Vector space P4 = set of all polynomials of degree Solution We start with a heur

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Vector space P4 = set of all polynomials of degree

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We start with a heuristic; to say that a polynomial x(t) is in Fmeans that if you flip the variable x to 1-x, the polynomialremains invariant. Thus whenever you see x, the term 1-x also (sortof appears). So for instance, the following polynomials are inF: f(t)=1; g(t)= t2 +(1-t)2 h(t)=t4+(1-t)4 Note that these are polynomials of degree 0,2,4 respectively, sothey are linearly independent. We now make another note: If X(t) = at+bt3 and X(t) isin F, then a=b=0. Indeed, if X(t) is in F, then at+bt3 =a(1-t)+b(1-t)3 => (equating coeffsof t3 on both sides) b=-b => b=0, so at=(1-t)a=>a=-a (equating coeff of t) => a=0 as claimed. What this just showed was the following: IF G is the vectorsubspace of all polynomials of the form at+bt3 then Gintersect F =(0) where 0 is the zero polynomial. We have seen that F has at least 3 independent vectors in it,namely f(t),g(t),h(t) and that it is disjoint with a vector spaceof dimension 2. But F,G are both vector subspaces of P4,a vector space of dimension 5, F intersect G =(0) and G hasdimension 2. Since for any two subspaces, dim F+ dim G= dim + dim we must have dim F +2 dim F

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