Proving W is a Subspace of P_3 & Finding a Vector Not in W

[Hiche]

Let W = \{ax^3 + bx^2 + cx + d : b + c + d = 0\} and P_3 be the set of all polynomials of degrees 3 or less.

So say we want to prove the W is a subspace of P_3. We let p(x) = a_1x^3 + a_2x^2 + a_3x + a_4 and g(x) = b_1x^3 + b_2x^2 + b_3x + b_4. So, we compute f(x) + kg(x) and the answer should be a polynomial of third degree or less? Is this enough?

And to find a basis for the subspace, we let b = -c - d and just replace into the polynomial and form a basis. How can we find a vector in P_3 but not in the subspace W?

 

[HallsofIvy]

Let W = \{ax^3 + bx^2 + cx + d : b + c + d = 0\} and P_3 be the set of all polynomials of degrees 3 or less.

So say we want to prove the W is a subspace of P_3. We let p(x) = a_1x^3 + a_2x^2 + a_3x + a_4 and g(x) = b_1x^3 + b_2x^2 + b_3x + b_4. So, we compute f(x) + kg(x) and the answer should be a polynomial of third degree or less? Is this enough?

No, that would just show that P_3 is a vector space itself. You have said nothing about W! To show that W is a subspace you need to show that if p(x) and q(x) are both members of W, then p+ kq, for any number k, is also a member of W. That is, you must have a_2+ a_3+ a_4= 0, b_2+ b_3+ b_4= 0, and then show that the corresponding coefficients for p+ kq also sum to 0.

And to find a basis for the subspace, we let b = -c - d and just replace into the polynomial and form a basis.

Yes, any "vector" in W can be written ax^3+ (-c-d)x^2+ cx+ d= ax^3+ c(x- x^2)+ d(1- x^2) so that \{x^3, x-x^2, 1- x^2\} is a basis.

How can we find a vector in P_3 but not in the subspace W?

Choose any numbers, a, b, c, d so that b+ c+ d is NOT equal to 0 and ax^3+ bx^2+ cx+ d will be in P_3 but not in W.