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For the linear transformation \( L: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} \) given by \[ L((x, y))=(x+2 y,-x, 0) \] find the matrix \( A_{B C} \) for \( L \) with respect to the bases \[ B=\{(1,3),(-2,4)\}, C=\{(1,1,1),(2,2,0),(3,0,0)\} \]

Ask by Mitchell Sherman. in Russia
Dec 16,2024

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The matrix \( A_{B C} \) for the linear transformation \( L \) is: \[ \boxed{\begin{bmatrix} 0 & 0 \\ -\frac{1}{2} & 1 \\ \frac{8}{3} & \frac{4}{3} \end{bmatrix}} \]

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To find the matrix \( A_{BC} \) for the linear transformation \( L \) with respect to the bases \( B \) and \( C \), we follow these steps: 1. **Write the Transformation**: The linear transformation is given by \[ L((x, y)) = (x + 2y, -x, 0). \] 2. **Compute the Images of the Basis Vectors**: We will apply \( L \) to each vector in the basis \( B \). - For \( \mathbf{b_1} = (1, 3) \): \[ L((1, 3)) = (1 + 2 \cdot 3, -1, 0) = (7, -1, 0). \] - For \( \mathbf{b_2} = (-2, 4) \): \[ L((-2, 4)) = (-2 + 2 \cdot 4, 2, 0) = (6, 2, 0). \] 3. **Express Images in Terms of the Basis \( C \)**: Next, we express the images in terms of the basis \( C = \{(1,1,1), (2,2,0), (3,0,0)\}\). - For \( L((1, 3)) = (7, -1, 0) \): \[ (7, -1, 0) = a(1, 1, 1) + b(2, 2, 0) + c(3, 0, 0). \] We set up the equations: 1. \( a + 2b + 3c = 7 \) 2. \( a + 2b = -1 \) 3. \( a = 0 \) (since the third component is 0) From the third equation, \( a = 0 \). Substituting \( a = 0 \) into the second equation gives: \[ 2b = -1 \implies b = -\frac{1}{2}. \] Substituting into the first equation: \[ 0 + 2\left(-\frac{1}{2}\right) + 3c = 7 \implies -1 + 3c = 7 \implies 3c = 8 \implies c = \frac{8}{3}. \] Therefore, in terms of the basis \( C \): \[ L((1, 3)) = 0 \cdot (1, 1, 1) - \frac{1}{2}(2, 2, 0) + \frac{8}{3}(3, 0, 0) \implies [0, -\frac{1}{2}, \frac{8}{3}]. \] - For \( L((-2, 4)) = (6, 2, 0) \): \[ (6, 2, 0) = a'(1, 1, 1) + b'(2, 2, 0) + c'(3, 0, 0). \] Setting up the equations: 1. \( a' + 2b' + 3c' = 6 \) 2. \( a' + 2b' = 2 \) 3. \( a' = 0 \) From the third equation, \( a' = 0 \). Using this in the second equation gives: \[ 2b' = 2 \implies b' = 1. \] Using this in the first equation: \[ 0 + 2(1) + 3c' = 6 \implies 2 + 3c' = 6 \implies 3c' = 4 \implies c' = \frac{4}{3}. \] Therefore, in terms of the basis \( C \): \[ L((-2, 4)) = 0 \cdot (1, 1, 1) + 1(2, 2, 0) + \frac{4}{3}(3, 0, 0) \implies [0, 1, \frac{4}{3}]. \] 4. **Construct the Matrix**: The resulting vectors from the transformation can be arranged as columns in the matrix: \[ A_{

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Other France Jan 22, 2025
Numéro d'étudiant : La qualité de la rédaction sera prise en compte. Exercice 1. Soit \[ \mathcal{B}=\left\{\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right],\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right]\right\} \] la base canonique de \( \operatorname{Mat}_{2}(\mathbb{R}) \) et soit \( f: \operatorname{Mat}_{2}(\mathbb{R}) \rightarrow \operatorname{Mat}_{2}(\mathbb{R}) \) l'endomorphisme de \( \operatorname{Mat}_{2}(\mathbb{R}) \) tel que, en base canonique, \[ f\left(\left[\begin{array}{ll} x_{1} & x_{2} \\ x_{3} & x_{4} \end{array}\right]\right)=\left(\left[\begin{array}{cc} x_{1}+2 x_{3} & 2 x_{1}-x_{2}+4 x_{3}-2 x_{4} \\ -x_{3} & -2 x_{3}+x_{4} \end{array}\right]\right) \] (a) Montrer que \[ A=\mu_{\mathcal{B}, \mathcal{B}}(f)=\left(\begin{array}{cccc} 1 & 0 & 2 & 0 \\ 2 & -1 & 4 & -2 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & -2 & 1 \end{array}\right) \] où \( \mu_{\mathcal{B}, \mathcal{B}}(f) \) est la matrice associée à \( f \) dans la base canonique. (b) Déterminer le polynôme caractéristique \( \chi_{f}(x) \). (c) Déterminer les valeurs propres de \( f \), leurs multiplicités algébriques et montrer que l'endomorphisme \( f \) est diagonalisable. (d) Déterminer une base \( \mathcal{B}^{\prime} \) de \( \operatorname{Mat}_{2}(\mathbb{R}) \) formée de vecteurs propres de \( \operatorname{Mat}_{2}(\mathbb{R}) \), la matrice de changement de base \( P:=\mu_{\mathcal{B}^{\prime}, \mathcal{B}}\left(\operatorname{Id}_{\operatorname{Mat}_{2}(\mathbb{R})}\right) \) et la matrice diagonale \( D:=\mu_{\mathcal{B}^{\prime}, \mathcal{B}^{\prime}}(f) \) telles que \[ \mu_{\mathcal{B}^{\prime}, \mathcal{B}^{\prime}}(f)=\left(\mu_{\mathcal{B}^{\prime}, \mathcal{B}}\left(\operatorname{Id}_{\operatorname{Mat}_{2}(\mathbb{R})}\right)\right)^{-1} \mu_{\mathcal{B}, \mathcal{B}}(f) \mu_{\mathcal{B}^{\prime}, \mathcal{B}}\left(\operatorname{Id}_{\operatorname{Mat}_{2}(\mathbb{R})}\right) \] Autrement dit, \[ D=P^{-1} A P \] où \( A=\mu_{\mathcal{B}, \mathcal{B}}(f) \).
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