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Chapter 2: Problem 3

For each of the following vector spaces \(V\) and bases \(\beta\), find explicit formulas for vectors of the dual basis \(\beta^{*}\) for \(\mathrm{V}^{*}\), as in Example 4 . (a) \(\mathrm{V}=\mathrm{R}^{3} ; \beta=\\{(1,0,1),(1,2,1),(0,0,1)\\}\) (b) \(\mathrm{V}=\mathrm{P}_{2}(R) ; \beta=\left\\{1, x, x^{2}\right\\}\)

Short Answer

Expert verified
(a) The dual basis for V* when V = R^3 and 尾={(1,0,1),(1,2,1),(0,0,1)} is: \(\beta^* = \\{(1, 0, -1),(-1/2, 1/2, 1/2),(0, 0, 1)\\}\) (b) The dual basis for V* when V = P_2(R) and 尾={1, x, x^2} is: \( \beta^* = \\{ f_1, f_2, f_3 \\} \) where \( f_1, f_2, f_3 \) are the linear functionals that satisfy the equations: \(f_1(1) = 1, \, f_1(x) = 0, \, f_1(x^2) = 0 \\ f_2(1) = 0, \, f_2(x) = 1, \, f_2(x^2) = 0 \\ f_3(1) = 0, \, f_3(x) = 0, \, f_3(x^2) = 1\)

Step by step solution

01

Find the matrix of the basis

Form a matrix B with 尾 as columns: \(B = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 1 & 1 & 1 \\ \end{bmatrix}\)
02

Compute the inverse of the matrix

Find the inverse of matrix B, called B_inv: \(B_{inv} = \begin{bmatrix} 1 & -1/2 & 0 \\ 0 & 1/2 & 0 \\ -1 & 1/2 & 1 \\ \end{bmatrix}\)
03

Find the dual basis

For each row in the inverse matrix B_inv, form the dual basis vector: \(\beta^* = \\{(1, 0, -1),(-1/2, 1/2, 1/2),(0, 0, 1)\\}\) So the dual basis for V* when V = R^3 and 尾={(1,0,1),(1,2,1),(0,0,1)} is: \(\beta^* = \\{(1, 0, -1),(-1/2, 1/2, 1/2),(0, 0, 1)\\}\) (b) Find explicit formulas for vectors of the dual basis when V = P_2(R) and 尾={1, x, x^2}
04

Define the basis vectors as polynomials

Basis vectors are given as: \(p_1(x) = 1, p_2(x) = x, p_3(x) = x^2\)
05

Define the dual basis vectors as linear functionals

Define the dual basis vectors f_1, f_2, f_3 as linear functionals such that: \(f_i(p_j(x)) = \delta_{ij}; \, \text{where } 1 \leq i, j \leq 3 \\ \text{and } \delta_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}\)
06

Find the explicit formulas for the dual basis vectors

Solve for f_i: \(f_1(p_1(x)) = f_1(1) = 1 \\ f_1(p_2(x)) = f_1(x) = 0 \\ f_1(p_3(x)) = f_1(x^2) = 0\) \(f_2(p_1(x)) = f_2(1) = 0 \\ f_2(p_2(x)) = f_2(x) = 1 \\ f_2(p_3(x)) = f_2(x^2) = 0\) \(f_3(p_1(x)) = f_3(1) = 0 \\ f_3(p_2(x)) = f_3(x) = 0 \\ f_3(p_3(x)) = f_3(x^2) = 1\) So the dual basis for V* when V = P_2(R) and 尾={1, x, x^2} is: \( \beta^* = \\{ f_1, f_2, f_3 \\} \) where \( f_1, f_2, f_3 \) are the linear functionals that satisfy the above equations.

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