advanced39 min studyLesson 907 in Mathematics

Identifiability and Kruskal conditions

Check a sufficient CP uniqueness certificate using column independence and distinguish identifiability from numerical stability.

On this page

What you’ll understand

  • Define Kruskal column rank and essential uniqueness.
  • Apply the three-factor sufficient inequality with exact hypotheses.
  • Explain why failure of the certificate or poor conditioning has a different meaning.

A low residual does not establish that recovered tensor components are uniquely determined by the data. CP identifiability asks whether another decomposition could represent exactly the same tensor after unavoidable rescalings and permutations are removed. Kruskal’s theorem gives a powerful sufficient answer using the independence of column subsets. The test is algebraic and exact; it neither supplies a necessary condition in every tensor size nor guarantees that noisy component estimates will be well conditioned.

Remove unavoidable ambiguities before asking for uniqueness

Let A, B, and C be real factor matrices with R nonzero columns, forming a three-way CP decomposition. Reordering the columns simultaneously leaves the tensor unchanged. Scaling a component’s three factors by nonzero scalars whose product is one also leaves its pure product unchanged. Essential uniqueness means that all minimal decompositions differ only through these shared permutations and reciprocal component rescalings.

For a matrix with R columns, its Kruskal rank k is the largest integer for which every set of k columns is linearly independent. This is stronger than ordinary matrix rank, which requires only some independent set of that size. A repeated or proportional nonzero column makes Kruskal rank one, even if other columns span a larger space. Identifiability tests using ordinary ranks in place of Kruskal ranks can therefore certify a false conclusion.

kA=max⁡{k:every set of k columns of A is independent},kA≤rank⁡(A)≤Rk_A=\max\{k:\text{every set of }k\text{ columns of }A\text{ is independent}\},\qquad k_A\le\operatorname{rank}(A)\le R
All factors are real and columns are nonzero. Kruskal rank tests every subset, not merely existence of one independent subset.↔ Scroll wide equations horizontally.

The inequality is a sufficient theorem

For a three-way decomposition with R ≥ 2 nonzero terms, Kruskal’s condition k_A + k_B + k_C ≥ 2R + 2 guarantees that the tensor has CP rank R and that the R-term decomposition is essentially unique. The proof uses structured dependence relations among slices and column subspaces; checking the inequality is not a substitute for the theorem’s proof. A one-term nonzero tensor is separately unique up to its factor scalings, although the displayed inequality cannot hold when R = 1.

The criterion is sufficient, not a classification of all identifiable tensors. If the sum is too small, the correct conclusion is that this test is inconclusive, rather than that a competing decomposition exists. Stronger size-specific or generic identifiability theorems have their own hypotheses. The guarantee here also concerns exact full tensors, not completion from a subset of entries or reliable recovery in the presence of noise.

kA+kB+kC≥2R+2⟹rank⁡CP(T)=R and essential uniquenessk_A+k_B+k_C\ge2R+2\quad\Longrightarrow\quad \operatorname{rank}_{CP}(T)=R\ \text{and essential uniqueness}
Assume a real three-factor CP expression with R ≥ 2 nonzero components. The implication is sufficient; its converse is not asserted.↔ Scroll wide equations horizontally.

Exact independence can be arbitrarily poorly conditioned

Two columns may remain independent while becoming nearly proportional. Their Kruskal rank stays two until exact dependence occurs, but small perturbations can then cause large changes in a factor recovery calculation. Thus identifiability and stability must be analyzed separately. Singular values of relevant column submatrices or the derivative of the parameter-to-tensor map provide quantitative information that the integer Kruskal ranks omit.

Normalizing factors removes some scale freedom but does not eliminate nearly dependent components or CP degeneracy. When interpreting latent variables, report the ambiguity conventions, the exact certificate used, and a perturbation or conditioning assessment appropriate to the data. A unique exact decomposition can coexist with wide uncertainty in finite-precision estimates. Conversely, several equivalent scaled factor triples should not be mistaken for distinct latent explanations.

Meet the threshold with one short factor matrix

Take R = 3, let A = C = I₃, and let B have columns (1, 0), (1, 1), and (1, 2). Check the Kruskal ranks and state exactly what follows.

  1. The three identity columns are independent, and so are all their subsets. Therefore k_A = k_C = 3.

  2. Any two B columns have determinant equal to the difference of their second coordinates, which is nonzero for the values zero, one, and two. Three columns in real two-space cannot be independent, so k_B = 2.

  3. The sum is 3 + 2 + 3 = 8, equal to 2R + 2 = 8. Kruskal’s theorem proves CP rank three and essential uniqueness of this three-term decomposition, despite B having only two rows.

The result

The sufficient inequality holds at equality. Full column rank of every factor is unnecessary; uniform independence of the relevant subsets is what matters.

Explore the model

Interactive lab

Interactive lab · matrix

Independent columns need not have good conditioning

For each run Pause and Reset, then set both controls. Choose a = 1 and b = 0; predict condition number 1. In a fresh run use a = 1 and b = 1; its singular values are the golden ratio and its reciprocal, giving condition number (3 + √5)/2, about 2.618. Both matrices still have two independent columns. The lab reports this small matrix’s conditioning, not a complete CP identifiability or perturbation analysis.

Independent columns need not have good conditioningThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn unchanged exact independence count can conceal a changed sensitivity to perturbations.

Check your understanding

What follows when a three-factor Kruskal-rank sum is below 2R + 2?

Try it yourself

Let R = 3 and A = [[1, 0, 1], [0, 1, 0]], while B = C = I₃. Compute the ordinary rank and Kruskal rank of A, then decide whether the sufficient uniqueness test applies.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The first and third A columns coincide, although the first two span real two-space.

Compare with the solution

The ordinary rank of A is two because its first two columns are independent. Every single column is nonzero, but the first and third form a dependent pair, so k_A = 1. The two identity factors have Kruskal ranks three, giving total 1 + 3 + 3 = 7, below the required eight. The test does not apply. Replacing k_A by its ordinary rank would incorrectly reach eight and claim a certificate that the hypotheses do not supply. This conclusion is deliberately only inconclusive about essential uniqueness.

Take it with you

  • Essential CP uniqueness permits shared component permutations and product-one rescalings.
  • Kruskal ranks test every column subset and give a sufficient three-factor uniqueness certificate.
  • Exact identifiability, noise stability, and observation-based recoverability are separate questions.

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