research33 min studyLesson 728 in Mathematics

Matrix completion and nuclear norms

Recover missing entries only when low-rank geometry and sampling information support a convex certificate.

On this page

What you’ll understand

  • Define nuclear-norm completion and identify missing-information obstructions.
  • State a deterministic strict dual-certificate theorem with tangent-space injectivity.
  • Distinguish unique low-rank completion from successful nuclear-norm minimization.

A low-rank matrix can contain far fewer free parameters than entries, suggesting that missing values might be recoverable. That suggestion is not a guarantee. The observation pattern can omit an entire direction of variation, and a convex surrogate for rank can prefer a different completion. Nuclear-norm methods succeed when the sampling operator and low-rank tangent geometry satisfy concrete conditions, often established probabilistically under additional incoherence assumptions.

Low rank alone does not reveal unseen entries

Let the sampling operator retain entries indexed by a set Ω\Omega and zero the rest. Noiseless nuclear-norm completion minimizes the sum of singular values subject to matching those observed entries. A rank-rr matrix with dimensions mm by kk has r(m+k−r)r(m+k-r) local degrees of freedom at a regular rank-rr point, but observing that many arbitrary entries is not sufficient. A completely unobserved row or a missed isolated nonzero entry can remain ambiguous despite a small rank.

Incoherence measures how concentrated the singular vector spaces are on coordinate axes. For an orthonormal left singular basis UU, a common measure is (m/r)max⁡i∥U⊤ei∥22(m/r)\max_i\|U^\top e_i\|_2^2, with an analogous right measure. Concentrated directions are easy to miss under entry sampling. Random-sampling recovery theorems therefore impose specified incoherence and rank conditions and a sufficient sample size with logarithmic factors. Their conclusion is probabilistic over an observation model, not a promise that every observation pattern with the same number of entries works.

min⁡Z∥Z∥∗subject to PΩZ=PΩM,∥Z∥∗=∑jσj(Z).\min_Z\|Z\|_*\quad\text{subject to }P_\Omega Z=P_\Omega M,\qquad \|Z\|_* =\sum_j\sigma_j(Z).
The nuclear norm is a convex surrogate for rank. The observations are exact and noiseless in this formulation.↔ Scroll wide equations horizontally.

A deterministic certificate for unique recovery

Write a compact singular value decomposition M=UDV⊤M=UDV^\top and define the tangent space T={UA⊤+BV⊤}T=\{UA^\top+BV^\top\} with the natural compatible matrix dimensions. Its orthogonal complement is taken in the Frobenius inner product. Suppose the sampling map is injective on this tangent space. Also suppose there is a matrix supported only on observed entries whose tangent projection is UV⊤UV^\top and whose orthogonal projection has operator norm strictly below one. Under these two conditions, the target is the unique minimizer of the noiseless convex program.

To see the role of each condition, consider a feasible perturbation whose observed entries vanish. Its inner product with the supported certificate is zero. The nuclear-norm subgradient inequality then forces a strictly positive objective increase whenever its orthogonal tangent component is nonzero, with margin proportional to one minus the certificate’s orthogonal operator norm. If that component is zero, the perturbation lies in the tangent space, and injectivity forces it to vanish. The strict certificate is sufficient; its absence alone is not a converse impossibility theorem.

Y=PΩY,PTY=UV⊤,∥PT⊥Y∥op<1,ker⁡(PΩ)∩T={0}.Y=P_\Omega Y,\qquad P_TY=UV^\top,\qquad\|P_{T^\perp}Y\|_{\mathrm{op}}<1,\qquad \ker(P_\Omega)\cap T=\{0\}.
Together these are sufficient for unique exact nuclear-norm recovery; they are not automatically implied by a small rank or a large count of observations.↔ Scroll wide equations horizontally.

A convex relaxation must still be checked

For a real two-by-two matrix, the squared nuclear norm equals its squared Frobenius norm plus twice the absolute determinant. This follows because the product of its two singular values is the absolute determinant. It makes small completion problems exactly analyzable without numerical optimization. Such examples can show both success and failure of the convex surrogate, even when a missing entry is uniquely determined by a rank-one constraint. A rank constraint and a nuclear penalty are different optimization problems.

Noisy observations require relaxing the equality constraint or using a penalized loss with an appropriate noise level. Exact recovery certificates must then be replaced or supplemented by stability bounds. Missingness that depends on unobserved values also differs from uniform entry sampling. These distinctions matter in applications: a clean theorem for a noiseless incoherent matrix and independent sampling does not justify arbitrary imputation, nor does a visually plausible completion establish uniqueness or uncertainty calibration.

An exactly solvable convex completion

Observe three entries equal to one in Z(x)=(111x)Z(x)=\begin{pmatrix}1&1\\1&x\end{pmatrix}. Find the rank-one completion and verify directly that it also minimizes nuclear norm.

  1. Rank one requires the determinant x−1x-1 to vanish, so the only rank-one completion is x=1x=1. This alone does not establish what the convex program chooses.

  2. The squared nuclear norm is 3+x2+2∣x−1∣3+x^2+2|x-1|. For x≤1x\leq1 this equals (x−1)2+4(x-1)^2+4, uniquely minimized in that region at one.

  3. For x≥1x\geq1 it equals (x+1)2(x+1)^2, also minimized on that region at one. Thus the unique global nuclear-norm minimizer is the all-ones matrix, with nuclear norm two. This is a direct verification, not a claim that the strict certificate has been constructed for this particular pattern.

The result

The rank-one and nuclear-norm completions agree at the missing value one in this example, with uniqueness verified across all real completions.

Explore the model

Interactive lab

Interactive lab · matrix

Matching observed entries can hide different matrices

Pause/Reset, set diagonal scale 1 and shear 0, and predict determinant 1 and nuclear norm 2 from the two singular values. For a fresh comparison Pause/Reset, set both scale 1 and shear 1. The diagonal and determinant still agree, but the nuclear norm becomes sqrt(5) because the singular values change.

Matching observed entries can hide different matricesThe 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 forObserved diagonal entries do not identify an unobserved shear, even with fixed determinant. This full-rank matrix family illustrates missing information; it is not a matrix-completion solver or a rank-one recovery experiment.

Check your understanding

Which combination certifies unique exact nuclear-norm recovery in the stated theorem?

Try it yourself

For Z(x)=(243x)Z(x)=\begin{pmatrix}2&4\\3&x\end{pmatrix}, find the rank-one completion and the nuclear-norm-minimizing completion. Use the two-by-two identity to decide whether they agree.

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 determinant is two x minus twelve. Split the squared nuclear norm at the point where that determinant changes sign.

Compare with the solution

Rank one requires x=6x=6. The squared nuclear norm is 29+x2+2∣2x−12∣29+x^2+2|2x-12|. On x≤6x\leq6 it is (x−2)2+49(x-2)^2+49, minimized at two with nuclear norm seven. On x≥6x\geq6 it is x2+4x+5x^2+4x+5, increasing there and minimized at six with squared value sixty-five. The global convex optimum is therefore x=2x=2, not six. A uniquely identified rank-one completion can fail nuclear-norm recovery when the required sufficient geometry is not available.

Take it with you

  • Low-rank degrees of freedom do not make arbitrary missing-entry patterns identifiable.
  • A strict dual certificate and tangent injectivity give a precise sufficient recovery theorem.
  • Unique low-rank completion and nuclear-norm success are different claims, as small exact examples can reveal.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.