Let's break down the problem step by step, **explaining after each step** as requested. --- ### **(a) Find \([T]_{\mathcal{C} \leftarrow \mathcal{B}}\) and \([S]_{\mathcal{E} \leftarrow \mathcal{C}}\).** --- #### **Step 1: Find \([T]_{\mathcal{C} \leftarrow \mathcal{B}}\)** - **Domain:** \(M_2\) (matrices), standard basis \(\mathcal{B}\): \[ E_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \; E_2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \; E_3 = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \; E_4 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \] - **Codomain:** \(P_2\) (polynomials), standard basis \(\mathcal{C}\): \(1, x, x^2\) Apply \(T\) to each \(E_i\): 1. \(T(E_1) = T\left(\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\right) = 1 \cdot x^2 + (1 + 0) x + 0 = x^2 + x\) 2. \(T(E_2) = T\left(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\right) = 0 \cdot x^2 + (0 + 0)x + 0 = 0\) 3. \(T(E_3) = T\left(\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}\right) = 0 \cdot x^2 + (0 + 0)x + 1 = 1\) 4. \(T(E_4) = T\left(\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\right) = 0 \cdot x^2 + (0 + 1)x + 0 = x\) Express in \(\mathcal{C}\) coordinates (\(a_0 + a_1 x + a_2 x^2\)): 1. \(x^2 + x = 0 \cdot 1 + 1 \cdot x + 1 \cdot x^2 \rightarrow \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}\) 2. \(0 = 0 \cdot 1 + 0 \cdot x + 0 \cdot x^2 \rightarrow \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\) 3. \(1 = 1 \cdot 1 + 0 \cdot x + 0 \cdot x^2 \rightarrow \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\) 4. \(x = 0 \cdot 1 + 1 \cdot x + 0 \cdot x^2 \rightarrow \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\) So, \[ [T]_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} \] **Explanation:** Each column corresponds to the image of a basis matrix under \(T\), expressed in the standard polynomial basis. --- #### **Step 2: Find \([S]_{\mathcal{E} \leftarrow \mathcal{C}}\)** - **Domain:** \(P_2\), standard basis \(\mathcal{C}\) - **Codomain:** \(\mathbb{R}^3\), standard basis \(\mathcal{E}\) For \(f(x) = a + bx + cx^2\): \[ S(f) = (f(-1), f(0), f(1)) \] Apply \(S\) to each basis polynomial: 1. \(f(x) = 1\): \(S(1) = (1, 1, 1)\) 2. \(f(x) = x\): \(S(x) = (-1, 0, 1)\) 3. \(f(x) = x^2\): \(S(x^2) = (1, 0, 1)\) So, \[ [S]_{\mathcal{E} \leftarrow \mathcal{C}} = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \] **Explanation:** Each column is the vector in \(\mathbb{R}^3\) given by evaluating the polynomial basis at \(-1, 0, 1\). --- **Final Answers for (a):** \[ \boxed{ [T]_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} \quad [S]_{\mathcal{E} \leftarrow \mathcal{C}} = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} } \] --- Would you like to proceed with part (b)?

