Talk:Tensor rank decomposition

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Plagia ?

Some portions of the article seem directly copied from http://www.models.life.ku.dk/~rasmus/presentations/parafac_tutorial/paraf.htm without appropriate quotations. — Preceding unsigned comment added by 216.165.125.249 (talk) 15:45, 15 April 2016 (UTC)

I removed the affected sections, as they indeed overlapped significantly with the source you mentioned. Ntheazk (talk) 19:41, 14 November 2016 (UTC)

Uniqueness of Minimality

In the section on field dependence, the article claims that the rank of a tensor depends on the field, over which the vector space is defined, on which the tensor acts. So I have stricken this section be stricken altogether, as it is based solely on an incorrect digression in paper focussing on other topics and cannot be corrected without (at the very least) original research and a rewrite for tone.

I include below my reasoning on this subject:

The article explains this claim with the following example, which is also how the cited article defends the claim: let ${\mathcal {A}}:\mathbb {R} ^{6}\to (\mathbb {R} ^{2})^{\otimes 3}$;

${\mathcal {A}}(\mathbf {x} _{1},\mathbf {y} _{1},\mathbf {x} _{2},\mathbf {y} _{2},\mathbf {x} _{3},\mathbf {y} _{3})=\mathbf {x} _{1}\otimes \mathbf {x} _{2}\otimes \mathbf {x} _{3}+\mathbf {x} _{1}\otimes \mathbf {y} _{2}\otimes \mathbf {y} _{3}-\mathbf {y} _{1}\otimes \mathbf {x} _{2}\otimes \mathbf {y} _{3}+\mathbf {y} _{1}\otimes \mathbf {y} _{2}\otimes \mathbf {x} _{3},$

Some elementary calculation confirms that ${\mathcal {A}}$ is of rank 3. But $\mathbb {R} ^{6}$ is of even dimension, so it is isomorphic to the vector space over the reals given by $\mathbb {C} ^{3}$. Then, for all complex numbers z₁, z₂, and z₃,

${\mathcal {A}}(\mathbf {z} _{1},\mathbf {z} _{2},\mathbf {z} _{3})={\frac {1}{2}}({\bar {\mathbf {z} }}_{1}\otimes \mathbf {z} _{2}\otimes {\bar {\mathbf {z} }}_{3}+\mathbf {z} _{1}\otimes {\bar {\mathbf {z} }}_{2}\otimes \mathbf {z} _{3}),$ (1)

where _· denotes complex conjugation. "So clearly, $\operatorname {rank} \limits _{\mathbb {C} }({\mathcal {A}})\leq 2<3=\operatorname {rank} \limits _{\mathbb {R} }({\mathcal {A}})$."

There are two problems with this argument.

1. The bolded conclusion makes no sense, because neither term in the right-hand side of 1 is truly an order-3, rank-1 tensor. Originally, we considered $\mathbb {C} ^{3}$ as a vector space of over the reals. But uniqueness of rank for a fixed base field guarantees that we must now interpret each element of the RHS as a tensor in $\mathbb {C} ^{3}$ as a complex vector space. But then we need our tensors to be complex-linear, which complex conjugation is not!
2. One could imagine a similar argument being made that sidesteps this issue: construct a particular order-3 tensor on $\mathbb {C} ^{3}$ of rank r named ${\mathcal {A}}$. Now, the complex numbers are naturally a vector space over the reals, so, as algebras over the reals, $\operatorname {End} \limits _{\mathbb {C} }{(\mathbb {C} ^{3}\to \mathbb {C} ^{3})}$ is naturally included in $\operatorname {End} {(\mathbb {R} ^{6}\to \mathbb {C} ^{3})}$. Then show that it just so happens that the rank of ${\mathcal {A}}$ in the latter space is s>r.
   I do not know if such an ${\mathcal {A}}$ exists to make this argument, but if one does, it would be correct.
   However, this sort of argument is not at all what the first sentence of this section suggests. The most natural interpretation of the phrase "change of field" is not the complexification inclusion, but rather the inclusion induced by a field extension. And as the cited article correctly notes, the latter inclusion preserves rank.

76.98.88.73 (talk) 20:30, 20 December 2017 (UTC)

I'm sorry for the confusion that my original formulation may have caused. The tensor ${\mathcal {A}}$ should not be interpreted as the map $\mathbb {R} ^{6}\to (\mathbb {R} ^{2})^{\otimes 3}$, but rather as an element of the latter space. Note specifically that the $\mathbf {x} _{i},\mathbf {y} _{i}$ are vectors of length 2 (which I failed to specify), rather than numbers. In this light, ${\mathcal {A}}={\frac {1}{2}}({\bar {\mathbf {z} }}_{1}\otimes \mathbf {z} _{2}\otimes {\bar {\mathbf {z} }}_{3}+\mathbf {z} _{1}\otimes {\bar {\mathbf {z} }}_{2}\otimes \mathbf {z} _{3})$ is a real tensor in $(\mathbb {C} ^{2})^{\otimes 3}$ that can be written as a linear combination of at most 2 complex rank-1 tensors, while one can prove that its real rank is 3. This shows that field extension does in fact not preserve tensor rank for higher-order tensors. Also note that by "change of field" I do indeed mean that tensor rank can decrease by field extension from $\mathbb {R}$ to $\mathbb {C}$, which is not the case for order-2 tensors (i.e. matrices). Further observe that Proposition 7.4 in de Silva and Lim (2008) deals with the multilinear rank rather than the tensor rank. Since the multilinear rank is essentially the matrix rank, its value does not change under field extensions. Ntheazk (talk) 23:33, 25 December 2017 (UTC)

Correctness of formula in Application

The equation in the Application section does not seem to be correct. It is currently written as:

$T=\sum _{i=1}^{k}Pr(h=k)E[x|h=k]^{\otimes 3}=k\ (Pr(h=k)E[x|h=k]^{\otimes 3})$

I presume it should be written as the following as it seems to correspond to the third moment:

$T=\sum _{i=1}^{k}Pr(h=i)E[x|h=i]^{\otimes 3}=E[E[x|h]^{\otimes 3}]]$

I cannot find the answer in the reference paper "Tensor decompositions for learning latent variable models" but the correction seems to make more sense given the text in the section. — Preceding unsigned comment added by 83.250.144.8 (talk) 11:47, 21 January 2018 (UTC)