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 $p(1) = p(-2)$?
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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 $q(x) (x-1)(x+2)$. $\endgroup$– Jair TaylorMay 24 at 19:52
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$\begingroup$ 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, $p(x) = x$. $\endgroup$– blamocurMay 24 at 19:52
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$\begingroup$ @blamocur How do you know $p(x) = x$ is in the span? As you show, it cannot be. $\endgroup$– Jair TaylorMay 24 at 19:54
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$\begingroup$ @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. $\endgroup$– blamocurMay 24 at 19:57
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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 $p(1) = p(-2)$. $\endgroup$– Jair TaylorMay 24 at 21:05
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3 Answers
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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).
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$\begingroup$ For anyone wondering the so-referred to easy verification, your best bet is probably to use repeated differentiation. $\endgroup$ May 25 at 3:58
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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: $\{(x - 1)(x + 2), x(x - 1)(x + 2), 1\}$
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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.
