Let $T$ be an order-$k$, rank-$m$ symmetric tensor, that is, $T=\sum\_{j=1}^m v\_j\otimes v\_j \otimes \cdots \otimes v\_j$, where the Segre outer product is taken $k$ times, with $v\_j\in\mathbb{R}^d$ for all $j$. That is to say, for every $i\_1,\dots,i\_k\in\{1,2,\dots,d\}$, it holds that, $T\_{i\_1,\dots,i\_k}=\sum\_{j=1}^m v\_{ji\_1}v\_{ji\_2}\cdots v\_{ji\_k}$. 


My question is as follows.


1) Given a $T$, that is known to be symmetric and order $k$, can such a decomposition be found, with the same rank? 


2) Assume, the same rank condition is removed. Can such a decomposition be found? That is, are there $v\_1',\dots,v\_M'\in\mathbb{R}^d$, such that, $\sum\_{j=1}^M v\_i'\otimes v\_i' \otimes \cdots \otimes v\_i'$ where the outer product is $k$ times can be found? 


3) How about uniqueness? Under what conditions on $m$ and $d$, we can say the recovery is unique?



Symmetric tensor decomposition
A recent introduction is [Carlini, et al, *Four lectures on secant varieties*](https://mathscinet.ams.org/mathscinet-getitem?mr=3213518). Adam mentioned [Landsberg, *Tensors: Geometry and Applications*](https://mathscinet.ams.org/mathscinet-getitem?mr=2865915).


In brief: 


1(a). If $T$ is a symmetric tensor of tensor rank $m$, does $T$ have a symmetric decomposition with $m$ terms?


Probably not. If $T$ is symmetric and has rank $m$, it may not be the case that there exists a symmetric decomposition with $m$ terms. In the lingo, the symmetric rank of a symmetric tensor may not be equal to the rank; this is known to happen over $\mathbb{C}$, and expected (at least by me) to also happen over $\mathbb{R}$. See [Waring rank vs tensor rank of symmetric tensors?](https://mathoverflow.net/questions/284079/waring-rank-vs-tensor-rank-of-symmetric-tensors), <https://mathoverflow.net/a/322467/88133>.


1(b). Suppose that $T$ is a symmetric tensor with both tensor rank and symmetric tensor rank equal to $m$. In this case, is it possible to find, in some sense, a symmetric decomposition with $m$ terms? That is, is there some algorithm or numerical method to find an $m$ term symmetric decomposition?


Yes and no. There are algorithms and methods. First of all, there is simple brute force: just write $T = v\_1^{\otimes k} + \dotsb + v\_m^{\otimes k}$, and solve for the entries of the $v\_i$, using methods such as Gröbner bases. Certainly this is quite inefficient, but it counts as a method. There are better methods. See for example [Brachat, et al, *Symmetric tensor decomposition*](https://mathscinet.ams.org/mathscinet-getitem?mr=2736103). There are quite a few more recent papers that refer to that one. You can use MathSciNet to find them. Some of them are iterative and numerical, others are exact (if the tensor $T$ was given exactly).


2. Can some symmetric decomposition be found?


Yes, a symmetric decomposition can be found, by polarization identities. I will describe a wildly non-minimal symmetric decomposition. Finding a *minimal* symmetric decomposition is a huge problem---see the algorithms discussed above; and even in the "theoretical" approach, it's still difficult.


The polarization identities, or at least one version, are the identities
$$
 v\_1 \cdot \dotsm \cdot v\_k
 = \frac{1}{2^{k-1}} \sum \pm (v\_1 \pm \dotsb \pm v\_k)^{\otimes k},
$$
where $\cdot$ is the symmetric tensor product (sum over all permutations),
the sum is taken over all $2^{k-1}$ choices of $\pm$ coefficients inside the parentheses, and the coefficient in front of the parentheses is the product of the ones inside. For example,
$$
 v\_1 \cdot v\_2 = v\_1 \otimes v\_2 + v\_2 \otimes v\_1 = \frac{1}{2}((v\_1+v\_2)^{\otimes 2} - (v\_1 - v\_2)^{\otimes 2}),
$$
and
$$
\begin{split} v\_1 \cdot v\_2 \cdot v\_3
 &= v\_1 \otimes v\_2 \otimes v\_3 + v\_1 \otimes v\_3 \otimes v\_2 + \dotsb + v\_3 \otimes v\_2 \otimes v\_1 \\
 &= \frac{1}{4}((v\_1+v\_2+v\_3)^{\otimes 3}-(v\_1+v\_2-v\_3)^{\otimes 3}-(v\_1-v\_2+v\_3)^{\otimes 3} \\
 &\qquad \qquad +(v\_1-v\_2-v\_3)^{\otimes 3}). \\
\end{split}
$$
These hold for *all* $v\_i$, even if they are linearly dependent, or even equal (although if $v\_1=v\_2$, say, then terms like $(v\_1-v\_2+\dotsb)$ will certainly simplify). So every symmetric tensor can be decomposed "term by term". This is explicit.


Even for a single term (symmetric product) the above decomposition is usually not minimal. A minimal decomposition over $\mathbb{C}$ is described in [Buczynska, et al, *Waring decompositions of monomials*](https://mathscinet.ams.org/mathscinet-getitem?mr=3017012). But that's just for monomials; beyond that, practically everything is open. Even for binomials, nobody knows the answer. Over $\mathbb{R}$, even the monomial case is open, see [Carlini, et al, *On the real rank of monomials*](https://mathscinet.ams.org/mathscinet-getitem?mr=3648510). As Adam said in the comments, these are very complicated questions and not much is known!


On MathOverflow, see [Efficient way to express a symmetric tensor in terms of rank one elements](https://mathoverflow.net/questions/321570/efficient-way-to-express-a-symmetric-tensor-in-terms-of-rank-one-elements); [Generalization of the Polarisation Formula for Symmetric Bilinear Forms to Symmetric multilinear Forms](https://mathoverflow.net/questions/98714/generalization-of-the-polarisation-formula-for-symmetric-bilinear-forms-to-symme); [polarization formula for homogeneous polynomials](https://mathoverflow.net/questions/61884/polarization-formula-for-homogeneous-polynomials/61891#61891); and more, if you search (in fact, if you browse through my answers, I think a lot of them are on this topic).


3. How about uniqueness?


A (symmetric) tensor is called **identifiable** if it has a unique shortest decomposition (uniqueness up to permuting terms and rescaling factors in terms). [Mella](https://mathscinet.ams.org/mathscinet-getitem?mr=2238925) and [Galuppi-Mella](https://arxiv.org/abs/1606.06895) have recently solved the problem of identifiability of general symmetric tensors. There are other results, e.g., identifiability of "most" tensors of less than generic rank, or [identifiability over the reals versus identifiability over the complex numbers](https://mathscinet.ams.org/mathscinet-getitem?mr=3781595). There is a well-known **Kruskal's criterion** for identifiability (which you can find by searching).


