Linear Elementary Algebra

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 Pi(x). Then each vector p in Pl(x) looks like, p(I) = ax + B, where o and B are any real number (including 0). Assuming that multiplication and addition are lefined as usual, use the definition from section 4.1, show that P (r) is a vector space (some of the properties have been done for you) Let p(x) = ar +b, 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(x) + 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(x) + q(x) = q(x) + p(x). (c) Show that [p(x) + q(x)] + r(x) = p(x) + [q(x) +r(x)]. (d) Let a = 0 and b = 0, then p(x) = 0 (that is, p(x) is zero for all values of x). Now for any q(x) in Pi(x), p(x) + q(x) = q(x), so there is a zero vector, which is just the zero polynomial. (e) Find a polynomial -p(r) such that p(x) + [-p(x)] = 0. (f) We consider sp(x). We get, sp(x) = s(ax + b) = (sa)x + sb. We see that sp(I) is just another polynomial of degree at most 1