research29 min studyLesson 722 in Mathematics

Sparse recovery and restricted eigenvalues

Separate sparse identification, null-space recovery, and stable curvature on an error cone.

On this page

What you’ll understand

  • Use the null-space property to certify exact basis-pursuit recovery.
  • Define a restricted-eigenvalue condition with its cone and normalization.
  • Explain why sparse uniqueness alone does not guarantee stability.

Restricting the true vector to be sparse is only the beginning of a recovery argument. The measurement matrix must preserve enough information about sparse perturbations, and a convex estimator must prefer the intended representative among all exact fits. Null-space conditions address exact recovery by an absolute-value penalty. Restricted eigenvalues address stable estimation when the error is not itself sparse but belongs to a larger cone.

A convex program sees more than sparse candidates

Basis pursuit minimizes the sum of absolute coefficient values subject to exact noiseless measurements. Although the target is sparse, the optimization searches all coefficient vectors. Excluding a sparse null vector establishes uniqueness among sparse candidates, but does not by itself prove that no dense candidate has a smaller absolute-value sum. For a fixed support, the null-space property supplies exactly the comparison needed: any nonzero invisible displacement must have more absolute mass outside that support than inside it.

To prove sufficiency, compare a supported target with that target plus a null displacement. The triangle inequality bounds the penalty decrease on the support by the displacement mass there, while outside the support the penalty increases by its full mass. The strict inequality makes the total increase positive. Conversely, if it fails, take a target equal to minus the displacement on the support. Its alternative equals the displacement off the support and has no larger penalty. Thus strictness characterizes recovery of every target on that support, not merely one favorable target.

β^∈arg⁡min⁡Xb=y∥b∥1,∥hS∥1<∥hSc∥1for every 0≠h∈ker⁡X.\widehat\beta\in\arg\min_{Xb=y}\|b\|_1,\qquad \|h_S\|_1<\|h_{S^c}\|_1\quad\text{for every }0\ne h\in\ker X.
The strict null-space property for S guarantees that every S-supported target is the unique noiseless basis-pursuit solution.↔ Scroll wide equations horizontally.

Errors live in a cone

With noise and regularization, an estimator generally has small but nonzero coefficients outside the true support. Its error is therefore not guaranteed sparse. A basic inequality can instead place the error in a cone where the outside absolute mass is at most three times the inside mass. The factor three is a consequence of a particular noise-to-penalty comparison; it is not an intrinsic geometric constant. A restricted-eigenvalue condition must specify which cone and which norm appear, since several conventions are used in the literature.

Here use the stronger full-error convention in the display: on the cone, the normalized prediction norm is at least a fixed positive constant times the full Euclidean coefficient norm. A wide matrix can satisfy this restricted lower bound even though its unrestricted smallest singular value is zero. The null space is allowed, but it must avoid the cone except at zero. This geometry converts prediction control into coefficient control precisely for error directions generated by the estimator’s penalty argument.

C(S,3)={h:∥hSc∥1≤3∥hS∥1},∥Xh∥22n≥κ2∥h∥22(h∈C(S,3)),κ>0.\mathcal C(S,3)=\{h:\|h_{S^c}\|_1\leq3\|h_S\|_1\},\qquad \frac{\|Xh\|_2^2}{n}\geq\kappa^2\|h\|_2^2\quad(h\in\mathcal C(S,3)),\quad\kappa>0.
The factor n normalizes the design; this lesson defines kappa through the full Euclidean error norm.↔ Scroll wide equations horizontally.

Identification and stability are different demands

An injective map on a collection of sparse vectors need not separate their images by a useful quantitative margin. Nearly repeated columns can make distinct sparse coefficients produce almost identical responses. Noise then hides their difference even though exact noiseless measurements would distinguish them. Restricted lower bounds rule out this instability on specified directions. They are assumptions about the actual design or high-probability consequences of a stated random-design model, rather than automatic consequences of having many observations.

Restricted isometry is another, usually stronger, sufficient framework: it controls both upper and lower distortion of every vector up to a specified sparsity. Suitable orders and constants can imply stable convex recovery, but a number quoted without its order or convention is not a theorem. For present purposes, the null-space property and the explicit cone bound isolate two separate roles cleanly. One certifies which exact fit minimizes a penalty; the other controls how much coefficient error can hide behind a small prediction error.

One design, two support sizes

Use X=(101011)X=\begin{pmatrix}1&0&1\\0&1&1\end{pmatrix} to check basis pursuit for support {1}\{1\} and show why the guarantee fails for support {1,2}\{1,2\}.

  1. Every null vector is a multiple of (−1,−1,1)(-1,-1,1). For support {1}\{1\}, the absolute mass on the support is one unit of that multiple, while the mass outside is two units. The strict null-space property holds.

  2. For the target (1,0,0)(1,0,0), every exact fit is (1−t,−t,t)(1-t,-t,t). Its penalty is ∣1−t∣+2∣t∣|1-t|+2|t|, uniquely minimized at zero, confirming the support-specific certificate directly.

  3. For support {1,2}\{1,2\}, the same null vector has inside mass two and outside mass one, so the condition fails. Indeed, target (1,1,0)(1,1,0) has penalty two, while its exact-fit alternative (0,0,1)(0,0,1) has penalty one. A guarantee for a smaller support cannot be transferred to a larger one.

The result

The matrix recovers every target on the first coordinate by basis pursuit, yet fails for a particular two-coordinate target.

Explore the model

Interactive lab

Interactive lab · matrix

Curvature through a smallest singular value

Pause/Reset, set diagonal scale 0.5 and shear 0, and inspect singular values 1 and 0.5. Predict minimum squared stretch 0.25. For a fresh comparison Pause/Reset, set both scale 1 and shear 0, giving minimum squared stretch 1. Readouts describe the selected final matrix even during playback.

Curvature through a smallest singular valueThe 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 forSquared singular values quantify curvature. This invertible two-dimensional family illustrates the lower-bound calculation; it does not test a high-dimensional restricted cone.

Check your understanding

Why is checking the smallest singular value only on sparse vectors insufficient for the displayed Lasso error-cone argument?

Try it yourself

Suppose a normalized Gram matrix is G=(11/21/21)G=\begin{pmatrix}1&1/2\\1/2&1\end{pmatrix}. Find a valid global kappa in the displayed convention and check it on the vector (1,−1)(1,-1).

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 eigenvectors along the diagonal and the opposite diagonal give the two eigenvalues. Kappa is the square root of the smaller eigenvalue.

Compare with the solution

The eigenvalues are 3/23/2 and 1/21/2, so h⊤Gh≥(1/2)∥h∥22h^\top Gh\geq(1/2)\|h\|_2^2 for every vector and κ=1/2\kappa=1/\sqrt2 works on every cone as well. For h=(1,−1)h=(1,-1), the quadratic form is one while the squared norm is two, giving equality in that bound. This small full-rank example illustrates the normalization; in a genuinely wide design only a restricted, rather than global, positive lower bound can hold.

Take it with you

  • Basis-pursuit recovery requires comparing all null displacements, including dense alternatives.
  • Specify the cone, normalization, and norm convention when stating a restricted eigenvalue.
  • Exact identifiability and quantitatively stable recovery are different properties.

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