frac{1}{4} \\ y = \frac{1}{4} \\ z = \frac{1}{4} \end{matrix} \end{matrix}

6

\begin{matrix} V is a vector space over F \\ Let u, v, w \in V, u \neq 0 \\ s p ({u}) \overset{?}{=} s p ({u, v}) \cap s p ({u, w}) \\ s p ({u, v}) = {α u + β v | α, β \in F} = {α u + β v + 0 w | α, β \in F} \\ s p ({u, w}) = {γ u + δ w | γ, δ \in F} = {γ u + δ w + 0 v | α, β \in F} \\ Let v \neq w \\ ⟹ s p ({u, v}) \cap s p ({u, w}) = {α_{1} u + α_{2} v + α_{3} w | \begin{matrix} α_{1}, α_{2}, α_{3} \in F \\ {\begin{matrix} α_{1} = α = γ \\ α_{2} = β = 0 \\ α_{3} = δ = 0 \end{matrix} \end{matrix}} = \\ = {α_{1} u | α_{1} \in F} = s p ({u}) \\ Let v = w \\ s p ({u, v}) \cap s p ({u, w}) = s p ({u, v}) \cap s p ({u, v}) = s p ({u, v}) \\ ⟹ {\begin{cases} s p ({u}) = s p ({u, v}) \cap s p ({u, w}) & v \neq w \lor u = v = w \\ s p ({u}) \neq s p ({u, v}) \cap s p ({u, w}) & u \neq v = w \end{cases} \end{matrix}

7

\begin{matrix} V is a vector space over F \\ S_{1}, S_{2} \subseteq V \\ Prove: S_{1} \subseteq S_{2} ⟹ s p (S_{1}) \subseteq s p (S_{2}) \\ Proof: \\ Let v_{1} \in s p (S_{1}) \\ ⟹ \exists {α_{1}, \dots, α_{n}} \subseteq F, {s_{1}, \dots, s_{n}} \subseteq S_{1} : v_{1} = \sum_{i = 1}^{n} α_{i} s_{i} \\ {s_{1}, \dots, s_{n}} \subseteq S_{1} \subseteq S_{2} ⟹ {s_{1}, \dots, s_{n}} \subseteq S_{2} \\ ⟹ \exists {α_{1}, \dots, α_{n}} \subseteq F, {s_{1}, \dots, s_{n}} \subseteq S_{2} : v_{1} = \sum_{i = 1}^{n} α_{i} s_{i} ⟺ v_{1} \in s p (S_{2}) \\ ⟹ s p (S_{1}) \subseteq s p (S_{2}) \end{matrix}

8

\begin{matrix} V is a vector space over F \\ A, B \subseteq V \\ W \subseteq V - vector subspace of V \end{matrix}

8a

\begin{matrix} Prove or disprove: s p (A \cup W \cup B) = s p (A) \cup s p (W) \cup s p (B) \\ Disproof: \\ V = R^{2} \\ A = {(\begin{matrix} 1 \\ 0 \end{matrix})}, B = {(\begin{matrix} 0 \\ 1 \end{matrix})} . W = {(\begin{matrix} 0 \\ 0 \end{matrix})} \\ A \cup B \cup W = {(\begin{matrix} 0 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})} ⟹ s p (A \cup W \cup B) = R^{2} \\ s p (A) = {(\begin{matrix} a \\ 0 \end{matrix}) | a \in R}, s p (B) = {(\begin{matrix} 0 \\ b \end{matrix}) | b i n R}, s p (W) = {(\begin{matrix} 0 \\ 0 \end{matrix})} \\ ⟹ s p (A) \cup s p (W) \cup s p (B) = {(\begin{matrix} a \\ 0 \end{matrix}), (\begin{matrix} 0 \\ b \end{matrix}), (\begin{matrix} 0 \\ 0 \end{matrix}) | a, b \in R} \\ (\begin{matrix} 1 \\ 1 \end{matrix}) \notin s p (A) \cup s p (W) \cup s p (B) \\ ⟹ s p (A) \cup s p (W) \cup s p (B) \neq R^{2} = s p (A \cup W \cup B) \\ ⟹ s p (A \cup W \cup B) \neq s p (A) \cup s p (W) \cup s p (B) \end{matrix}

8b

\begin{matrix} Prove or disprove: s p (A \cup W \cup B) = (s p (A) + s p (B)) \cup s p (W) \\ Disproof: \\ V = R^{3} \\ A = {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})}, B = {(\begin{matrix} 0 \\ 1 \\ 0 \end{matrix})}, W = {(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ x \end{matrix}) | x \in R} \\ A \cup W \cup B = {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ x \end{matrix}) | x \in R} ⟹ s p (A \cup W \cup B) = R^{3} \\ s p (A) = {(\begin{matrix} a \\ 0 \\ 0 \end{matrix})}, s p (B) = {(\begin{matrix} 0 \\ b \\ 0 \end{matrix}) | b \in R} ⟹ s p (A) + s p (B) = {(\begin{matrix} a \\ b \\ 0 \end{matrix}) | a, b \in R} \\ s p (W) = {(\begin{matrix} 0 \\ 0 \\ x \end{matrix}) | x \in R} ⟹ (s p (A) + s p (B)) \cup s p (W) = {(\begin{matrix} a \\ b \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ x \end{matrix}) | a, b, x \in R} \\ (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}) \notin (s p (A) + s p (B)) \cup s p (W) \\ ⟹ (s p (A) + s p (B)) \cup s p (W) \neq R^{3} = s p (A \cup W \cup B) \\ ⟹ s p (A \cup W \cup B) \neq (s p (A) + s p (B)) \cup s p (W) \end{matrix}