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Let $\mathcal{P}_n[x]$ be the space of polynomials of degree $\leq n$ with real coefficients. I want to find a basis for the subspace of $\mathcal{P}_n[x]$ where the coefficents sum to zero, that is, a space $W$ such that $$ W = \left\lbrace \sum^n a_k x^k \Bigg| \sum^n a_k = 0 \right\rbrace. $$

Can one simply use a basis for $\mathcal{P}_n[x]$, for example $\lbrace 1, x, \dots, x^n \rbrace$? My reasoning is that if we have a $u \in W$, then we can write $u$ as a linear combination of elements from $\lbrace 1, x, \dots, x^n \rbrace$ by using the cofficients in $u$.

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  • $\begingroup$ No. Because it gives lots of other polynomials which are not in $W$. Indeed none of $1,x,\dots$ are in $W$. You want things like $1-x$ $\endgroup$ – almagest Sep 4 '14 at 15:01
  • $\begingroup$ @almagest yes but does the definition of a basis restrict the elements to only yield elements in the target space? $\endgroup$ – Slug Pue Sep 4 '14 at 15:03
  • $\begingroup$ It does, since you require the basis vectors to span just the subspace. In your case, the vectors span the entire space. $\endgroup$ – Semiclassical Sep 4 '14 at 15:13
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The condition $\sum a_k = 0$ is a single (linear) constraint on an $(n + 1)$-dimensional vector space, so the space $W$ of vectors that satisfies it has dimension $n$. Then, the ordered set

$(1 - x, x - x^2, \ldots, x^{n - 1} - x^n)$

obviously (1) is a subset of $W$, (2) is linearly independent, and (3) has $n$ elements, so it must be a basis for $W$.

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