Linear-1 12

Properties of representation matrix #lemma

\begin{matrix} Let T_{2}, T_{1} : V \to U linear transformations \\ S : U \to W linear transformation \\ B, C, E bases of V, U, W \\ α \in F \\ Then: \\ 1. [T_{2} + T_{1}]_{C}^{B} = [T_{2}]_{C}^{B} + [T_{1}]_{C}^{B} \\ 2. [α T_{1}]_{C}^{B} = α [T_{1}]_{C}^{B} \\ 3. [\underset{S \circ T_{1}}{\underset{⏟}{S T_{1}}}]_{E}^{B} = [S]_{E}^{C} \cdot [T_{1}]_{C}^{B} \\ Proof for 1. \\ Let v \in V \\ ([T_{2}]_{C}^{B} + [T_{1}]_{C}^{B}) [v]_{B} = [T_{2}]_{C}^{B} [v]_{B} + [T_{1}]_{C}^{B} [v]_{B} = [T_{2} (v)]_{C} + [T_{1} (v)]_{C} = [T_{2} (v) + T_{1} (v)]_{C} = \\ = [(T_{2} + T_{1}) (v)]_{C} \\ ⟹ [T_{2}]_{C}^{B} + [T_{1}]_{C}^{B} = [T_{2} + T_{1}]_{C}^{B} \\ Proof for 2. \\ Let v \in V \\ (α [T_{1}]_{C}^{B}) [v]_{B} = α ([T_{1}]_{C}^{B} [v]_{B}) = α [T_{1} (v)]_{C} = [(α T_{1}) (v)]_{C} \\ ⟹ α [T_{1}]_{C}^{B} = [α T_{1}]_{C}^{B} \\ Proof for 3. \\ Let v \in V \\ ([S]_{E}^{C} \cdot [T_{1}]_{C}^{B}) [v]_{B} = [S]_{E}^{C} ([T_{1}]_{C}^{B} \cdot [v]_{B}) = [S]_{E}^{C} \cdot [T_{1} (v)]_{C} = [S (T_{1} (v))]_{E} = [(S T_{1}) (v)]_{E} \\ ⟹ [S]_{E}^{C} \cdot [T_{1}]_{C}^{B} = [S T_{1}]_{E}^{B} \end{matrix}

\begin{matrix} Let V be a finitely generated vector space over F \\ Let B basis of V \\ Then [I]_{B}^{B} = I \\ Proof: \\ Let v \in V \\ [I]_{B}^{B} [v]_{B} = [I (v)]_{B} = [v]_{B} ⟹ [I]_{B}^{B} = I \\ [I]_{B}^{B} = (\begin{matrix} ⋮ & ⋮ \\ [I (v_{1})]_{B} & \dots & [I (v_{n})]_{B} \\ ⋮ & ⋮ \end{matrix}) = (\begin{matrix} ⋮ & ⋮ \\ [v_{1}]_{B} & \dots & [v_{n}]_{B} \\ ⋮ & ⋮ \end{matrix}) \\ \forall i \in [1, n] : [v_{i}]_{B} = e_{i} \\ ⟹ [I]_{B}^{B} = (\begin{matrix} ⋮ & ⋮ & ⋮ \\ e_{1} & e_{2} & \dots & e_{n} \\ ⋮ & ⋮ & ⋮ \end{matrix}) = I_{n} \end{matrix}

\begin{matrix} Let T : V \to U linear transformation \\ Let B, C bases of V, U \\ Then T is invertible ⟺ [T]_{C}^{B} is invertible \\ Proof: \\ Let T be invertible \\ ⟹ d i m (V) = d i m (U) = n ⟹ [T]_{C}^{B} \in F^{n \times n} \\ [T^{- 1}]_{B}^{C} [T]_{C}^{B} = [T^{- 1} T]_{B}^{B} = [I]_{B}^{B} = I \\ ⟹ [T^{- 1}]_{B}^{C} is an inverse of [T]_{C}^{B} \\ Let [T]_{C}^{B} be invertible \\ ⟹ [T]_{C}^{B} \in F^{n \times n} ⟹ d i m (V) = d i m (U) = n \\ ⟹ [T is surjective ⟺ T is injective] \\ Let v \in k e r (T) \\ ⟹ T (v) = 0 ⟹ [T (v)]_{C} = [0]_{C} = 0 \\ ⟹ [T (v)]_{C} = [T]_{C}^{B} [v]_{B} = 0 \\ [T]_{C}^{B} is invertible ⟹ N ([T]_{C}^{B}) = {0} ⟹ [v]_{B} = 0 ⟹ v = 0 \\ ⟹ k e r (T) = {0} ⟹ T is injective ⟹ T is bijective \\ ⟹ T is invertible \end{matrix}

