Linear-1 11

Coordinate vectors #definition

\begin{matrix} Let V = R^{2} \\ Let B_{V} = {(\binom{1}{0}), (\binom{0}{1})} \\ ⟹ (\binom{4}{3}) = 4 (\binom{1}{0}) + 3 (\binom{0}{1}) \\ ⟹ {[(\binom{4}{3})]}_{B_{V}} = (\binom{4}{3}) \\ Now let B_{V} = {(\binom{1}{1}), (\binom{1}{2})} \\ ⟹ (\binom{4}{3}) = 5 (\binom{1}{1}) - 1 (\binom{1}{2}) \\ ⟹ {[(\binom{4}{3})]}_{B_{V}} = (\binom{5}{- 1}) \end{matrix}

\begin{matrix} Let V be a vector space over F \\ Let B = {v_{1}, v_{2}, \dots, v_{n}} basis of V \\ Let v \in V \\ \exists α_{1}, \dots, α_{n} : v = \sum_{i = 1}^{n} α_{i} v_{i} \\ Then [v]_{B} = (\begin{matrix} α_{1} \\ ⋮ \\ α_{n} \end{matrix}) is called a coordinate vector \\ Coordinate vector is a function: \\ []_{B} : V \to F^{n} \\ This function is injective: \\ Let [v]_{B} = [u]_{B} \\ ⟹ v = α_{1} v_{1} + α_{2} v_{2} + \dots + α_{n} v_{n} = u \\ ⟹ v = u \end{matrix}

\begin{matrix} Let V be a finitely generated vector space over F \\ Let B = {v_{1}, v_{2}, \dots, v_{n}} basis of V \\ Then \\ 1. \forall v, u \in V : [v + u]_{B} = [v]_{B} + [u]_{B} \\ 2. \forall v \in V, α \in F : [α v]_{B} = α [v]_{B} \\ 3. [v]_{B} = 0 ⟺ v = 0 \\ Proof for 1: \\ Let v, u \in V \\ s p (B) = V ⟹ v = \sum_{i = 1}^{n} α_{i} v_{i}, u = \sum_{i = 1}^{n} β_{i} v_{i} \\ ⟹ [v]_{B} + [u]_{B} = (\begin{matrix} α_{1} + β_{1} \\ ⋮ \\ α_{n} + β_{n} \end{matrix}) \\ [v + u]_{B} = {[\sum_{i = 1}^{n} α_{i} v_{i} + \sum_{i = 1}^{n} β_{i} v_{i}]}_{B} = (\begin{matrix} α_{1} + β_{1} \\ ⋮ \\ α_{n} + β_{n} \end{matrix}) \\ ⟹ [v + u]_{B} = [v]_{B} + [u]_{B} \end{matrix}

\begin{matrix} Proof for 2: \\ Let v \in V, α \in F \\ s p (B) = V ⟹ v = \sum_{i = 1}^{n} α_{i} v_{i} \\ α [v]_{B} = α (\begin{matrix} α_{1} \\ ⋮ \\ α_{n} \end{matrix}) \\ [α v]_{B} = [α \sum_{i = 1}^{n} α_{i} v_{i}] = (\begin{matrix} α α_{1} \\ ⋮ \\ α α_{n} \end{matrix}) = α (\begin{matrix} α_{1} \\ ⋮ \\ α_{n} \end{matrix}) \\ ⟹ [α v]_{B} = α [v_{B}] \\ Proof for 3: \\ Let v \in V \\ Let [v]_{B} = 0 \\ ⟹ 0 v_{1} + 0 v_{2} + \dots + 0 v_{n} = v ⟹ v = 0 \\ Let v = 0 \\ s p (B) = V ⟹ v = α_{1} v_{1} + \dots + α_{n} v_{n} = 0 \\ B is a linear independence ⟹ α_{1} = \dots = α_{n} = 0 ⟹ [v]_{B} = 0 \\ ⟹ [v]_{B} = 0 ⟺ v = 0 \end{matrix}

\begin{matrix} []_{B} is invertible (an isomorphism) \\ Proof: \\ 1. and 2. ⟹ []_{B} is a linear transformation \\ 3. ⟹ k e r ([]_{B}) = {0} ⟹ []_{B} is injective \\ d i m (F^{n}) = n = d i m (V) ⟹ []_{B} is an isomorphism \\ Isomorphism is denoted as V ≅ F^{n} \end{matrix}

LD is linear dependence
LID is linear independence

\begin{matrix} Let V, U finitely generated vector spaces \\ Let T : V \to U be an isomorphism \\ Then \\ 1. \forall v_{1}, \dots, v_{n} \in V : {v_{1}, \dots, v_{n}} is a LID ⟺ {T (v_{1}), \dots, T (v_{n})} is a LID \\ 2. \forall v_{1}, \dots, v_{n}, w \in V : w \in s p ({v_{1}, \dots, v_{n}}) ⟺ T (w) \in s p ({T (v_{1}), \dots, T (v_{n})}) \\ Proof for 1: \\ T O D O B Y Y O U R S E L F \\ Proof for 2: \\ Let w, v_{1}, \dots, v_{n} \in V \\ w \in s p ({v_{1}, \dots, v_{n}}) ⟺ w = \sum_{i = 1}^{n} α_{i} v_{i} \\ \underset{T is an isomorphism ⟹ T is bijective}{\underset{⏟}{⟺}} T (w) = T (\sum_{i = 1}^{n} α_{i} v_{i}) = \sum_{i = 1}^{n} T (α_{i} v_{i}) = \sum_{i = 1}^{n} α_{i} T (v_{i}) \\ ⟺ T (w) \in s p ({T (v_{1}), \dots, T (v_{n})}) \end{matrix}

