Linear-1 11

1

\begin{matrix} Let V = R_{2} [x] over R \\ Let S be a standard basis of V \\ Let B = {1, 1 + x, 1 + x + x^{2}} basis of V \end{matrix}

\begin{matrix} Find [I]_{S}^{B} \\ Solution: \\ [I]_{S}^{B} = (\begin{matrix} \overset{|}{\underset{|}{[1]_{S}}} & \overset{|}{\underset{|}{[1 + x]_{S}}} & \overset{|}{\underset{|}{[1 + x + x^{2}]_{S}}} \end{matrix}) \\ ⟹ [I]_{S}^{B} = (\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}) \end{matrix}

\begin{matrix} Find [v]_{B} for all v \in V \\ Solution: \\ [I]_{B}^{S} = ([I]_{S}^{B})^{- 1} = {(\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix})}^{- 1} \overset{R_{1} - R_{2}}{\underset{R_{2} - R_{3}}{=}} (\begin{matrix} 1 & - 1 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{matrix}) \\ ⟹ [1]_{B} = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), [x]_{B} = (\begin{matrix} - 1 \\ 1 \\ 0 \end{matrix}), [x^{2}]_{B} = (\begin{matrix} 0 \\ - 1 \\ 1 \end{matrix}) \\ ⟹ [v]_{B} = [a + b x + c x^{2}]_{B} = a [1]_{B} + b [x]_{B} + c [x^{2}]_{B} \\ ⟹ [v]_{B} = a (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + b (\begin{matrix} - 1 \\ 1 \\ 0 \end{matrix}) + c (\begin{matrix} 0 \\ - 1 \\ 1 \end{matrix}) = (\begin{matrix} a - b \\ b - c \\ c \end{matrix}) \end{matrix}

\begin{matrix} Let V be a vector space over F \\ Let B be a basis of V \\ Let {v_{1}, \dots, v_{n}} \subseteq V \\ Prove: \\ {v_{1}, \dots, v_{n}} is a linear independence ⟺ {[v_{1}]_{B}, \dots, [v_{n}] B} is a linear independence \\ Proof: \\ Let {v_{1}, \dots, v_{n}} be a linear dependence \\ ⟹ \exists {α_{1}, \dots, α_{n}} \neq {0} : \sum_{i = 1}^{n} α_{i} v_{i} = 0 \\ \sum_{i = 1}^{n} α_{i} v_{i} = 0 ⟹ {[\sum_{i = 1}^{n} α_{i} v_{i}]}_{B} = 0 ⟹ \sum_{i = 1}^{n} α_{i} [v_{i}]_{B} = 0 \\ ⟹ {[v_{1}]_{B}, \dots, [v_{n}]_{B}} is a linear dependence \\ Let {[v_{1}]_{B}, \dots, [v_{n}]_{B}} be a linear dependence \\ ⟹ \exists {α_{1}, α_{2}, \dots, α_{n}} \neq {0} : \sum_{i = 1}^{n} α_{i} [v_{i}]_{B} = 0 \\ \sum_{i = 1}^{n} α_{i} [v_{i}]_{B} = 0 ⟹ {[\sum_{i = 1}^{n} α_{i} v_{i}]}_{B} = 0 ⟹ \sum_{i = 1}^{n} α_{i} v_{i} = 0 \\ ⟹ {v_{1}, \dots, v_{n}} is a linear dependence \\ ⟹ {v_{1}, \dots, v_{n}} is a linear dependence ⟺ {[v_{1}]_{B}, \dots, [v_{n}]_{B}} is a linear dependence \\ {v_{1}, \dots, v_{n}} is a linear independence ⟺ {[v_{1}]_{B}, \dots, [v_{n}]_{B}} is a linear independence \\ Another way to prove it: \\ []_{B} is an invertible linear transformation \\ ⟹ {v_{1}, \dots, v_{n}} is linearly independent ⟺ {[v_{1}]_{B}, \dots, [v_{n}]_{B}} is linearly independent \end{matrix}

\begin{matrix} Let V = R^{3} over R \\ Let B = {(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix})}, C = {(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 0 \end{matrix})} be bases of V \\ Find [I]_{C}^{B} \\ Solution: \\ [I]_{C}^{B} = [I]_{C}^{S} [I]_{S}^{B} \\ [I]_{S}^{B} = (\begin{matrix} \overset{|}{\underset{|}{b_{1}}} & \overset{|}{\underset{|}{b_{2}}} & \overset{|}{\underset{|}{b_{3}}} \end{matrix}) = (\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}) \\ [I]_{S}^{C} = (\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{matrix}) \\ (\begin{array}{cccccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 \end{array}) \to (\begin{array}{cccccc} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & - 1 \end{array}) \to (\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & - 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 & - 1 \end{array}) \\ [I]_{C}^{S} = ([I]_{S}^{C})^{- 1} = {(\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{matrix})}^{- 1} = (\begin{matrix} 0 & 0 & 1 \\ - 1 & 1 & 1 \\ 1 & 0 & - 1 \end{matrix}) \\ ⟹ [I]_{C}^{B} = (\begin{matrix} 0 & 0 & 1 \\ - 1 & 1 & 1 \\ 1 & 0 & - 1 \end{matrix}) (\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}) = (\begin{matrix} 0 & 0 & 1 \\ - 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}) \end{matrix}

