Linear-1 8

Exercise

\begin{matrix} B = {1 + x, 3 + x^{2}, x} \\ Prove: B is basis of R_{2} [x] \\ Proof: \\ α (1 + x) + β (3 + x^{2}) + γ x = 0 \\ {\begin{matrix} α + 3 β = 0 \\ α + γ = 0 \\ β = 0 \end{matrix} ⟹ {\begin{matrix} α = 0 \\ β = 0 \\ γ = 0 \end{matrix} ⟹ B is a linear independence \\ a + b x + c x^{2} = α (1 + x) + β (3 + x^{2}) + γ x \\ (\begin{array}{cccc} 1 & 3 & 0 & a \\ 1 & 0 & 1 & b \\ 0 & 1 & 0 & c \end{array}) ⟹ {\begin{matrix} α = a - 3 c \\ β = c \\ γ = b - a + 3 c \end{matrix} ⟹ s p (B) = R_{2} [x] \end{matrix}

\begin{matrix} U, W \subseteq V - vector subspaces of V \\ U \subseteq W \land d i m (U) = d i m (W) ⟹ U = W \end{matrix}

\begin{matrix} U, W \subseteq V - vector subspaces of V \\ d i m (U + W) = d i m (U \cap W) + 1 ⟹ U \subseteq W \lor W \subseteq W \\ Proof: \\ Let d i m (U + W) = d i m (U \cap W) + 1 \\ Let U ⊈ W \land W ⊈ U \\ U \cap W \subseteq W \\ W ⊈ U ⟹ \exists w \in W : w \notin U ⟹ w \notin U \cap W ⟹ U \cap W \neq W \\ ⟹ U \cap W \subset W \\ W \subseteq U + W \\ U ⊈ W ⟹ \exists u \in U : u \notin W \\ u = u + 0 ⟹ u \in U + W ⟹ W \neq U + W \\ ⟹ W \subset U + W \\ d i m (U + W) > d i m (W) ⟹ d i m (U + W) \geq d i m (W) + 1 \geq d i m (U \cap W) + 2 \\ ⟹ d i m (U + W) > d i m (U \cap W) + 1 - Condtradiction! \end{matrix}

\begin{matrix} U, W \subseteq V - vector subspaces of V \\ d i m (V) < d i m (U) + d i m (W) ⟹ U \cap W \neq \emptyset {0} \\ Proof: \\ Let U \cap W = {0} ⟹ U \oplus W = V \\ ⟹ d i m (V) = d i m (U) + d i m (V) - Contradiction! \end{matrix}