A vector space is a set V, together with a field F that satisfies the following

Question

A vector space is a set V, together with a field F that satisfies the following properties: u + v = v + u for all u, v V. (u + v) + w = u + (v + w) for all u, r, u V. There exists ail element a V such that v + a v for all v V. For every v V, there exists a w V such that v + w = 0. For the multiplicative identity element 1 F, it follows that 1v = v for all v V. a(u + v) = au + av and (a + b)u = au + bu for all a, b F and all u, v V. Without defining what a field is, let's just say that R is one. Therefore, for this exercise, think of F and R as interchangeable. Consider the differential equation y" + 3y' + 2y = 0. Find two linearly independent solutions y_1, y_2 to this differential equation, setting their coefficients equal to 1. Show that the set S = {a_1 y_1 + a_2 y_2, a_1 a_2 R}, when paired with the field R. is a vector space. Show that an arbitrary element a_1 y_1 + a_2 y_2 with a_1, a_2 R is a solution of the differential equation y" + 3y' + 2y = 0. To show that S is a vector space, simply work through the six properties of a vector space. For several of the properties, the process of checking might seem trivial at first, but challenge yourself to give as much detail as possible as to why a property holds. Since this may be the student's first exposure to abstract mathematics, an example Is provided below where we check that the Additive Inverse property holds for the set of solutions of the differential equation y" - 2y + y = 0. We wish to prove that the set T = {a_1 e^x + a_2 x e^x : a_1, a_2 R}, together with the field R, satisfies the Additive Inverse property of a vector space. Let a_1 e^x + a_2 x e^x be an arbitrary element of V. Since a_1, a_2, are real numbers, there are additive inverses -a_1, -a_2 R. Hence, the element -a_1 e^x - a^2 xe^x is in V (fitting the description of elements in the set V). Then as desired.

Explanation / Answer

YOur proof is absolutely right and every one uses the same method to prove

a1ex+a2ex has additive inverse as -(a1ex+a2ex)

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