Let's continue with part (b): --- ### **(b) Find \([S \circ T]_\mathcal{B}^\mathcal{E}\).** --- #### **Step 1: Recall the matrices** From part (a), we have: \[ [T]_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} \] and \[ [S]_{\mathcal{E} \leftarrow \mathcal{C}} = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \] --- #### **Step 2: Find \([S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}}\)** - The matrix representation of the composite \(S \circ T\) with respect to basis \(\mathcal{B}\) (domain) and \(\mathcal{E}\) (codomain) is: \[ [S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}} = [S]_{\mathcal{E} \leftarrow \mathcal{C}} \times [T]_{\mathcal{C} \leftarrow \mathcal{B}} \] - Perform the matrix multiplication: \[ \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \times \begin{bmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} \] Calculations: - First row: \[ (1)(0) + (-1)(1) + (1)(1) = 0 - 1 + 1 = 0 \] \[ (1)(0) + (-1)(0) + (1)(0) = 0 + 0 + 0 = 0 \] \[ (1)(1) + (-1)(0) + (1)(0) = 1 + 0 + 0 = 1 \] \[ (1)(0) + (-1)(1) + (1)(0) = 0 - 1 + 0 = -1 \] - Second row: \[ (1)(0) + (0)(1) + (0)(1) = 0 + 0 + 0 = 0 \] \[ (1)(0) + (0)(0) + (0)(0) = 0 + 0 + 0 = 0 \] \[ (1)(1) + (0)(0) + (0)(0) = 1 + 0 + 0 = 1 \] \[ (1)(0) + (0)(1) + (0)(0) = 0 + 0 + 0 = 0 \] - Third row: \[ (1)(0) + (1)(1) + (1)(1) = 0 + 1 + 1 = 2 \] \[ (1)(0) + (1)(0) + (1)(0) = 0 + 0 + 0 = 0 \] \[ (1)(1) + (1)(0) + (1)(0) = 1 + 0 + 0 = 1 \] \[ (1)(0) + (1)(1) + (1)(0) = 0 + 1 + 0 = 1 \] --- ### **Final matrix:** \[ [S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}} = \boxed{ \begin{bmatrix} 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 \\ 2 & 0 & 1 & 1 \end{bmatrix} } \] --- ### **Explanation:** - The columns of this matrix give the images of the basis vectors in \(\mathcal{B}\) under \(S \circ T\), expressed in the standard basis \(\mathcal{E}\). **Final answer for (b):** \[ \boxed{ [S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}} = \begin{bmatrix} 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 \\ 2 & 0 & 1 & 1 \end{bmatrix} } \] --- Would you like to proceed with part (c)?

Let's proceed with part (c): --- ### **(c) Given \(u = [A]\) (the coordinate vector of \(A\) in basis \(\mathcal{B}\)), and using part (b), find \([S \circ T](u)_{\mathcal{E}}\).** --- #### **Step 1: Recall what the problem states** - \(u\) is the coordinate vector of some matrix \(A\) in the basis \(\mathcal{B}\). - We want to find the image of \(A\) under \(S \circ T\), expressed in the basis \(\mathcal{E}\). - The relation is: \[ [S \circ T](u)_{\mathcal{E}} = [S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}} \times u \] - From part (b), we have the matrix: \[ [S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}} = \begin{bmatrix} 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 \\ 2 & 0 & 1 & 1 \end{bmatrix} \] --- #### **Step 2: Compute \([S \circ T](u)_{\mathcal{E}}\)** - For a specific \(u\), the result is obtained by multiplying the matrix by \(u\): \[ [S \circ T](u)_{\mathcal{E}} = \begin{bmatrix} 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 \\ 2 & 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{bmatrix} = \begin{bmatrix} 0 \cdot u_1 + 0 \cdot u_2 + 1 \cdot u_3 + (-1) \cdot u_4 \\ 0 \cdot u_1 + 0 \cdot u_2 + 1 \cdot u_3 + 0 \cdot u_4 \\ 2 \cdot u_1 + 0 \cdot u_2 + 1 \cdot u_3 + 1 \cdot u_4 \end{bmatrix} \] which simplifies to: \[ \boxed{ \begin{bmatrix} u_3 - u_4 \\ u_3 \\ 2 u_1 + u_3 + u_4 \end{bmatrix} } \] **Explanation:** The expression computes the image of the vector \(u\) under the matrix representing \(S \circ T\), giving the coordinates of the resulting vector in basis \(\mathcal{E}\). --- ### **Final answer to (c):** \[ \boxed{ [S \circ T](u)_{\mathcal{E}} = \begin{bmatrix} u_3 - u_4 \\ u_3 \\ 2 u_1 + u_3 + u_4 \end{bmatrix} } \] --- Would you like to continue with part (d)?

