Linear-1 11

Properties of linear transformation matrix

\begin{matrix} 1. [T + S]_{C}^{B} = [T]_{C}^{B} + [S]_{C}^{B} \\ 2. T is invertible ⟹ [T^{- 1}]_{B}^{C} = ([T]_{C}^{B})^{- 1} \\ 3. [S T]_{D}^{B} = [S]_{D}^{C} \cdot [T]_{C}^{B} \end{matrix}

\begin{matrix} S, S^{'} are standard bases of R^{2}, R_{2} [x] \\ T : R_{2} [x] \to R^{2} \\ T (a_{0} + a_{1} x + a_{2} x^{2}) = (\begin{matrix} 2 a_{2} + a_{1} \\ a_{0} \end{matrix}) \\ E = {1, 1 + x, x + x^{2}}, F = {(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 1 \end{matrix})} \\ Find [T]_{S^{'}}^{S} \\ Find [T]_{F}^{E} \\ Solution: \\ S = {1, x, x^{2}}, S^{'} = {(\begin{matrix} 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 1 \end{matrix})} \\ T (1) = (\begin{matrix} 0 \\ 1 \end{matrix}), T (x) = (\begin{matrix} 1 \\ 0 \end{matrix}), T (x^{2}) = (\begin{matrix} 2 \\ 0 \end{matrix}) \\ ⟹ [T]_{S^{'}}^{S} = (\begin{matrix} 0 & 1 & 2 \\ 1 & 0 & 0 \end{matrix}) \\ [T]_{F}^{E} = [I]_{F}^{S^{'}} \cdot [T]_{S^{'}}^{S} \cdot [I]_{S}^{E} \\ [I]_{F}^{S^{'}} = (\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}) \\ [T]_{F}^{E} = (\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}) (\begin{matrix} 0 & 1 & 2 \\ 1 & 0 & 0 \end{matrix}) (\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}) = (\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix}) (\begin{matrix} 0 & 1 & 3 \\ 1 & 1 & 0 \end{matrix}) = (\begin{matrix} - 1 & 0 & 3 \\ 1 & 1 & 0 \end{matrix}) \\ [T]_{F}^{E} [v]_{E} = [T (v)]_{F} \end{matrix}

\begin{matrix} [k e r (T)]_{B} = N ([T]_{C}^{B}) \\ [I m (T)]_{C} = C ([T]_{C}^{B}) \end{matrix}

\begin{matrix} E = {(\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 0 \\ 2 \end{matrix})}, F = {(\begin{matrix} 1 \\ 1 \end{matrix}), (\begin{matrix} 1 \\ 2 \end{matrix})} \\ T : R^{3} \to R^{2} \\ [T]_{F}^{E} = (\begin{matrix} 1 & - 2 & 3 \\ 2 & - 4 & 6 \end{matrix}) \\ C F ([T]_{F}^{E}) = (\begin{matrix} 1 & - 2 & 3 \\ 0 & 0 & 0 \end{matrix}) \\ ⟹ N ([T]_{F}^{E}) = s p ({(\begin{matrix} 2 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} - 3 \\ 0 \\ 1 \end{matrix})}) \\ ⟹ C ([T_{F}^{E}]) = s p ({(\begin{matrix} 1 \\ 2 \end{matrix})}) \end{matrix}

\begin{matrix} V = R^{n} over R \\ T : V \to V \\ A \in R^{n \times n} \\ 1. \forall B, C bases of V : [T]_{C}^{B} = A \\ 2. \forall B basis of V : [T]_{B}^{B} = A \\ Solution for 1: \\ T = 0 \\ Solution for 2: \\ T = 0 \\ T = I \\ T = α I \end{matrix}

\begin{matrix} T = D + D^{2} + D^{3} \\ ⟹ [T]_{S} = [D]_{S} + [D^{2}]_{S} + [D^{3}]_{S} = [D]_{S} + ([D]_{S})^{2} + ([D]_{S})^{3} \end{matrix}

\begin{matrix} T : R^{3} \to R^{3} \\ T ((\begin{matrix} x \\ y \\ z \end{matrix})) = (\begin{matrix} x \\ x + y \\ x + y + z \end{matrix}) \\ A = (\begin{matrix} 1 & 1 & 0 \\ 2 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}) \\ B basis of R^{3} \\ 1. Find [T]_{S}^{S} \\ 2. [T]_{B}^{S} = A, find [I]_{S}^{B} \\ 3. Find B \\ 4. Find basis E of R^{3} such that [T]_{E}^{E} is upper-triangle \\ Solution: \\ [T]_{S}^{S} = (\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{matrix}) \\ [T]_{B}^{S} = [I]_{B}^{S} [T]_{S}^{S} \\ [I]_{S}^{B} A = [I]_{S}^{B} [I]_{B}^{S} [T]_{S}^{S} \\ ⟹ [I]_{S}^{B} = [T]_{S}^{S} A^{- 1} \\ B = {C_{1} ([I]_{S}^{B}), C_{2} ([I]_{S}^{B}), C_{3} ([I]_{S}^{B})} \\ E = {(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}), (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})} \end{matrix}

\begin{matrix} V is a vector space of dimension 2 over R \\ T : V \to V \\ T^{2} = - I \\ 1. Prove: \forall v \neq 0 \in V : {v, T (v)} is a linear independence \\ 2. Find basis E of V such that: [T]_{E}^{E} = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}) \\ Solution: \\ \dots \end{matrix}

\begin{matrix} T : R^{2} \to R^{2} \\ T ((\begin{matrix} x \\ y \end{matrix})) = (\begin{matrix} x + 2 y \\ x + y \end{matrix}) \\ If there exists basis E such that [T]_{E}^{E} is diagonal \\ Solution: \\ [T]_{S}^{S} = (\begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix}) = A \\ (A - λ I) v = 0 \\ A - λ I = (\begin{matrix} 1 - λ & 2 \\ 1 & 1 - λ \end{matrix}) \\ (\begin{matrix} 1 - λ & 2 \\ 1 & 1 - λ \end{matrix}) \to (\begin{matrix} - λ & 1 + λ \\ 1 & 1 - λ \end{matrix}) \underset{λ \neq 0}{\to} (\begin{matrix} 1 & \frac{- 1 - λ}{λ} \\ 0 & 1 - λ + \frac{1 + λ}{λ} \end{matrix}) \end{matrix}