\begin{matrix} Let V be a vector space \\ Let B = {v_{1}, \dots, v_{n}} be a basis of V \\ Let C = {w_{1}, \dots, w_{n}} be a basis of V such that \\ \forall i \in [1, n] : w_{i} = \sum_{k = 1}^{i} v_{k} \\ Find [I]_{B}^{C} \\ Solution: \\ [I]_{B}^{C} = (\begin{matrix} \overset{|}{\underset{|}{[I (w_{1})]_{B}}} & \dots & \overset{|}{\underset{|}{[I (w_{n})]_{B}}} \end{matrix}) \\ I (w_{1}) = I (\sum_{k = 1}^{1} v_{k}) = I (v_{1}) = v_{1} \\ ⟹ [I (w_{1})]_{B} = [v_{1}]_{B} = e_{1} \\ I (w_{2}) = I (\sum_{k = 1}^{2} v_{k}) = I (v_{1} + v_{2}) = v_{1} + v_{2} \\ ⟹ [I (w_{2})]_{B} = [v_{1} + v_{2}]_{B} = e_{1} + e_{2} \\ I (w_{i}) = I (\sum_{k = 1}^{i} v_{k}) ⟹ [I (w_{i})]_{B} = \sum_{k = 1}^{i} e_{k} \\ ⟹ [I]_{B}^{C} = (\begin{matrix} e_{1} & \sum_{k = 1}^{2} e_{k} & \dots & \sum_{k = 1}^{n} e_{k} \end{matrix}) = (\begin{matrix} 1 & 1 & 1 & \dots & 1 \\ 0 & 1 & 1 & \dots & 1 \\ 0 & 0 & 1 & \dots & 1 \\ ⋮ & ⋮ & ⋱ & ⋱ & ⋮ \\ 0 & 0 & \dots & 0 & 1 \end{matrix}) = {\begin{cases} 1 & i \leq j \\ 0 & i > j \end{cases} \end{matrix}

\begin{matrix} Let V be a vector space over F, d i m (V) = n \\ Let A \in F^{n \times n} be invertible \\ Prove: \exists B, C bases of V : [I]_{C}^{B} = A \\ Proof: \\ Let A = (\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & a_{n 2} & \dots & a_{n n} \end{matrix}) \\ Let C = {v_{1}, \dots, v_{n}} be a basis of V \\ Let B = {w_{1}, \dots, w_{n}} such that \forall i \in [1, n] : w_{i} = \sum_{k = 1}^{n} a_{k i} v_{k} \\ ⟹ \forall i \in [1, n] : [w_{i}]_{C} = {[\sum_{k = 1}^{n} a_{k i} v_{k}]}_{C} = (\begin{matrix} a_{1 i} \\ a_{2 i} \\ ⋮ \\ a_{n i} \end{matrix}) = C_{i} (A) \\ A is invertible ⟹ {C_{1} (A), \dots, C_{n} (A)} is a linear independence \\ ⟹ {[w_{1}]_{C}, \dots, [w_{n}]_{C}} is a linear independence ⟹ B is a linear independence \\ B \subseteq V, B is a linear independence, | B | = d i m (V) ⟹ B is a basis of V \\ and [I]_{C}^{B} = (\begin{matrix} \overset{|}{\underset{|}{[w_{1}]_{C}}} & \dots & \overset{|}{\underset{|}{[w_{n}]_{C}}} \end{matrix}) = (\begin{matrix} \overset{|}{\underset{|}{C_{1} (A)}} & \dots & \overset{|}{\underset{|}{C_{n} (A)}} \end{matrix}) = A \end{matrix}

\begin{matrix} Let V be a vector space over F, d i m (V) = n \\ Let B, C, D be bases of V \\ Prove or disprove: [I]_{C}^{B} = [I]_{C}^{D} ⟹ B = D \\ Proof: \\ [I]_{D}^{B} = [I]_{D}^{C} [I]_{C}^{B} = ([I]_{C}^{D})^{- 1} [I]_{C}^{B} = ([I]_{C}^{B})^{- 1} [I]_{C}^{B} = I \\ ⟹ [I]_{B}^{B} = I = [I]_{D}^{B} \\ Transformation matrix is unique \\ ⟹ B = D \end{matrix}