Second response is correct: To find the basis of P₁(R) for which {f₁, f₂} is the dual basis, we need to construct a basis {p₁(t), p₂(t)} such that f₁(p₂(t)) = 0, f₂(p₁(t)) = 0, f₁(p₁(t)) = 1, and f₂(p₂(t)) = 1. Given: - P₁(R) is the vector space of polynomials of degree at most 1 over the real numbers R. - f₁[p(t)] = p(1) - f₂[p(t)] = p(2) Step 1: Find a polynomial p₁(t) such that f₁(p₁(t)) = 1 and f₂(p₁(t)) = 0. Let p₁(t) = t - 2. Then, f₁(p₁(t)) = p₁(1) = 1 - 2 = -1 f₂(p₁(t)) = p₁(2) = 2 - 2 = 0 To make f₁(p₁(t)) = 1, we need to divide p₁(t) by -1: p₁(t) = (t - 2) / (-1) = 2 - t Step 2: Find a polynomial p₂(t) such that f₁(p₂(t)) = 0 and f₂(p₂(t)) = 1. Let p₂(t) = t - 1. Then, f₁(p₂(t)) = p₂(1) = 1 - 1 = 0 f₂(p₂(t)) = p₂(2) = 2 - 1 = 1 Step 3: Verify that {p₁(t), p₂(t)} is a basis for P₁(R). To show that {p₁(t), p₂(t)} is a basis for P₁(R), we need to prove that: 1. p₁(t) and p₂(t) are linearly independent. 2. Every polynomial p(t) in P₁(R) can be expressed as a linear combination of p₁(t) and p₂(t). Proof of (1): Assume α(2 - t) + β(t - 1) = 0 for some scalars α and β. Substituting t = 1, we get α + 0 = 0, which implies α = 0. Substituting t = 2, we get -2β + β = 0, which implies β = 0. Since the only scalars that satisfy the equation are α = 0 and β = 0, p₁(t) and p₂(t) are linearly independent. Proof of (2): Let p(t) = a + bt be a polynomial in P₁(R). We can express p(t) as a linear combination of p₁(t) and p₂(t) as follows: p(t) = a + bt = a(2 - t) + (b - a)(t - 1) = a(2 - t) + (b - a)p₂(t) Therefore, {p₁(t), p₂(t)} = {2 - t, t - 1} is a basis for P₁(R). Step 4: Verify that {f₁, f₂} is the dual basis of {p₁(t), p₂(t)}. By construction, we have: f₁(p₁(t)) = f₁(2 - t) = 1 f₂(p₂(t)) = f₂(t - 1) = 1 f₁(p₂(t)) = f₁(t - 1) = 0 f₂(p₁(t)) = f₂(2 - t) = 0 Therefore, {f₁, f₂} is indeed the dual basis of {p₁(t), p₂(t)} = {2 - t, t - 1}. --- To find the basis of P₁(R) for which {f₁, f₂} is the dual basis, we need to construct a basis {p₁(t), p₂(t)} such that f₁(p₂(t)) = 0, f₂(p₁(t)) = 0, f₁(p₁(t)) = 1, and f₂(p₂(t)) = 1. Given: - P₁(R) is the vector space of polynomials of degree at most 1 over the real field R. - f₁[p(t)] = p(1) and f₂[p(t)] = p(2) are linear functionals on P₁(R). - The basis {p₁(t), p₂(t)} we want to find should satisfy the dual basis conditions with respect to {f₁, f₂}. Step 1: Choose a suitable polynomial p₁(t) such that f₁(p₁(t)) = 1 and f₂(p₁(t)) = 0. Let p₁(t) = t - 1. Then, f₁(p₁(t)) = p₁(1) = (1 - 1) = 0 and f₂(p₁(t)) = p₁(2) = (2 - 1) = 1 Step 2: Choose a suitable polynomial p₂(t) such that f₁(p₂(t)) = 0 and f₂(p₂(t)) = 1. Let p₂(t) = t - 2. Then, f₁(p₂(t)) = p₂(1) = (1 - 2) = -1 and f₂(p₂(t)) = p₂(2) = (2 - 2) = 0 Step 3: Verify that {p₁(t), p₂(t)} is a basis for P₁(R). To show that {p₁(t), p₂(t)} is a basis, we need to prove that it is linearly independent and spans P₁(R). Proof of linear independence: Assume α(t - 1) + β(t - 2) = 0 for some scalars α and β. Substituting t = 1, we get α = 0. Substituting t = 2, we get β = 0. Therefore, α = β = 0, which implies that {p₁(t), p₂(t)} is linearly independent. Proof of spanning P₁(R): Let p(t) = a + bt be any polynomial in P₁(R). We can express p(t) as a linear combination of p₁(t) and p₂(t) as follows: p(t) = a + bt = a + b(t - 1) + b(t - 2) = (a + b)(t - 1) + b(t - 2) Therefore, {p₁(t), p₂(t)} spans P₁(R). Step 4: Verify that {f₁, f₂} is the dual basis of {p₁(t), p₂(t)}. We need to check the dual basis conditions: f₁(p₂(t)) = f₁(t - 2) = (1 - 2) = -1 = 0 f₂(p₁(t)) = f₂(t - 1) = (2 - 1) = 1 = 0 f₁(p₁(t)) = f₁(t - 1) = (1 - 1) = 0 = 1 f₂(p₂(t)) = f₂(t - 2) = (2 - 2) = 0 = 1 Therefore, {f₁, f₂} is the dual basis of {p₁(t), p₂(t)}. Hence, the basis of P₁(R) for which {f₁, f₂} is the dual basis is {p₁(t), p₂(t)} = {t - 1, t - 2}.