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