I don't understand how to do problem where you have to write a basis for a polynomial.

For a example a typical problem would be something like:

Let $U = \{p \in P_n(F): p(2) = p(5) \text{ or } p''(1) = 0\}$ . Find a basis of U. Then extend that basis to $P_n(F)$ with a subspace W such that $P_4(F) = U \oplus W$.

My book doesn't have a very good explanation on how to go about problems like these, could someone maybe do an example and explain the steps?? I haven't been able to find an example of finding the basis of polynomials when it has conditions like that. Thank you!


Here is an example of such a problem:

Find a basis for the subspace $U=\{p\in P_{4}(\mathbb{R}): p(4)=3p(2)\}$.

Let $p(x)=a_0+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4$. Then

$p(4)=3p(2)\iff a_0+4a_1+16a_2+64a^3+256a^4=3(a_0+2a_1+4a_2+8a_3+16a_4)$

$\iff 2a_0+2a_1-4a_2-40a_3-208a_4=0 \iff a_0+a_1-2a_2-20a_3-104a_4=0$

$\iff a_0=-a_1+2a_2+20a_3+104a_4$.

Then $p(x)=(-a_1+2a_2+20a_3+104a_4)+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4$


Therefore $\{x-1, x^2+2, x^3+20, x^4+104\}$ forms a basis for U.

(Notice that if we chose to solve the above equation for $a_1$ instead of $a_0$, we would have obtained the basis $\{1-x, x^2+2x, x^3+20x, x^4+104x\}$ for U instead.)

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  • $\begingroup$ Thank you! So I am able to solve this problem now, but the second half of the problem says extend this basis to find a basis of $P_4(F)$. My question is, isn't this already a basis for $P_4(F)$? Or do you need to add a constant, like maybe a one to the basis? $\endgroup$ – Soaps Jul 14 '14 at 23:42
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    $\begingroup$ Good question - Since $P_4(\mathbb{R})$ has dimension 5, we just need to choose any polynomial in $P_4(\mathbb{R})$ which is not a linear combination of the basis vectors for U that we have found, and taking our 5th vector to be 1 as you suggest is the simplest choice. $\endgroup$ – user84413 Jul 14 '14 at 23:48
  • $\begingroup$ So W = {1} is a subspace of $P_4(F)$ and is not a linear combination of the rest of the basis vectors in U, if we said $P_4(F)$ = U + U, could we say the sum of those two sub-spaces is a direct sum follows directly from the fact that they are all linearly independent? $\endgroup$ – Soaps Jul 14 '14 at 23:52
  • $\begingroup$ Yes, that's right; since the 5 vectors are linearly independent, we can conclude that $P_4(\mathbb{R})=U\oplus W$. $\endgroup$ – user84413 Jul 15 '14 at 0:05

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