Let P = {f elementof F |df(x)/dx = 2x}. Determine whether or not P is a subspace

Question

Let P = {f elementof F |df(x)/dx = 2x}. Determine whether or not P is a subspace of F. Let Q = { f elementof F |df(x)/dx = 2f(x)}. Determine whether or not Q is a subspace of F. (c) Let u and v be linearly independent vectors in a vector space V. Let S = Span{u, v}. Prove that {u + v, u - v} is a basis for S. (d) Suppose A is an n times n matrix, and that lambda_1 and lambda_2 are eigenvalues of A such that lambda_1 notequalto lambda_2. Let v_1 be an eigenvector corresponding to lambda_1 and lambda_2 be an eigenvector corresponding to lambda_2. Prove that v_1 and v_2 are linearly independent.

Explanation / Answer

(c) given u and v be linearly independent vectors in a vector space V

TO Prove that {u+v,u-v} is a basis for S i.e Span S is closure under addition and scalar multiplication

span(S) is a subspace of V, we only need to show that span(S) is closed under addition and scalar multiplication. So, let u, v be two elements in span(S). We can write u = c1v1 + c2v2 + · · · + ckvk and v = d1v1 + d2v2 + · · · + dkvk where c1, c2, . . . , ck, d1, d2, . . . , dk are scalars. It follows u + v = (c1 + d1)v1 + (c2 + d2)v2 + · · · + (ck + dk)vk and for a scalar c, we have cu = (cc1)v1 + (cc2)v2 + · · · + (cck)vk. So, both u + v and cu are in span(S), because the are linear combination of elements in S. So, span(S) is closed under addition and scalar multiplication, hence a subspace of V. hence {u+v} is a basis of S. SIMILARLY u-v is nothing but replacing +v with -v. therefore by the theorem we can prove that{u+v,u-v} are in base S.

(d)the above question is a theorem and the proof to it is given by

Suppose c1v1 + c2v2 = 0, where one of the coefficients, say c1 is nonzero. Then v1 = v2, for some 6= 0. (If = 0, then v1 = 0 and v1 by definition is not an eigenvector.) Multiplying both sides on the left by A gives Av1 = 1v1 = Av2 = 2v2. On the other hand, multiplying the same equation by 1 and then subtracting the two equations gives 0 = (2 1)v2 which is impossible, since neither nor (1 2) = 0, and v2 6= 0. It follows that if A2×2 has two distinct real eigenvalues, then it has two linearly independent eigenvectors and can be diagonalized. In a similar way A matrix with n*n has n distinct eigen vectors and they are linearly independent.

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