MATH 240 Lecture 19: A4sol
25 Jan 2019
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I"ll give two solutions, the rst one works with the polynomials and the second one uses co-ordinate vectors in r4. The standard basis for p3 is {1, t, t2, t3} so dim p3 = 4. If the four polynomials are linearly independent, then dim s = 4. So let"s consider solving c1(1 + t) + c2(1 + t2) + c3(t2 t) + c4t3 = 0 for c1, c2, c3, c4. To simplify this notice that there is only one polynomial with t3 in it so obviously t3 is linearly independent of the others. So if there is a linear dependency, there will be a non-zero solution to c1(1 + t) + c2(1 + t2) + c3(t2 t) = 0. Regrouping terms we have (c1 + c2)1 + (c1 c3)t + (c2 + c3)t2 = 0. (1) Now, key argument: the standard basis for p2 is {1, t, t2} which are linearly independent.
