In the paper [1], it is proven in Theorem 2 that any $n$-tensor $\mathcal{A}\in\mathbb{R}^{d_1\times...\times d_n}$ can be decomposed as

$$ \mathcal{A}=\mathcal{S} \times_1 U_1 ...\times_n U_n $$ where $\times_n$ is the mode-$n$ product defined as $\mathcal{T}\times M=\sum_{i_k}\mathcal{T}_{i_1...i_k...i_n}M_{i_kj}$ and $U_n$ are appropriately shaped matrices. More interesting is the condition imposed on $\mathcal{S}$ by the theorem to ensure uniqueness of the decomposition, which the authors label all-orthogonality and which prescribes that

$$ \sum_{i_1,...,i_{k-1},\\i_{k+1},...,i_n}\mathcal{S}_{i_1,...,i_k,...,i_n}\mathcal{S}^{i_1,...,j_k,...,i_n}=\delta_{i_k}^{j_k}a_{k, i_k} \quad \forall k $$ Where $\delta$ is the kronecker delta and $a_k$ is a set of weights particular to each $k$, so that the right-hand side is a diagonal matrix. My question is a bit broad, as I am looking for any possible characterization of the family of all-orthogonal tensors. I would largely be happy with results for $3$-tensors, especially if it's easier to say something concrete. A parameterization or further decomposition into simpler matrix/tensor components would be the ideal, but I'd also appreciate some ideas on the space/manifold they inhabit and whether they possess any invariances. Results in $\mathbb{C}$ are also welcome.

I have been trying to attack the problem from different angles, but without much luck. By QR decomposition, it would seem that the QR of an arbitrary matricization $\mathcal{A}_{(k)}\in \mathbb{R}^{D\times d_k}$ separating one mode from the rest (where $D=\prod_{j\neq k}d_j$),

$$ \mathcal{A}_{(k)}=QR $$

has $R$ being diagonal, since the all-orthogonality condition corresponds to orthogonality of the matricizations. So all-orthogonality means that every matricization of the type above splits into an orthogonal and a diagonal matrix.

Some observations of varying utility:

  • the all-orthogonality property plays a parallel role to the diagonality property in matrix SVD, which is also what makes me assume there is some underlying simplicity to the condition.
  • Empirically, there does exist tensors where $R$ in the QR above is the identity for all choices of $k$, i.e. tensors where all matricizations are orthogonal.
  • The Levi-Civita antisymmetric tensor is all-orthogonal after appropriate scaling, along with any transformations of it where each mode is independently transformed by orthogonal matrices.

[1] "L. De Lathauwer, B. De Moor, and J. Vandewalle, “A Multilinear Singular Value Decomposition,” SIAM J. Matrix Anal. Appl., vol. 21, no. 4, pp. 1253–1278, Jan. 2000."


1 Answer 1


There is another paper by L. De Lathauwer et al. which might be interesting for you: On the Largest Multilinear Singular Values of Higher-Order Tensors

In this paper, they describe when certain all-orthogonal tensors exist, depending on the collection of weights you mentioned. In some cases, this even allows for an explicit construction.

Within the paper A Geometric Description of Feasible Singular Values in the Tensor Train Format, there are further references to papers that have dealt with similar problems - all depending on statements of existence depending on whats similar to above mentioned weights (usually referred to as singular values), though only sometimes explicit constructions. In particular, due to the relation to the quantum marginal problem, there is surprisingly much literature on this.

Given that the literature on even the principle existence of such objects is elaborate, a further decomposition of all-orthogonal tensors might be too much to ask in general, with the exception of some special cases. These cases are a bit similar to decomposition of large matrices into block matrices, but only on a very rough level.


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