0
$\begingroup$

I understand that the monomial basis proposed in this answer: $\{1,x,x^2,x^3,\ldots,x^n\}$ spans a regular polynomial vector space, but what process would I use to create a basis when there is additional criteria regarding the polynomials? I.e. how would I construct the basis of a vector space of polynomials of degree 3 or less where $p(1) = p(-2)$?

$\endgroup$
5
  • 3
    $\begingroup$ Hint, if $p(x)$ is such a polynomial then $p(x) - p(1)$ has zeros at $x = 1$, $x = -2$, so you can factor it in the form $q(x) (x-1)(x+2)$. $\endgroup$ May 24 at 19:52
  • $\begingroup$ Do you mean that $p(1) = p(-2)$ for every $p$ from the basis? If that is the case then I believe it is impossible. Indeed, if the equation holds for every polynomial in the basis then it must hold for every polynomial of degree 3 or less, but we know that it is not true for, among others, $p(x) = x$. $\endgroup$
    – blamocur
    May 24 at 19:52
  • $\begingroup$ @blamocur How do you know $p(x) = x$ is in the span? As you show, it cannot be. $\endgroup$ May 24 at 19:54
  • $\begingroup$ @JairTaylor because $x$ has degree 3 or less, and the span of the basis is supposed to include all polynomials of degree 3 or less. $\endgroup$
    – blamocur
    May 24 at 19:57
  • 1
    $\begingroup$ No, I don't think it has to include all polynomials of degree $3$ or less. Just all polynomials $p(x)$ of degree $3$ or less such that $p(1) = p(-2)$. $\endgroup$ May 24 at 21:05

3 Answers 3

2
$\begingroup$

If $p(x) = a_0+a_1x+a_2x^2+a_3x^3$ is a such polynomial, then $$\begin{align} a_0+a_1+a_2+a_3 &= p(1) \\ &= p(-2) = a_0-2a_1+4a_2-8a_3, \end{align}$$ which is equivalent to $3a_1-3a_2+9a_3=0$, or $a_2=a_1+3a_3$. Hence $$\begin{align} p(x) &= a_0+a_1x+(a_1+3a_3)x^2+a_3x^3 \\ &= a_0 \color{red}1 + a_1\color{red}{(x+x^2)} + a_3 \color{red}{(3x^2+x^3)}. \end{align}$$ Thus, we see that every polynomial $p(x)$ with $p(1)=p(-2)$ is a linear combination of the polynomials $1$, $x+x^2$ and $3x^2+x^3$. One can easily verify that these two indeed form a basis for that vector space (i.e. they are linearly independent).

$\endgroup$
1
  • $\begingroup$ For anyone wondering the so-referred to easy verification, your best bet is probably to use repeated differentiation. $\endgroup$ May 25 at 3:58
1
$\begingroup$

Using @Jair Taylor's hint:

If $p(x)$ is a polynomial that is in $V$, we can "shift" the function down so that $p(1)$ and $p(-2)$, would be zeroes, i.e. $p(x) - p(1)$ or equivalently $p(x) - p(-2)$. This can then be factored into the form: $q(x)(x - 1)(x + 2)$ so then

$p(x) = q(x)(x - 1)(x + 2) + p(1)$,

which is a set of polynomials (of degree 3 or less) spanned by: $\{(x - 1)(x + 2), x(x - 1)(x + 2), 1\}$

$\endgroup$
0
$\begingroup$

1 will also come in the basis because, p(x) =C also satisfy the condition. Therefore all constant polynomial will come in the span. Here dimension of basis will be 3.

$\endgroup$

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.