Let's proceed with part (d): --- ### **(d) Is \(T\) an isomorphism? If yes, find \([T]^{-1}_{\mathcal{C} \leftarrow \mathcal{B}}\), and if not, explain.** --- #### **Step 1: Determine whether \(T\) is an isomorphism** - To be an isomorphism between finite-dimensional vector spaces, \(T\) must be **bijective**, i.e., its matrix representation \([T]_{\mathcal{C} \leftarrow \mathcal{B}}\) must be invertible. - Recall from part (a): \[ [T]_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} \] - **Dimensions:** - Domain: \(M_2\) (dimension 4) - Codomain: \(P_2\) (dimension 3) Since the matrix is \(3 \times 4\), the linear transformation cannot be invertible (since invertible matrices are square). **Conclusion:** \(T\) is **not** an isomorphism because the matrix is not square and cannot have an inverse in the usual sense. --- #### **Step 2: Summary** - Because the matrix is not square, \(T\) is **not** an isomorphism. **Final answer:** \[ \boxed{ \text{No, } T \text{ is not an isomorphism because } [T]_{\mathcal{C} \leftarrow \mathcal{B}} \text{ is not square and thus not invertible.} } \] --- ### **(e) Is \(S\) an isomorphism? If yes, find \([S]^{-1}_{\mathcal{C} \leftarrow \mathcal{E}}\).** --- #### **Step 1: Check if \(S\) is an isomorphism** - From part (a): \[ [S]_{\mathcal{E} \leftarrow \mathcal{C}} = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \] - Since this is a \(3 \times 3\) square matrix, we can check if it is invertible by calculating its determinant. --- #### **Step 2: Compute the determinant** \[ \det \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \] Calculate: \[ \det = 1 \times \det \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} - (-1) \times \det \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} + 1 \times \det \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \] Compute minors: - \( \det \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} = 0 \times 1 - 0 \times 1 = 0 \) - \( \det \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = 1 \times 1 - 0 \times 1 = 1 \) Plug in: \[ \det = 1 \times 0 - (-1) \times 1 + 1 \times 1 = 0 + 1 + 1 = 2 \] Since the determinant is **non-zero (2)**, \(S\) **is invertible**. --- #### **Step 3: Find \([S]^{-1}_{\mathcal{C} \leftarrow \mathcal{E}}\)** - The inverse of a matrix \(A\) is \(\frac{1}{\det} \text{adj}(A)\). - Compute the adjugate matrix: Let \(A = [a_{ij}]\), then: \[ A = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \] Calculate cofactors: - \(C_{11} = \det \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} = 0\) - \(C_{12} = -\det \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = -1\) - \(C_{13} = \det \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = 1\) - \(C_{21} = -\det \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix} = -((-1)(1) - 1 \times 1) = -(-1 -1) = -(-2) = 2\) - \(C_{22} = \det \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = 1 \times 1 - 1 \times 1 = 0\) - \(C_{23} = -\det \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} = - (1 \times 1 - (-1) \times 1) = - (1 + 1) = -2\) - \(C_{31} = \det \begin{bmatrix} -1 & 1 \\ 0 & 0 \end{bmatrix} = (-1)(0) - 1 \times 0 = 0\) - \(C_{32} = -\det \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} = - (1 \times 0 - 1 \times 1) = - (0 - 1) = 1\) - \(C_{33} = \det \begin{bmatrix} 1 & -1 \\ 1 & 0 \end{bmatrix} = 1 \times 0 - (-1) \times 1 = 0 + 1 = 1\) Construct the cofactor matrix: \[ \text{Cofactor} = \begin{bmatrix} 0 & -1 & 1 \\ 2 & 0 & -2 \\ 0 & 1 & 1 \end{bmatrix} \] Transpose to get adjugate: \[ \text{adj}(A) = \begin{bmatrix} 0 & 2 & 0 \\ -1 & 0 & 1 \\ 1 & -2 & 1 \end{bmatrix} \] Finally, divide by \(\det=2\): \[ A^{-1} = \frac{1}{2} \begin{bmatrix} 0 & 2 & 0 \\ -1 & 0 & 1 \\ 1 & -2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & -1 & \frac{1}{2} \end{bmatrix} \] --- ### **Final answer:** \[ \boxed{ [S]^{-1}_{\mathcal{C} \leftarrow \mathcal{E}} = \begin{bmatrix} 0 & 1 & 0 \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & -1 & \frac{1}{2} \end{bmatrix} } \] --- ### **Summary:** - \(T\) is **not** an isomorphism. - \(S\) **is** an isomorphism, and its inverse matrix is as above. --- Would you like any further explanation or assistance?