\begin{matrix} A \in F^{m \times n} \\ T_{A} : F^{n} \to F^{m} \\ T_{A} (v) = A v \end{matrix}

\begin{matrix} Let T : V \to U \\ Let B basis of V \\ Let C basis of U \\ We want to find a matrix A such that \\ \forall v \in V : T (v) = A [v]_{B} = [T (v)]_{C} \\ ⟺ \forall v \in V : T (v) = T_{A} ([v]_{B}) = [T (v)]_{C} \end{matrix}

\begin{matrix} Let T : R_{3} [x] \to R_{2} [x] \\ T (p (x)) = p^{'} (x) \\ (a + b x + c x^{2} + d x^{3})^{'} = b + 2 c x + 3 d x^{2} \\ A \cdot (\begin{matrix} a \\ b \\ c \\ d \end{matrix}) = (\begin{matrix} b \\ 2 c \\ 3 d \end{matrix}) ⟹ A = (\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{matrix}) \end{matrix}

\begin{matrix} Let B = {v_{1}, \dots, v_{n}} basis of V \\ Let C = {u_{1}, \dots, u_{m}} basis of U \\ Let T : V \to U \\ d i m (V) = n \\ d i m (U) = m \\ Let A \in F^{m \times n} \\ Let v \in V \\ ⟹ v = \sum_{i = 1}^{n} α_{i} v_{i} \\ ⟹ A [v]_{B} = A {[\sum_{i = 1}^{n} α_{i} v_{i}]}_{B} = A \sum_{i = 1}^{n} α_{i} [v_{i}]_{B} = \sum_{i = 1}^{n} α_{i} A [v_{i}]_{B} = \sum_{i = 1}^{n} α_{i} T_{A} ([v_{i}]_{B}) = \\ = \sum_{i = 1}^{n} α_{i} [T (v_{i})]_{C} = {[\sum_{i = 1}^{n} α_{i} T (v_{i})]}_{C} = {[T (\sum_{i = 1}^{n} α_{i} v_{i})]}_{C} = [T (v)]_{C} \end{matrix}

\begin{matrix} [v_{i}]_{B} = e_{i} = (\begin{matrix} 0 \\ ⋮ \\ 1 \\ ⋮ \\ 0 \end{matrix}) \in F^{n} ⟹ A [v_{i}]_{B} = A e_{i} which is i -th column of A \\ ⟹ i -th column of A is equal to [T (v_{i})]_{C} \\ [T]_{C}^{B} is a notation for transformation matrix A \\ ⟹ A = [T]_{C}^{B} = (\begin{matrix} ⋮ & ⋮ \\ [T (v_{1})]_{C} & \dots & [T (v_{n})]_{C} \\ ⋮ & ⋮ \end{matrix}) \in F^{m \times n} \\ [T]_{C}^{B} = (\begin{matrix} ⋮ & ⋮ \\ [T (v_{1})]_{C} & \dots & [T (v_{n})]_{C} \\ ⋮ & ⋮ \end{matrix}) \in F^{m \times n} \\ For the previous example: \\ [T]_{S_{2}}^{S_{1}} = (\begin{matrix} ⋮ & ⋮ & ⋮ & ⋮ \\ [T (1)]_{S_{2}} & [T (x)]_{S_{2}} & [T (x^{2})]_{S_{2}} & [T (x^{3})]_{S_{2}} \\ ⋮ & ⋮ & ⋮ & ⋮ \end{matrix}) = (\begin{matrix} ⋮ & ⋮ & ⋮ & ⋮ \\ [0]_{S_{2}} & [1]_{S_{2}} & [2 x]_{S_{2}} & [3 x^{2}]_{S_{2}} \\ ⋮ & ⋮ & ⋮ & ⋮ \end{matrix}) \\ ⟹ [T]_{S_{2}}^{S_{1}} = (\begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{matrix}) \\ Now let I : V_{B} \to V_{C} \\ Meaning it doesn’t change the vector, but represents it via different basis \\ Then [I]_{C}^{B} = (\begin{matrix} [I (v_{1})]_{C} & \dots & [I (v_{n})]_{C} \end{matrix}) = (\begin{matrix} [v_{1}]_{C} & \dots & [v_{n}]_{C} \end{matrix}) \\ [I]_{C}^{B} \underset{[v]_{B}}{\underset{⏟}{[I]_{B}^{D} [v]_{D}}} = [I]_{C}^{B} [v]_{B} = [v]_{C} \\ ⟹ [I]_{C}^{B} [I]_{B}^{D} = [I]_{C}^{D} \end{matrix}

\begin{matrix} Let T_{A} : F^{n} \to F^{m} \\ T_{A} (v) = A v \\ T_{A} = []_{C} \circ T \circ []_{B}^{- 1} \\ ⟹ T_{A} \circ []_{B} = []_{C} \circ T \\ ⟹ T_{A} ([v]_{B}) = [T (v)]_{C} ⟹ A [v]_{B} = [T (v)]_{C} \\ [v_{i}]_{B} = e_{i} = (\begin{matrix} 0 \\ ⋮ \\ 1 \\ ⋮ \\ 0 \end{matrix}) \in F^{n} ⟹ A [v_{i}]_{B} = A e_{i} which is i -th column of A \\ ⟹ i -th column of A is equal to [T (v_{i})]_{C} \\ ⟹ A = [T]_{C}^{B} = (\begin{matrix} ⋮ & ⋮ \\ [T (v_{1})]_{C} & \dots & [T (v_{n})]_{C} \\ ⋮ & ⋮ \end{matrix}) \in F^{m \times n} \end{matrix}