# Polynomial as a sum of powers of linear forms (with restrictions)

Let $$p(x)\in\mathbb{R}[x]$$ be a polynomial in $$n$$ variables, where we let $$x=(x_1,\dots,x_n)$$. It is known that $$p(x)$$ can be written as a linear combination of powers of linear forms, this is, there exist linear forms $$L_1,\dots,L_N$$, real numbers $$a_1,\dots,a_N$$, and nonnegative integers $$k_1,\dots,k_N$$, such that $$p(x)=\sum_{i=1}^Na_iL_i(x)^{k_i}.\qquad(*)$$

Now, given nonzero vectors $$v_1,\dots,v_\ell\in\mathbb{R}^n$$, let $$V_j=\mathbb{R}v_j$$ be the one dimensional vector space generated by $$v_j$$, and let $$S=\left(\bigcup_{j=1}^\ell V_j\right)\setminus\{0\},$$ which is the union of $$\ell$$ lines wich passes through the origin, minus the origin.

Question: Is it true that for any polynomial $$p(x)\in\mathbb{R}[x]$$ we can find a decomposition as in $$(*)$$ but adding the restriction that each $$L_i$$ does not vanish on $$S$$? Note that this is the same as requiring that $$L_i(v_j)\ne0$$ for all $$i,j$$. Also, if true, is there any reference in the literature?

The case $$n=1$$ is trivial. I have checked the case $$n=2$$, but then it gets messy. Before working the general case, I prefer to ask for a reference here, since this seems to be a known result (if true, of course).

• Seems pretty reasonable. You can do a linear change of coordinates so that none of the $v$'s has any zero entries, which should be quite helpful. – Anthony Quas May 18 at 18:28
• Yes, it is true. I’ll try to write more later. There’s a paper about this by Joachim Jelisiejew, should be his earliest paper on the arxiv. – Zach Teitler May 18 at 19:14

Yes, it is true. Every polynomial does have a decomposition like you ask for, with restrictions.

To explain why this is the case, first suppose that $$p$$ is homogeneous of degree $$d$$, and we seek a decomposition where each linear form $$L_i$$ is a homogeneous linear form (no constant term) and each power $$k_i = d$$. The projective variety whose points are $$L^d$$ for linear forms $$L$$, up to scalar multiple, is the Veronese variety. It's an irreducible variety, and it's nondegenerate, meaning that it's not contained in any hyperplane. If remove any proper subvariety, then what's left is still not contained in any hyperplane because it's dense in the whole Veronese variety (assuming the ground field is infinite). Your restrictions correspond to proper subvarieties. So, this argument shows that you can impose any finite number of your nonvanishing restrictions and the $$L^d$$'s that remain are still sufficient to span all the degree $$d$$ homogeneous polynomials.

If $$p$$ isn't homogeneous you can simply decompose each of its homogeneous parts separately.

A reference for this is:

1. Białynicki-Birula, Schinzel, Representations of multivariate polynomials by sums of univariate polynomials in linear forms, Colloq. Math. 112 (2008), no. 2, 201–233. (MR2383331, Zbl 1154.11011)

A more general version (working with restrictions to any open set of the space of linear forms, not just complements of hyperplanes) is:

1. Jelisiejew, An upper bound for the Waring rank of a form, Arch. Math. (Basel) 102 (2014), no. 4, 329–336. (MR3196960, Zbl 1322.14079)

Both of these references are more concerned with giving an upper bound for the number of terms needed in a decomposition, so they don't spend much time on the existence of a decomposition.

A few minor notes in closing:

The number of terms in a decomposition of a nonhomogeneous polynomial can (at least sometimes) be decreased if we allow nonhomogeneous linear forms. The common examples are things like $$(x+1)^n-x^n$$, a polynomial of degree $$n-1$$, which requires much more than $$2$$ terms if it is decomposed in powers $$k_i \leq n-1$$.

Białynicki-Birula and Schinzel have a hypothesis that the leading form of the polynomial should depend essentially on all of the variables. That doesn't matter if all you're asking for is existence of a decomposition (it matters if you want decompositions with minimal number of terms). If the leading term were, say, $$x_1^d$$, and you wanted to decompose in terms $$L_i^d$$ where $$L_i(p) \neq 0$$, but $$x_1(p)=0$$, then you couldn't use $$x_1$$ as one of the $$L_i$$, and you'd have to use a very non-minimal decomposition involving stuff like $$(x_1 + a x_2)^d$$ for several values of $$a$$. Anyway, the point is, if you don't worry about minimality then don't worry about that hypothesis in their paper.

• Wow, quick and great answer, Zach. Thank you very much. I'm not well versed in algebraic geometry but I'll definitively take a look at the references you mention. A random question: When you posted your comment I actually came to the article by Jelisiejew, and searching from there, I ended reading about the "apolarity lemma". Is it related to your answer above? – EFinat-S May 19 at 3:59
• It’s related. The apolarity lemma gives a condition for a polynomial to lie in the span of some $L_i^d$s in terms of some ideals. It applies to any set of linear forms, more or less independently of any restrictions like nonvanishing at certain points. – Zach Teitler May 19 at 5:28