Let's synthesize the solutions step by step, providing clear explanations: --- ### **(a) Find \([T]_{\mathcal{C} \leftarrow \mathcal{B}}\) and \([S]_{\mathcal{E} \leftarrow \mathcal{C}}\)** **Step 1:** Determine \([T]_{\mathcal{C} \leftarrow \mathcal{B}}\) - **Basis \(\mathcal{B}\):** standard basis of \(M_2\): \[ E_1 = \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}, \quad E_2 = \begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}, \quad E_3 = \begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}, \quad E_4 = \begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix} \] - **Basis \(\mathcal{C}\):** standard basis of \(P_2\): \(\{1, x, x^2\}\). Apply \(T\) to each basis vector: - \(T(E_1) = x^2 + x\) → coefficients: \([0, 1, 1]\) - \(T(E_2) = 0\) → \([0, 0, 0]\) - \(T(E_3) = 1\) → \([1, 0, 0]\) - \(T(E_4) = x\) → \([0, 1, 0]\) **Matrix \([T]\):** \[ \boxed{ [T]_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} } \] --- **Step 2:** Find \([S]_{\mathcal{E} \leftarrow \mathcal{C}}\) - \(\mathcal{E}\): basis of \(\mathbb{R}^3\) Apply \(S\) to basis: - \(f(x) = 1\): \(S(1) = (1, 1, 1)\) - \(f(x) = x\): \(S(x) = (-1, 0, 1)\) - \(f(x) = x^2\): \(S(x^2) = (1, 0, 1)\) Coefficients form columns: \[ \boxed{ [S]_{\mathcal{E} \leftarrow \mathcal{C}} = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} } \] --- ### **(b) Find \([S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}}\)** - The matrix of the composition: \[ [S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}} = [S]_{\mathcal{E} \leftarrow \mathcal{C}} \times [T]_{\mathcal{C} \leftarrow \mathcal{B}} \] Multiplying: \[ \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} = \boxed{ \begin{bmatrix} 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 \\ 2 & 0 & 1 & 1 \end{bmatrix} } \] --- ### **(c) For \(u = [A]\), find \([S \circ T](u)_{\mathcal{E}}\)** - The image: \[ [S \circ T](u)_{\mathcal{E}} = [S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}} \times u \] which yields: \[ \boxed{ \begin{bmatrix} u_3 - u_4 \\ u_3 \\ 2 u_1 + u_3 + u_4 \end{bmatrix} } \] This expresses the transformed coordinates in \(\mathcal{E}\). --- ### **(d) Is \(T\) an isomorphism?** - The matrix \([T]_{\mathcal{C} \leftarrow \mathcal{B}}\) is \(3 \times 4\), so **not square**, hence **not invertible**. - **Conclusion:** \(T\) is **not** an isomorphism. --- ### **(e) Is \(S\) an isomorphism?** - The matrix: \[ \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \] - Determinant: \[ \det = 2 \neq 0 \] - **Therefore,** \(S\) **is invertible**. - Its inverse: \[ [S]^{-1}_{\mathcal{C} \leftarrow \mathcal{E}} = \begin{bmatrix} 0 & 1 & 0 \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & -1 & \frac{1}{2} \end{bmatrix} \] --- **Final notes:** - \(T\) is **not** an isomorphism. - \(S\) **is** an isomorphism, with the inverse matrix as above. Please let me know if you'd like further clarification!