\begin{matrix} T : \overset{B}{V} \to \overset{C, S}{U} \\ [T]_{C}^{B} = [I]_{C}^{S} \cdot [T]_{S}^{B} = ([I]_{S}^{C})^{- 1} [T]_{S}^{B} \\ Much easier to calculate [T]_{S}^{B} than [T]_{C}^{B} \\ Might be much easier to calculate ([I]_{S}^{C})^{- 1} than [I]_{C}^{S} \end{matrix}

\begin{matrix} Let T : V \to U linear transformation \\ Let B, D bases of V, U \\ Then \\ 1. [k e r (T)]_{B} = N ([T]_{D}^{B}) \\ 2. [I m (T)]_{D} = C ([T]_{D}^{B}) \\ Proof for 1. \\ [v]_{B} \in [k e r (T)]_{B} ⟺ v \in k e r (T) \\ ⟺ T (v) = 0 ⟺ [T (v)]_{D} = [0]_{D} = 0 \\ ⟺ [T]_{D}^{B} \underset{v \in V}{\underset{⏟}{[v]_{B}}} = [T (v)]_{D} = 0 ⟺ [v]_{B} \in N ([T]_{D}^{B}) \\ ⟹ [k e r (T)]_{B} = N ([T_{D}^{B}]) \\ Proof for 2. \\ [u]_{D} \in [I m (T)]_{D} ⟺ u \in I m (T) ⟺ \exists v \in V : T (v) = u \\ ⟺ T (v) \in I m (T) ⟺ [T (v)]_{D} \in [I m (T)]_{D} ⟺ [T]_{D}^{B} \underset{v \in V}{\underset{⏟}{[v]_{B}}} \in [I m (T)]_{D} \\ Note: [T]_{D}^{B} [v]_{B} is a linear combination of columns of [T]_{D}^{B} \\ ⟺ [T]_{D}^{B} [v]_{B} \in C ([T]_{D}^{B}) \\ ⟹ [I m (T)]_{D} = C ([T]_{D}^{B}) \end{matrix}

\begin{matrix} T : \overset{E}{R_{2} [x]} \to \overset{F}{R^{2 \times 2}} \\ E = {1, 1 + x, 1 + x^{2}} \\ F = {(\begin{matrix} 0 & 0 \\ 0 & - 1 \end{matrix}), (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 2 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix})} \\ [T]_{F}^{E} = (\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 1 & 1 & 5 \\ 1 & - 1 & - 1 \end{matrix}) \\ ⟹ [T (1)]_{F} = (\begin{matrix} 1 \\ 0 \\ 1 \\ 1 \end{matrix}), [T (1 + x)]_{F} = (\begin{matrix} 0 \\ 1 \\ 1 \\ - 1 \end{matrix}), [T (1 + x^{2})]_{F} = (\begin{matrix} 2 \\ 3 \\ 5 \\ - 1 \end{matrix}) \\ a + b x + c x^{2} = α (1) + β (1 + x) + γ (1 + x^{2}) \\ ⟹ {\begin{matrix} α = a - b - c \\ β = b \\ γ = c \end{matrix} \\ Let v \in R_{2} [x] \\ ⟹ [T]_{F}^{E} [v]_{E} = [T]_{F}^{E} \cdot [a + b x + c x^{2}]_{E} = \\ = (\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 3 \\ 1 & 1 & 5 \\ 1 & - 1 & - 1 \end{matrix}) \cdot (\begin{matrix} a - b - c \\ b \\ c \end{matrix}) = (\begin{matrix} a - b + c \\ b + 3 c \\ a - 4 c \\ a - 2 b - 2 c \end{matrix}) = [T (v)]_{F} \\ ⟹ T (v) = T (a + b x + c x^{2}) = \\ = (a - b + c) (\begin{matrix} 0 & 0 \\ 0 & - 1 \end{matrix}) + (b + 3 c) (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) + (a - 4 c) (\begin{matrix} 0 & 2 \\ 0 & 0 \end{matrix}) + (a - 2 b - 2 c) (\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}) = \\ = (\begin{matrix} b + 3 c & 2 a - 8 c \\ a - 2 b - 2 c & - b - 3 c \end{matrix}) \end{matrix}