Problem 1. Let V be the subspace of P3(R) that consists of all polynomials p(t) of degree less than or equal to 3, such that p(1) = p(2). Construct a basis of V and prove that it is indeed a basis. Claude: To solve this problem, we need to find a set of polynomials that span the subspace V and ensure that they are linearly independent. Step 1: Construct a set of polynomials that satisfy the condition p(1) = p(2). Let p(t) = a + bt + ct^2 + dt^3, where a, b, c, and d are real numbers. Substituting t = 1, we get: p(1) = a + b + c + d Substituting t = 2, we get: p(2) = a + 2b + 4c + 8d Since p(1) = p(2), we have: a + b + c + d = a + 2b + 4c + 8d b + 3c + 7d = 0 This equation represents a constraint on the coefficients b, c, and d. Step 2: Find a basis for V by selecting a set of linearly independent polynomials that satisfy the constraint. One possible basis for V is: B = {1, t - 1, (t - 1)^2, (t - 1)^3} We can verify that these polynomials satisfy the condition p(1) = p(2): 1(1) = 1(2) = 1 (t - 1)(1) = (t - 1)(2) = 0 (t - 1)^2(1) = (t - 1)^2(2) = 0 (t - 1)^3(1) = (t - 1)^3(2) = 0 Step 3: Prove that B is a basis for V. To prove that B is a basis for V, we need to show that: 1. Every polynomial p(t) in V can be expressed as a linear combination of the elements in B. 2. The elements in B are linearly independent. Proof of (1): Let p(t) be a polynomial in V, with p(t) = a + bt + ct^2 + dt^3. Since p(t) satisfies the condition p(1) = p(2), we have b + 3c + 7d = 0. We can express p(t) as a linear combination of the elements in B as follows: p(t) = a + b(t - 1) + c(t - 1)^2 + d(t - 1)^3 = a + b(t - 1) + c(t - 1)^2 + (-b/7 - 3c/7)(t - 1)^3 Therefore, every polynomial p(t) in V can be expressed as a linear combination of the elements in B. Proof of (2): Assume that there exist scalars α, β, γ, δ such that α + β(t - 1) + γ(t - 1)^2 + δ(t - 1)^3 = 0 for all t. Substituting t = 1, we get: α = 0 Substituting t = 2, we get: β = 0 Substituting t = 3, we get: γ = 0 Substituting t = 4, we get: δ = 0 Therefore, the only way for the linear combination to be equal to the zero polynomial is if all the scalars are zero, which implies that the elements in B are linearly independent. Hence, we have proved that B = {1, t - 1, (t - 1)^2, (t - 1)^3} is a basis for the subspace V. --- student note: substituting t=2 yields a + b+ c+d =0 Instead, multiply out: a + bt -b + ct^2 -2ct +c +dt^3 -3dt^2 +3dt -d = a -b +c -d + t(b -2c +3d ) + t^2(c -3d) + t^3(d) so d must =0, then c=0 then b = 0, then a = 0. so all coefficients must = 0 for polynomial to equal 0. --- use same basis as Frenkel proof: Yes, the set B = {1, t^2 - 3t, t^3 - 7t} is another valid basis for the subspace V of P₃(R) that consists of all polynomials p(t) of degree less than or equal to 3, such that p(1) = p(2). To verify that B is a basis for V, we need to check the following conditions: 1. Every element of B belongs to V. 2. The elements in B are linearly independent. 3. The set B spans V. Condition 1: Every element of B belongs to V. Substituting t = 1 and t = 2 in each element of B, we can verify that all polynomials satisfy the condition p(1) = p(2). 1(1) = 1(2) = 1 (t^2 - 3t)(1) = (t^2 - 3t)(2) = 1 - 3 = -2 (t^3 - 7t)(1) = (t^3 - 7t)(2) = 1 - 7 = -6 Condition 2: The elements in B are linearly independent. Suppose there exist scalars α, β, γ in R such that α + β(t^2 - 3t) + γ(t^3 - 7t) = 0 (the zero polynomial). Substituting t = 0, we get α = 0. Substituting t = 1, we get β = 0. Substituting t = 2, we get γ = 0. Therefore, the only way for the linear combination to be equal to the zero polynomial is if all the scalars are zero, which implies that the elements in B are linearly independent. Condition 3: The set B spans V. Let p(t) = a + bt + ct^2 + dt^3 be a polynomial in V, where a, b, c, and d are real numbers. Since p(t) satisfies the condition p(1) = p(2), we have a + b + c + d = a + 2b + 4c + 8d, which implies b + 3c + 7d = 0. We can express p(t) as a linear combination of the elements in B as follows: p(t) = a + bt + ct^2 + dt^3 = a + b(t^2 - 3t + 3t - 1) + c(t^2 - 3t) + d(t^3 - 7t) = a + (b + 3c)(t^2 - 3t) + (3b - d)(t^3 - 7t) Therefore, every polynomial p(t) in V can be expressed as a linear combination of the elements in B. Since all three conditions are satisfied, the set B = {1, t^2 - 3t, t^3 - 7t} forms a valid basis for the subspace V of P₃(R) that consists of all polynomials p(t) of degree less than or equal to 3, such that p(1) = p(2).