Let's summarize and conclude: **(a)** The matrix representation of \(T\) with respect to bases \(\mathcal{B}\) (matrices) and \(\mathcal{C}\) (polynomials): \[ [T]_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \end{bmatrix} \] The matrix of \(S\) with respect to bases \(\mathcal{C}\) (polynomials) and \(\mathcal{E}\) (\(\mathbb{R}^3\)): \[ [S]_{\mathcal{E} \leftarrow \mathcal{C}} = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \] --- **(b)** The matrix of \(S \circ T\) from \(\mathcal{B}\) to \(\mathcal{E}\): \[ [S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}} = [S]_{\mathcal{E} \leftarrow \mathcal{C}} \times [T]_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} 0 & 0 & 1 & -1 \\ 0 & 0 & 1 & 0 \\ 2 & 0 & 1 & 1 \end{bmatrix} \] --- **(c)** For a vector \(u = [A]\), the image under \(S \circ T\) is: \[ [S \circ T](u)_{\mathcal{E}} = \begin{bmatrix} u_3 - u_4 \\ u_3 \\ 2 u_1 + u_3 + u_4 \end{bmatrix} \] --- **(d)** Since \([T]_{\mathcal{C} \leftarrow \mathcal{B}}\) is a \(3 \times 4\) matrix (not square), **\(T\) is not an isomorphism**. --- **(e)** The matrix \([S]_{\mathcal{E} \leftarrow \mathcal{C}}\) is invertible (determinant = 2), so **\(S\) is an isomorphism**. Its inverse is: \[ [S]^{-1}_{\mathcal{C} \leftarrow \mathcal{E}} = \begin{bmatrix} 0 & 1 & 0 \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & -1 & \frac{1}{2} \end{bmatrix} \] --- If you need any further assistance, feel free to ask!

Let's analyze and solve each part step by step with detailed explanations. --- ### **(a) Find \([T]_{\mathcal{C} \leftarrow \mathcal{B}}\) and \([S]_{\mathcal{E} \leftarrow \mathcal{C}}\).** **Step 1: Find \([T]_{\mathcal{C} \leftarrow \mathcal{B}}\).** - **Basis \(\mathcal{B}\) for \(M_2\):** matrices \(E_1, E_2, E_3, E_4\): \[ E_1 = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}, \quad E_2 = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}, \quad E_3 = \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}, \quad E_4 = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} \] - **Basis \(\mathcal{C}\):** polynomials \(1, x, x^2\). Applying \(T\) to these basis matrices: 1. \(T(E_1) = a z^{2} + (a + d) z + c\) For \(E_1\): \(a=1, d=0, c=0\), so: \[ T(E_1) = 1 \cdot z^2 + (1 + 0) z + 0 = z^2 + z \] Coefficients: \([0, 1, 1]\) (constant, x, \(x^2\)). 2. \(E_2\): \(a=0, d=0, c=0\), so: \[ T(E_2) = 0 \cdot z^2 + (0 + 0) z + 0= 0 \] Coefficients: \([0, 0, 0]\). 3. \(E_3\): \(a=0, d=1, c=0\), so: \[ T(E_3) = 0 \cdot z^2 + (0 + 1) z + 0= z \] Coefficients: \([0, 1, 0]\). 4. \(E_4\): \(a=0, d=0, c=1\), so: \[ T(E_4) = 0 \cdot z^2 + (0+0) z + 1=1 \] Coefficients: \([1, 0, 0]\). **Construct the matrix \([T]_{\mathcal{C} \leftarrow \mathcal{B}}\):** Each column corresponds to the image of a basis matrix: \[ [T]_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{bmatrix} \text{coeffs of } T(E_1) & T(E_2) & T(E_3) & T(E_4) \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \] --- **Step 2: Find \([S]_{\mathcal{E} \leftarrow \mathcal{C}}\).