Show that $\mathrm{span}\{1, x^2 , −3x + x^3 , x^4 , −5x + x^5 \} ⊂ W$ . Where $W=\{p ∈ \Pi_5(Ω) : p'(−1) + p'(1) = 0\}.$

Question

Show that $\mathrm{span}\{1, x^2 , −3x + x^3 , x^4 , −5x + x^5 \} ⊂ W$ where $$W=\{p ∈ \Pi_5(Ω) : p'(−1) + p'(1) = 0\}.$$

Here is what i did so far but not sure if its right at all:

$$c_1(1) + c_2(x^2) + c_3(-3x + x^3) + c_4(x^4) + c_5(-5+x^5) = 0 \\ c_1 + x(-3c_3 -5c_5) + x^2(c_2) + x^3(c_3) + x^4(c_4) + x^5(c_5) = 0 $$ And then i just assumed that because it takes the form of a polynomial then it is a subset of $W$.

I also though of making $p(x)$ equal to each element in the set and then differentiating each on and use the conditions to see if they equal zero. Really not sure about this question and would really appreciate your help. Thank you so much!

Edit: ok this is what I ended up with

Answer 1

It seems like you're overcomplicating the question. First observe $W$ is a vector subspace of $\Pi_5(\Omega)$, and so to check that the span of the five vectors you've given is in $\Pi_5(\Omega)$, you just need to check that each of the vectors is in $W$. But this condition is easy to check.