Linear-1 6

\begin{matrix} V is a vector space \\ U, W are vector subspaces of V \\ U \cap W is a vector subspace of V \\ Proved on lectures \end{matrix}

Exercise

\begin{matrix} V = R_{2} [x] \\ U = {p (x) \in V | p (2) = 0} \\ W = {p (x) \in V | p (x) = x p^{'} (x)} \\ Show: U \cap W is a vector subspace \\ Solution: \\ U = {a_{0} + a_{1} x + a_{2} x^{2} | a_{0} + 2 a_{1} + 4 a_{2} = 0} \\ W = {a_{0} + a_{1} x + a_{2} x^{2} | a_{0} + a_{1} x + a_{2}^{2} = x (a_{1} + 2 a_{2} x) = a_{1} x + 2 a_{2} x^{2}} = \\ = {a_{0} + a_{1} x + a_{2} x^{2} | a_{0} - a_{2} x^{2} = 0} = {a_{0} + a_{1} x + a_{2} x^{2} | \begin{matrix} a_{0} = 0 \\ a_{2} = 0 \end{matrix}} \\ U \cap W = {a_{0} + a_{1} x + a_{2} x^{2} | \begin{matrix} a_{0} = 0 \\ a_{2} = 0 \\ a_{1} = 0 \end{matrix}} = {0} ⟹ U \cap W is a vector subspace of V \end{matrix}

\begin{matrix} V is a vector space \\ U, W are vector subspaces of V \\ U + W = {u + w | u \in U, w \in W} is a vector subspace of V \\ Proved in lectures \end{matrix}

\begin{matrix} V = R^{n \times n} \\ U = {A \in V | A = A^{T}} \\ W = {A \in V | A is a diagonal matrix} \\ Show: U + W is a vector subspace of V \\ Solution: \\ W \subseteq U ⟹ U + W = U ⟹ U + W is a vector subspace of V \end{matrix}

\begin{matrix} V = R^{2 \times 2} \\ U = {A \in V | A = A^{T}} \\ W = {A \in V | A is an upper-triangle matrix} \\ Show: U + W = V \\ Solution: \\ (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} 1 & c \\ c & 1 \end{matrix}) + (\begin{matrix} a - 1 & b - c \\ 0 & d - 1 \end{matrix}) \\ \forall A \in V : A = U_{1} + W_{1} \in U + W \\ ⟹ V \subseteq U + W \underset{U + W is a vector subspace of V}{⟹} U + W = V \end{matrix}

\begin{matrix} V is a vector space \\ U, W are vector subspaces of V \\ U \oplus W = V ⟺ U + W = V \land U \cap W = {0} \\ Proved in lectures \end{matrix}

\begin{matrix} V = R^{2 \times 2} \\ U = {A \in V | A = α I} \\ W = {A \in V | t r (A) = 0} \\ Show: U \oplus W = V \\ Solution: \\ Let A \in V \\ A = (\begin{matrix} a & b \\ c & d \end{matrix}) = \underset{= \frac{a + d}{2} \cdot I}{\underset{⏟}{(\begin{matrix} \frac{a + d}{2} & 0 \\ 0 & \frac{a + d}{2} \end{matrix})}} + (\begin{matrix} \frac{a - d}{2} & b \\ c & \frac{d - a}{2} \end{matrix}) \\ \forall A \in V : A = U_{1} + W_{1} ⟹ V \subseteq U + W ⟹ V = U + W \\ U \cap W = {A \in V | A = α I \land t r (A) = 0} = {0} ⟹ V = U \oplus W \end{matrix}

\begin{matrix} V is a vector space over F, S \subseteq V \\ S = {v_{1}, v_{2}, \dots, v_{n}} \\ Linear combination of S is a sum: \\ α_{1} v_{1} + α_{2} v_{2} + \dots + α_{n} v_{n} \\ Sometimes denoted as: (S, {α_{1}, α_{2}, \dots, α_{n}}) \\ s p (S) is called span of S and is a set of all linear combinations of S \\ s p (S) = {α_{1} v_{1} + \dots + α_{n} v_{n} | \begin{matrix} α_{1}, α_{2}, \dots, α_{n} \in F \\ v_{1}, v_{2}, \dots, v_{n} \in S \end{matrix}} \end{matrix}

\begin{matrix} V = R^{n \times n} \\ Define: s p (S = {(\begin{matrix} - 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 2 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 0 & - 4 \end{matrix})}) \\ Solution: \\ s p (S) = {(\begin{matrix} - α & 2 β \\ 0 & - 4 γ \end{matrix}) | α, β, γ \in R} \end{matrix}

\begin{matrix} V = R_{2} [x] \\ 8 - 3 x + x^{2} \overset{?}{\in} s p (S = {1 + x^{2}, x + x^{2}, 3 + x + 2 x^{2}}) \\ Solution: \\ {\begin{matrix} α + 3 γ = 8 \\ β + γ = - 3 \\ α + β + 2 γ = 1 \end{matrix} ⟺ {\begin{matrix} α + 3 γ = 8 \\ β + γ = - 3 \\ β - γ = - 7 \end{matrix} ⟺ {\begin{matrix} α + 3 γ = 8 \\ β + γ = - 3 \\ - 2 γ = - 4 \end{matrix} ⟺ {\begin{matrix} α = 2 \\ β = - 5 \\ γ = 2 \end{matrix} \\ (\begin{array}{ccccc} v_{1} & v_{2} & v_{3} & p \\ 1 & 1 & 0 & 3 & 8 \\ x & 0 & 1 & 1 & - 3 \\ x^{2} & 1 & 1 & 2 & 1 \end{array}) \\ ⟹ 8 - 3 x + x^{2} \in s p (S) \end{matrix}

\begin{matrix} V = R^{3} \\ W = s p ({(\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}), (\begin{matrix} 2 \\ 3 \\ - 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix})}) \\ Show: W = {(\begin{matrix} x \\ y \\ z \end{matrix}) | 2 x - y + z = 0} \\ Solution: \\ W = {(\begin{matrix} x \\ y \\ z \end{matrix}) | \exists α, β, γ : α (\begin{matrix} 1 \\ 2 \\ 0 \end{matrix}) + β (\begin{matrix} 2 \\ 3 \\ - 1 \end{matrix}) + γ (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}) = (\begin{matrix} x \\ y \\ z \end{matrix})} = \\ = {(\begin{matrix} x \\ y \\ z \end{matrix}) | (\begin{array}{cccc} 1 & 2 & 0 & x \\ 2 & 3 & 1 & y \\ 0 & - 1 & 1 & z \end{array}) has a solution} = \\ = {(\begin{matrix} x \\ y \\ z \end{matrix}) | (\begin{array}{cccc} 1 & 2 & 0 & x \\ 0 & - 1 & 1 & y - 2 x \\ 0 & 0 & 0 & 2 x - y + z \end{array}) has a solution} = \\ = {(\begin{matrix} x \\ y \\ z \end{matrix}) | 2 x - y + z = 0} \end{matrix}