# W a subspace of P3

1. May 28, 2012

### 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 $W$?

2. May 28, 2012

### HallsofIvy

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.

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.

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.

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