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I have a very simple question which seemingly falls into the category of multilinear algebra. I need this concept in one of my research papers, but unfortunately did not learn it before. Let me state a simpler version of the question first:

Suppose that $\mathbf{A} := ((A_{i_1,\ldots,i_k}))_{1\leq i_1,\ldots,i_k\leq n}$ be a $k$-tensorof dimension $n\times \ldots\times n$. Assume that $A_{i_1,\ldots,i_k} = 0$ if $i_s = i_t$ for some $1\leq s<t\leq k$, and also assume that $\mathbf{A}$ is symmetric, i.e. $A_{i_1,\ldots,i_k} = A_{\pi(i_1)},\ldots,A_{\pi(i_k)}$ for any $1\leq i_1 <\ldots < i_k \leq n$ and any permutation $\pi$ on $\{1,\ldots,n\}.$ For any $n$-dimensional vector $\mathbf{x}:= (x_1,\ldots,x_n)$, define:

$$Q(\mathbf{x}) := \sum_{1\leq i_1,\ldots,i_k\leq n}A_{i_1,\ldots,i_k} x_{i_1}\ldots x_{i_k}~.$$ My question is: Is there a way to write the multilinear form $Q(\mathbf{x})$ as a weighted sum of powers of the entries of some transformed vector of $\mathbf{x}$? That is, can $Q(\mathbf{x})$ be written as: $$Q(\mathbf{x}) = \sum_{j=1}^n a_j y_j^m$$ for some $m$ (I am not sure whether this $m$ should be $k$ or $2$, or even something else, at this moment), where $\mathbf{y}$ is some transformed vector obtained from $\mathbf{x}$? In other words, I am looking for a spectral-type decomposition of the symmetric tensor $\mathbf{A}$, that allows one to write such $k$-multilinear forms in a separated manner.

Now, my original question is essentially the same as the one I asked just now, just in functional terminologies. I had a symmetric function $f: [0,1]^k \mapsto [0,1]$ and a function $g: [0,1]\mapsto [0,1]$, and I was trying to look at a way of expressing $$\int_{[0,1]^k} f(x_1,\ldots,x_k) g(x_1)\ldots g(x_k) dx_1\ldots dx_k$$ as a separated form similar to the one above, possibly in terms of the eigenobjects of the function $f$. I believe the first answer would lead to the second.

By the way, I tried to study some papers which treat spectral theory for tensors (such as https://arxiv.org/pdf/1008.2923.pdf and https://www.stat.uchicago.edu/~lekheng/work/davis.pdf), but could not quite come up with a solution. Any help will be appreciated, and thanks in advance!

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    $\begingroup$ @Bugs Bunny the OP said the question came up in the course of a research project, so to insist this is not a research level question seems unreasonable. See Joe Silverman's comments at mathoverflow.net/questions/248466/…. $\endgroup$
    – KConrad
    Commented Aug 5, 2019 at 7:38
  • $\begingroup$ $m=k$ if I understand correctly. Symmetric tensors may be identified with homogeneous polynomials, and then you are asking for what is called a power sum decomposition or a Waring decomposition. They exist, but in general it’s not easy to compute a minimal decomposition (with $n$ minimal). Non-minimal decompositions are easier to find, eg, decompose each monomial separately. Your diagonal vanishing condition translates to polynomials supported on square-free monomials. I suggest as introductions arxiv.org/abs/1309.4145 or arxiv.org/abs/1812.10267. $\endgroup$ Commented Aug 5, 2019 at 13:29
  • $\begingroup$ In particular tensor eigenvectors don’t correspond to power sum decompositions, sorry. And I think that even less is known in the case of general functions instead of polynomials. $\endgroup$ Commented Aug 5, 2019 at 13:33
  • $\begingroup$ Thanks a lot @Zach Teitler! $\endgroup$
    – Somabha
    Commented Aug 5, 2019 at 13:35
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    $\begingroup$ The Newton identities do not seem relevant. The polynomial $Q$ is not necessarily symmetric, and even if it were, the Newton identities don’t produce a power sum decomposition (rather, a polynomial in the power sums). ... Having said that it occurs to me that OP left open what kind of transformation can produce $y$. Everything I’ve said is for $y$ a linear function of $x$. If you allow other things (eg, writing a degree $k=6$ polynomial as a combination of squares $m=2$ of degree $3$ polynomials) then... well, it’s still mysterious. The introductions above should have references. $\endgroup$ Commented Aug 5, 2019 at 13:52

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