Consider the set V of all polynomials with real coefficients

Consider the set V of all polynomials with real coefficients...

Consider the set V of all polynomials with real coefficients - this set is a vector space over R if operations of addition and scalar multiplication are defined in the \"standard\" way. (You don\'t have to prove that V is a vector space - take it as a given). Consider a subset W of V consisting of all polynomials of degree less or equal to 2. Is W a subspace of V? Can you give an example of a basis for W? What is the dimension of W? State clear reasons for all your answers.

Solution

W is a subset of V, consisting of all polynomials of degree less than or equal to 2. Then an arbitrary elemt of W is a polynomial of the form p(x) = ax² +bx +c, where a,b,c are real numbers.V is a vector space and apparently, W is a subset of V. Further, since a set of all polynomials with real coefficients is a vector space, the set of all polynomials of degree less thn or equal to 2 is also a vector space. we can prove this even otherwise. If p(x) = ax² +bx +c, and q(x) = dx² +ex + f are two arbitrary elements of W,then p(x) + q(x) = ax² +bx +c + dx² +ex + f = (a +d)x² + (b +e)x + (c + f) also belongs to W as a+d, b+e and c+f are real numbers. Thus W is closed under addition. Now, let be an arbitrary scalar. Then p(x) = [  ax² +bx +c] =  ax² + bx + c also belongs to W as a, b and c belong to R.Thus W is also closed under scalar multiplication. Apparently, a basis for W is { x² , x , 1} as all polynomials of degree less than or equal to 2 are linear combinations of x², x, and 1. The dimension of W is as the basis for W contains 3 elements.