research30 min studyLesson 853 in Mathematics

Wishart matrices and Marchenko–Pastur law

Normalize rectangular covariance spectra and keep the aspect ratio and zero atom attached to the chosen matrix dimension.

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What you’ll understand

  • Specify the observation shape and covariance normalization in a Wishart model.
  • Calculate the Marchenko–Pastur support and zero-atom mass.
  • Relate feature and companion empirical measures without changing their nonzero eigenvalues.

Sample covariance matrices are positive semidefinite products of rectangular data matrices, so their limiting spectrum is not a symmetric semicircle. The Marchenko–Pastur law depends on how the number of measured coordinates compares with the number of samples. Its zero atom is not a cosmetic addition: it records an exact rank obstruction that changes when one switches between the two associated Gram matrices.

Keep the rectangular dimensions visible

Let XX have pp rows and nn columns, with independent standard real Gaussian entries. The columns represent independent mean-zero observations in feature dimension pp. Define the feature covariance S=XX⊤/nS=XX^\top/n and normalize its empirical measure by pp. This is a real Wishart model; independent circular complex Gaussian entries with expected absolute square one give its complex counterpart. The division is by sample size, not feature dimension, and the mean is known rather than estimated from the columns.

Let p,np,n tend to infinity with p/n→γ∈(0,∞)p/n\to\gamma\in(0,\infty). The empirical measure converges almost surely weakly to the displayed Marchenko–Pastur law. Gaussianity is a sufficient assumption used here, while more general independent-entry theorems require their own moment or truncation conditions. Positivity follows already from the finite identity u⊤Su=∥X⊤u∥22/nu^\top Su=\|X^\top u\|_2^2/n. No spectral limit is needed to establish that all finite covariance eigenvalues are nonnegative.

μγ=(1−1γ)+δ0+(b−x)(x−a)2πγx1[a,b](x) dx,a=(1−γ)2,b=(1+γ)2.\mu_\gamma=\left(1-\frac1\gamma\right)_+\delta_0+\frac{\sqrt{(b-x)(x-a)}}{2\pi\gamma x}\mathbf1_{[a,b]}(x)\,dx,\qquad a=(1-\sqrt\gamma)^2,\quad b=(1+\sqrt\gamma)^2.
This is the p-normalized spectral law of XX transpose divided by n, with gamma equal to features over samples.↔ Scroll wide equations horizontally.

The zero atom records missing rank

When features outnumber samples, the feature covariance has at least p−np-n zero eigenvalues. Gaussian data have maximal possible rank almost surely, so this zero count is exact. Dividing by feature dimension produces limiting mass 1−1/γ1-1/\gamma. The continuous part consequently has mass 1/γ1/\gamma in that regime. When the aspect ratio is below or equal to one, there is no limiting zero atom in this feature-normalized measure, though the density reaches the origin at the square case.

At aspect ratio one, the lower edge is zero and the density has an integrable inverse-square-root singularity there; that is not an atom. More generally the limiting first moment is one under unit entry variance, even when most eigenvalues are zero. A small continuous fraction must then carry a correspondingly larger average nonzero eigenvalue. This separates total spectral mass from conditional mass among nonzero eigenvalues, a distinction that matters for rank estimation, inverse covariance calculations, and comparisons of differently shaped data matrices.

The companion matrix changes multiplicities

The companion matrix is X⊤X/nX^\top X/n, now of dimension nn. Its positive eigenvalues agree with those of the feature matrix, including multiplicity, because both are squared singular values of the same scaled rectangular matrix. Their empirical measures nevertheless differ: one divides the shared eigenvalue list by feature dimension and the other by sample dimension, and each adds its own number of zeros. The displayed transform identity is exact at finite dimensions and remains valid in the limit.

If the aspect ratio is below one, the companion has zero mass 1−γ1-\gamma, even though the feature matrix has no zero atom. For aspect ratio above one, the signed atom term in the measure relation cancels the feature matrix’s excess zeros and the companion has no zeros in the Gaussian full-rank case. The relation should therefore be interpreted as an algebraic identity between complete measures, not as a convex mixture with nonnegative weights for every aspect ratio. Confusing the two normalizations can move an apparent zero atom to the wrong spectral law.

μ‾n=pnμp+(1−pn)δ0,m‾n(z)=pnmp(z)+p/n−1z.\underline\mu_n=\frac pn\mu_p+\left(1-\frac pn\right)\delta_0,\qquad \underline m_n(z)=\frac pn m_p(z)+\frac{p/n-1}{z}.
The feature and companion transforms both use denominator x minus z. The zero atom contributes minus one over z.↔ Scroll wide equations horizontally.

More features than observations

Take four hundred Gaussian features and one hundred samples. Calculate the exact zero fraction, limiting continuous support, and average nonzero eigenvalue scale.

  1. The aspect ratio is four. The feature covariance has rank one hundred almost surely, so exactly three hundred of its four hundred eigenvalues vanish. The zero fraction is 3/43/4.

  2. The limiting support edges are (1−2)2=1(1-2)^2=1 and (1+2)2=9(1+2)^2=9. These are asymptotic edges for the continuous law, not a deterministic finite-sample assertion that every nonzero eigenvalue lies between them.

  3. The continuous component carries mass 1/41/4 while the limiting first moment of the full measure is one. Thus its conditional mean among nonzero eigenvalues is four. The companion matrix contains just the nonzero list and has no zero eigenvalues almost surely in this Gaussian example.

The result

Three quarters of the feature spectrum is exactly zero; the remaining quarter converges to a continuous law supported from one to nine.

Explore the model

Interactive lab

Interactive lab · matrix

Covariance eigenvalues are squared singular values

Pause/Reset, set diagonal scale 2 and shear 0. Predict Gram eigenvalues 4 and 1 from singular values 2 and 1. For a fresh comparison Pause/Reset, set both scale 0.5 and shear 0, giving Gram eigenvalues 0.25 and 1. Matrix readouts describe the chosen final transformation during playback.

Covariance eigenvalues are squared singular valuesThe 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 forGram spectra square singular values and stay nonnegative. This supported matrix family is always invertible; it cannot model a rectangular rank deficit or generate Marchenko–Pastur data.

Check your understanding

For feature-over-sample ratio one quarter, which zero-atom statement is correct?

Try it yourself

For one hundred features and four hundred Gaussian samples, compute the limiting feature support, the feature zero mass, and the companion zero fraction. State which matrix uses dimension one hundred in its empirical normalization.

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

Give me a hint

Use gamma equal to one quarter and distinguish the two matrix sizes before counting zeros.

Compare with the solution

The feature support runs from (1−1/2)2=1/4(1-1/2)^2=1/4 to (1+1/2)2=9/4(1+1/2)^2=9/4. Its limiting zero mass is zero. The four-hundred-dimensional companion has three hundred zero eigenvalues almost surely, hence zero fraction 3/43/4. The feature matrix is the one-hundred-dimensional matrix normalized by one hundred in its empirical measure, although both covariance products themselves are divided by sample size four hundred. These two different divisions serve different purposes and must not be swapped.

Take it with you

  • Feature-over-sample ratio and sample-size scaling determine the Marchenko–Pastur convention.
  • A zero atom records a rank deficit; an integrable density singularity is not an atom.
  • Feature and companion matrices share nonzero eigenvalues but have different normalized measures.

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