# Complete Solution ## Part (a): Show that \( B \) and \( B' \) are two bases of \( \mathbb{R}^2 \) and draw those basis vectors ### Step 1: Vectors in Standard Basis Given: \[ b_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \quad b_2 = \begin{bmatrix} -1 \\ -1 \end{bmatrix} \] \[ b_1' = \begin{bmatrix} 2 \\ -2 \end{bmatrix}, \quad b_2' = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \] ### Step 2: Check Linear Independence for \( B = (b_1, b_2) \) Form matrix \( M_B = \begin{bmatrix} b_1 & b_2 \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 1 & -1 \end{bmatrix} \). Calculate the determinant: \[ \det(M_B) = (2)(-1) - (1)(-1) = -2 + 1 = -1 \neq 0 \] Hence, \( b_1 \) and \( b_2 \) are linearly independent and form a basis. ### Step 3: Check Linear Independence for \( B' = (b_1', b_2') \) Form matrix \( M_{B'} = \begin{bmatrix} b_1' & b_2' \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ -2 & 1 \end{bmatrix} \). Calculate the determinant: \[ \det(M_{B'}) = (2)(1) - (-2)(1) = 2 + 2 = 4 \neq 0 \] Hence, \( b_1' \) and \( b_2' \) are linearly independent and form a basis. ### Step 4: Drawing the Vectors (Description for drawing: Plot the vectors in \( \mathbb{R}^2 \) starting from the origin: - \( b_1 \) points to (2,1) - \( b_2 \) points to (-1,-1) - \( b_1' \) points to (2,-2) - \( b_2' \) points to (1,1) ) --- ## Part (b): Compute the matrix \( P_1 \) that performs a basis change from \( B' \) to \( B \) ### Step 1: Express \( b_1' \) and \( b_2' \) in the \( B \) basis Let \( [x\ y]^T_B \) be the coordinates in basis \( B \) such that: \[ b_1' = a_{11}b_1 + a_{21}b_2 \] \[ b_2' = a_{12}b_1 + a_{22}b_2 \] So, \[ \begin{bmatrix} 2 \\ -2 \end{bmatrix} = a_{11} \begin{bmatrix} 2 \\ 1 \end{bmatrix} + a_{21} \begin{bmatrix} -1 \\ -1 \end{bmatrix} \] \[ \begin{bmatrix} 1 \\ 1 \end{bmatrix} = a_{12} \begin{bmatrix} 2 \\ 1 \end{bmatrix} + a_{22} \begin{bmatrix} -1 \\ -1 \end{bmatrix} \] #### For \( b_1' \): Set up the system: \[ 2 = 2a_{11} - a_{21} \] \[ -2 = a_{11} - a_{21} \] Subtract the second from the first: \[ 2 - (-2) = (2a_{11} - a_{21}) - (a_{11} - a_{21}) \implies 4 = a_{11} \] Plug back: \[ -2 = 4 - a_{21} \implies a_{21} = 6 \] So, \( b_1' = 4b_1 + 6b_2 \) #### For \( b_2' \): \[ 1 = 2a_{12} - a_{22} \] \[ 1 = a_{12} - a_{22} \] Subtract: \[ 1 - 1 = (2a_{12} - a_{22}) - (a_{12} - a_{22}) \implies 0 = a_{12} \] So, \( a_{12} = 0 \) Plug back: \[ 1 = 0 - a_{22} \implies a_{22} = -1 \] So, \( b_2' = 0b_1 - 1b_2 \) ### Step 2: The Change of Basis Matrix \( P_1 \): \[ P_1 = \begin{bmatrix} 4 & 0 \\ 6 & -1 \end{bmatrix} \] This matrix transforms coordinates from \( B' \) to \( B \). --- ## Part (c): Basis for \( \mathbb{R}^3 \), and Change of Basis Matrix ### (i) Show that \( C = (c_1, c_2, c_3) \) is a basis of \( \mathbb{R}^3 \) Given: \[ c_1 = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}, \quad c_2 = \begin{bmatrix} 0 \\ -1 \\ 2 \end{bmatrix}, \quad c_3 = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \] Form matrix: \[ M_C = \begin{bmatrix} 1 & 0 & 1 \\ 2 & -1 & 0 \\ -1 & 2 & -1 \end{bmatrix} \] Calculate the determinant: \[ \det(M_C) = 1 \cdot (-1 \cdot -1 - 0 \cdot 2) - 0 \cdot (2 \cdot -1 - 0 \cdot -1) + 1 \cdot (2 \cdot 2 - (-1) \cdot -1) \] \[ = 1 \cdot (1 - 0) - 0 \cdot (-2 - 0) + 1 \cdot (4 - 1) \] \[ = 1 + 0 + 3 = 4 \neq 0 \] Therefore, \( c_1, c_2, c_3 \) are linearly independent and form a basis. --- ### (ii) Change of Basis Matrix \( P_2 \) from \( C \) to Standard Basis \( C' \) The matrix that changes from \( C \) to \( C' \) is the matrix whose columns are \( c_1, c_2, c_3 \): \[ P_2 = \begin{bmatrix} 1 & 0 & 1 \\ 2 & -1 & 0 \\ -1 & 2 & -1 \end{bmatrix} \] --- # Explanation of Each Part ## (a) Basis Verification - A set of vectors is a basis if it is linearly independent and spans the space. - For \( \mathbb{R}^2 \), two vectors are a basis if the determinant of the matrix formed by placing them as columns is nonzero. - Both \( B \) and \( B' \) satisfy this condition. ## (b) Change of Basis Matrix - To convert coordinates from basis \( B' \) to \( B \), express each \( B' \) vector as a linear combination of \( B \) vectors. - The coefficients form the columns of the change of basis matrix. ## (c) Basis for \( \mathbb{R}^3 \), and Change of Basis Matrix - Three vectors in \( \mathbb{R}^3 \) are a basis if the determinant of the matrix with them as columns is nonzero. - The change of basis matrix from \( C \) to the standard basis is simply the matrix with \( C \)'s vectors as columns. --- # Final Conclusion Statement - Both \( B \) and \( B' \) are valid bases for \( \mathbb{R}^2 \), as shown by their nonzero determinants. - The change of basis matrix from \( B' \) to \( B \) is \( P_1 = \begin{bmatrix} 4 & 0 \\ 6 & -1 \end{bmatrix} \). - The set \( C = (c_1, c_2, c_3) \) forms a basis for \( \mathbb{R}^3 \) due to nonzero determinant. - The change of basis matrix from \( C \) to the standard basis \( C' \) is \( P_2 = \begin{bmatrix} 1 & 0 & 1 \\ 2 & -1 & 0 \\ -1 & 2 & -1 \end{bmatrix} \). - All steps confirm the bases and the basis change matrices as required.