question archive Now that we have formally defined vector spaces, we can show that certain sets of polynomials are vector spaces

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Now that we have formally defined vector spaces, we can show that certain sets of polynomials are vector spaces. Consider the set of all polynomials in a of degree at most 1. We denote this space as Pl(x). Then each vector p in Pl(x) looks like, p(z) = ar + 8, where a and # are any real number (including 0). Assuming that multiplication and addition are defined as usual, use the definition from section 4.1, show that Pl(x) is a vector space (some of the properties have been done for you). Let p(x) = ar +h, q(x) = cr +d, r(x) = fx +g be polynomials in Pl(x), with a, b, c, d, f, g real numbers and s, t real numbers. (a) p(z) + q(x) = ar + b+ cr +d = (a + c)x + (b+d). We see the sum of two polynomials of order at most one is still a polynomial of order at most one. (b) Show that p(z) + q(z) = q(z) + p(z). (c) Show that [p(z) + q(x)] + r(x) = p(z) + [q(z) + r(z)]. (d) Let a = 0 and b = 0, then p(z) = 0 (that is, p(x) is zero for all values of x). Now for any q(r) in Pi(x), p(x) + q(z) = q(x), so there is a zero vector, which is just the zero polynomial. (e) Find a polynomial -p(r) such that p(z) + [-p(x)] = 0. (f) We consider sp(z). We get, sp(z) = s(ar + b) = (sa)r + sb. We see that sp(x) is just another polynomial of degree at most 1. (g) Show that s(p(z) + q(z)) = sp(x) + sq(z). (h) We consider (s + ()p(s). We have, (s + 1)(ar +b) = (so)x + ab+ (ta)x+th = s(ar + b) + far + b) = sp(x) + tp(z). as desired. (i) Show that s[tp(z)] = (st)p(z).