1
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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

$\;\;\;\;\;\;\;\;=a_{1}(x-1)+a_{2}(x^2+2)+a_{3}(x^3+20)+a_{4}(x^4+104)$.

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.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.