** - \(\mathcal{E}\): standard basis of \(\mathbb{R}^3\) Apply \(S\) to basis polynomials: 1. \(f(x) = 1\): \[ S(1) = (1, 1, 1) \] 2. \(f(x) = x\): \[ S(x) = (-1, 0, 1) \] 3. \(f(x) = x^2\): \[ S(x^2) = (1, 0, 1) \] Construct the matrix with these as columns: \[ [S]_{\mathcal{E} \leftarrow \mathcal{C}} = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \] --- ### **(b) Find \([S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}}\).** This is the composition matrix: \[ [S \circ T] = [S] \times [T] \] Compute: \[ \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \] Multiplying: - First row: \[ 1*0 + (-1)*1 + 1*1 = 0 - 1 + 1= 0 \] \[ 1*0 + (-1)*0 + 1*0= 0 \] \[ 1*0 + (-1)*1 + 1*0= 0 -1 + 0= -1 \] \[ 1*1 + (-1)*0 + 1*0= 1 + 0 + 0= 1 \] - Second row: \[ 1*0 + 0*1 + 0*1= 0 \] \[ 1*0 + 0*0 + 0*0= 0 \] \[ 1*0 + 0*1 + 0*0= 0 \] \[ 1*1 + 0*0 + 0*0= 1 \] - Third row: \[ 1*0 + 1*1 + 1*1= 0 + 1 + 1= 2 \] \[ 1*0 + 1*0 + 1*0= 0 \] \[ 1*0 + 1*1 + 1*0= 0 + 1 + 0=1 \] \[ 1*1 + 1*0 + 1*0= 1 \] **Resulting matrix:** \[ [S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}} = \begin{bmatrix} 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 1 \\ 2 & 0 & 1 & 1 \end{bmatrix} \] --- ### **(c) Given \(u = [A]\), find \([S \circ T](u)_{\mathcal{E}}\).** This is: \[ [S \circ T](u) = [S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}} \times u \] which yields: \[ \boxed{ \begin{bmatrix} - u_3 + u_4 \\ u_4 \\ 2 u_1 + u_3 + u_4 \end{bmatrix} } \] (Note: sign correction based on previous calculations: since the first entry is \(0 \times u_1 + 0 \times u_2 + (-1) \times u_3 + 1 \times u_4 = -u_3 + u_4\).) --- ### **(d) Is \(T\) an isomorphism?** - The matrix \([T]\) is \(3 \times 4\), so not square, hence **not invertible**. - **Answer:** No, \(T\) is **not** an isomorphism. --- ### **(e) Is \(S\) an isomorphism?** - The matrix \([S]\) is \(3 \times 3\). Calculate determinant: \[ \det = 1*(0*1 - 0*1) - (-1)*(1*1 - 0*1) + 1*(1*1 - 0*1) = 0 - (-1)*1 + 1*1= 0 + 1 + 1= 2 \neq 0 \] - Since determinant ≠ 0, \(S\) **is** an isomorphism. - Its inverse: \[ [S]^{-1}_{\mathcal{C} \leftarrow \mathcal{E}} = \frac{1}{2} \times \text{adjugate of } [S] \] Calculating the inverse yields: \[ \boxed{ [S]^{-1} = \begin{bmatrix} 0 & 1 & 0 \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & -1 & \frac{1}{2} \end{bmatrix} } \] --- **Final Summary:** - \([T]_{\mathcal{C} \leftarrow \mathcal{B}}\): \[ \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \] - \([S]_{\mathcal{E} \leftarrow \mathcal{C}}\): \[ \begin{bmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix} \] - \([S \circ T]_{\mathcal{E} \leftarrow \mathcal{B}}\): \[ \begin{bmatrix} 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 1 \\ 2 & 0 & 1 & 1 \end{bmatrix} \] - \([S \circ T](u)\): \[ \begin{bmatrix} -u_3 + u_4 \\ u_4 \\ 2 u_1 + u_3 + u_4 \end{bmatrix} \] - \(T\) **not** an isomorphism (matrix not square). - \(S\) **is** an isomorphism; inverse as above. Let me know if further clarification is needed!