Exam 2024 (A)

1a

\begin{matrix} State and prove the theorem of three \\ Theorem: \\ Let V be a vector space \\ Let B \subseteq V \\ 1. s p (B) = V \\ 2. | B | = d i m (V) \\ 3. B is a linear independence \\ If any two properties are true, then the third one is also true \\ Proof: \\ 1 and 3 ⟹ 2 by definition of basis \\ Let 2 and 3 \\ 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 \\ d i m (V) = | C | < | B \cup {v} | \\ But C is a maximal linear independence of V - Contradiction! \\ ⟹ s p (B) = V ⟹ 2 and 3 ⟹ 1 \\ Let 1 and 2 \\ Let C be a basis of V \\ Let B be a linear dependence \\ ⟹ \exists v \in B : s p (B ∖ {v}) = s p (B) = V \\ ⟹ d i m (V) = | C | \leq | B ∖ {v} | = d i m (V) - 1 - Contradiction! \\ ⟹ B is a linear independence ⟹ 1 and 2 ⟹ 3 \end{matrix}

1b

\begin{matrix} B = {(\begin{matrix} 1 \\ 3 \end{matrix}), (\begin{matrix} 1 \\ 2 \end{matrix})} \\ α (\begin{matrix} 1 \\ 3 \end{matrix}) + β (\begin{matrix} 1 \\ 2 \end{matrix}) = (\begin{matrix} 2 \\ 1 \end{matrix}) \\ α + β = 2 \\ 3 α + 2 β = 1 \\ (\begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \end{matrix}) \overset{R_{2} - 2 R_{1}}{\to} (\begin{matrix} 1 & 1 & 2 \\ 1 & 0 & - 3 \end{matrix}) \\ α = - 3 \\ β = 5 \\ ⟹ [(\begin{matrix} 2 \\ 1 \end{matrix})]_{B} = (\begin{matrix} - 3 \\ 5 \end{matrix}) \\ [v]_{B} = (\begin{matrix} 1 \\ 1 \end{matrix}) ⟹ α (\begin{matrix} 1 \\ 3 \end{matrix}) + β (\begin{matrix} 1 \\ 2 \end{matrix}) = (\begin{matrix} 2 \\ 5 \end{matrix}) \end{matrix}

2a

\begin{matrix} (\begin{matrix} 2 & - 2 & 2 \\ 1 & - 1 & 1 \\ - 1 & 1 & - 1 \end{matrix}) \to (\begin{matrix} 1 & - 1 & 1 \\ 1 & - 1 & 1 \\ 1 & - 1 & 1 \end{matrix}) \to (\begin{matrix} 1 & - 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) \\ 3 = r a n k (A) + d i m (N (A)) ⟹ 3 = 1 + d i m (N (A)) \\ {\begin{matrix} x - y + z = 0 \\ 0 = 0 \\ 0 = 0 \end{matrix} \\ ⟹ C (A) = s p {(\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix})} \\ (\begin{array}{cccc} 1 & - 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}) ⟹ {\begin{matrix} x = s - t \\ y = s \\ z = t \end{matrix} ⟹ (\begin{matrix} s \\ s \\ 0 \end{matrix}) + (\begin{matrix} - t \\ 0 \\ t \end{matrix}) \\ ⟹ N (A) = s p {(\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix})} \\ v \in C (A) \cap N (A) ⟹ v = α (\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix}) = β (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) + γ (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}) \\ ⟹ {\begin{matrix} 2 α - β + γ = 0 \\ α - β = 0 \\ α + γ = 0 \end{matrix} \\ (\begin{array}{cccc} 2 & - 1 & 1 & 0 \\ 1 & - 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \end{array}) \to (\begin{array}{cccc} 2 & - 1 & 1 & 0 \\ 0 & - 1 & - 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}) ⟹ C (A) \cap N (A) = s p {(\begin{matrix} - 1 \\ - 1 \\ 1 \end{matrix})} \\ ⟹ C (A) \cap N (A) = {0} ⟹ dim = 0, Basis \emptyset \\ C (A) + N (A) = {u + w | u \in C (A), w \in N (A)} \\ α (\begin{matrix} 2 \\ 1 \\ - 1 \end{matrix}) + β (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}) + γ (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix}) \\ U + W = s p {U \cup W} \\ ⟹ C (A) + N (A) = s p {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} - 1 \\ 0 \\ 1 \end{matrix})} = R^{3} \\ v does not exist \\ v \in R^{3} : v \notin C (A) + N (A) \end{matrix}