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A best rank-two CP approximation need not exist

Statement refuted

Every finite tensor has a best Frobenius approximation among tensors of CP rank at most two.

Facts & Assumptions

Given: Real three-way tensors in $R^{2 \times 2 \times 2}$ , with standard basis $e_{1}, e_{2}$ in each mode.

[F1]

Canonical polyadic rank: CP rank is the least number of elementary tensors whose sum is the tensor; a rank-at-most-two tensor is a sum of at most two elementary tensors.

Counterexample

1.1givenF1constructalgebra

Set $W = e_{2} \otimes e_{1} \otimes e_{1} + e_{1} \otimes e_{2} \otimes e_{1} + e_{1} \otimes e_{1} \otimes e_{2}$ . It has an explicit three-term CP expression, so its CP rank is at most three. Its first-mode slices are $W_{1} = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ and $W_{2} = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ . They are linearly independent, and $\det (x W_{1} + y W_{2}) = - x^{2}$ .

2.1F1step 1.1algebra

Suppose $W$ had a two-term expression $a \otimes b \otimes c + a^{'} \otimes b^{'} \otimes c^{'}$ . The first-mode slices being independent forces $a, a^{'}$ to be independent. If $b, b^{'}$ or $c, c^{'}$ were dependent, every linear combination of the two slices would have matrix rank at most one, so its determinant would vanish identically, contrary to $\det (x W_{1} + y W_{2}) = - x^{2}$ . Thus both pairs are independent. For any $x, y$ , the pencil slice is $(x a_{1} + y a_{2}) b c^{T} + (x a_{1}^{'} + y a_{2}^{'}) b^{'} c^{' T}$ . Its determinant is a nonzero constant times $(x a_{1} + y a_{2}) (x a_{1}^{'} + y a_{2}^{'})$ . Independence of $a, a^{'}$ makes the two linear factors nonproportional, so this determinant has two distinct zero lines; it cannot equal $- x^{2}$ , which vanishes on the repeated line $x = 0$ . This contradiction proves ${rank}_{C P} (W) > 2$ , hence it is exactly three.

3.1F1step 2.1algebra∎

For every $ε > 0$ , define $W_{ε} = ε^{- 1} ((e_{1} + ε e_{2})^{\otimes 3} - e_{1}^{\otimes 3})$ . This is a difference of two elementary tensors, with the scalar absorbed into one factor, so its CP rank is at most two. Expanding gives $W_{ε} = W + ε (e_{2} \otimes e_{2} \otimes e_{1} + e_{2} \otimes e_{1} \otimes e_{2} + e_{1} \otimes e_{2} \otimes e_{2}) + ε^{2} e_{2}^{\otimes 3}$ , so $∥ W_{ε} - W ∥_{F} \to 0$ . Thus the infimal rank-two approximation error is zero. Since step 2.1 shows $W$ itself has rank three, no rank-at-most-two tensor attains that infimum.

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Sources

Tamara G. Kolda and Brett W. Bader, Tensor Decompositions and Applications, SIAM Review 51(3) (2009), §3.3 (standard reference, not scraped)
Mi-Song Dupuy, Tensor trains for high-dimensional problems, full lecture notes, Chapter 1, §1.2.3 (2026) (standard reference, not scraped)