$\begingroup$

Let $V=\mathscr{P}_{3}$ be the vector space of polynomials of degree 3. Let W be the subspace of polynomials p(x) such that p(0)= 0 and p(1)= 0. Find a basis for W. Extend the basis to a basis of V.

Here is what I've done so far.

$$p(x) = ax^3 + bx^2 + cx + d$$

$$p(0) = 0 = ax^3 + bx^2 + cx + d\\\text{d = 0}\\ p(1) = 0 = ax^3 + bx^2 + cx + 0 => a + b + c = 0\\ c = -a - b\\ p(x)= ax^3 + bx^2 + (-a-b)x = 0\\ = a(x^3-x) + b(x^2-x)\\ \text{Basis is {(x^3-x),(x^2-x)}} $$

Would this be a correct basis for W, and how would I extend it to the vector space V?

$\endgroup$ 2

2 Answers

$\begingroup$

I believe you're correct.

Specifying $p(0) = p(1) = 0$ means that any polynomial in $W$ must be divisible by $x$ and $(x-1)$. That is $W = \{ x(1-x)p(x) \, | \, p(x) \in \mathscr{P}_{1}\}$. Since $\mathscr{P}_{1}$ has dimension $2$, $W$ must have dimension $2$.

Extending $W$ to a basis for $V$ just requires picking any two other polynomials of degree $3$ which are linearly independent from the others. So in particular, you might choose $p_{0}(x) = 1$ and $p_{1}(x) = x$ to throw in.

$\endgroup$ $\begingroup$

Yes, that's fine. Notice also that the polynomials you have factorise as $$ x(x-1)q(x), $$ where you have $q(x)=1$ and $q(x)=1+x$.

An obvious initial way to extend is to throw in a constant function, since we clearly can't make one of those out of things that have roots at $0$ and $1$. Similarly, we can't make a polynomial of degree $1$ out of this basis, so we can add $x$. The original space has dimension $4$ (being spanned by $\{1,x,x^2,x^3\}$, so we can't find more linearly independent vectors. Therefore the final basis is $$ \{ 1,x,x(1-x),x(1-x)(1+x) \}. $$

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy