advanced42 min studyLesson 810 in Mathematics

Multiplicative ergodic theorem and Lyapunov spectra

State the invertible Oseledets theorem precisely and compute growth along invariant directions rather than coordinate axes.

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What you’ll understand

  • State finite-dimensional invertible Oseledets hypotheses and the equivariant splitting conclusion.
  • Distinguish finite-time singular stretches, invariant growth directions, and asymptotic exponents.
  • Compute an explicit triangular spectrum and a periodic cocycle spectrum with the correct time normalization.

A matrix product can stretch different directions at different exponential rates. Kingman's theorem identifies a top norm-growth rate under suitable hypotheses, but a full spectrum needs more structure. Oseledets' multiplicative ergodic theorem supplies measurable invariant directions and their asymptotic exponents. Its invertible formulation requires integrability for both forward matrices and their inverses. A triangular example shows why reading diagonal entries or following a coordinate axis can misidentify a vector's rate, even when the matrix is constant and all products are explicit.

Specify the base and both integrability conditions

Let T be an invertible bimeasurable ergodic probability-preserving transformation on a standard Borel probability space, and let A be a measurable map into invertible real d-by-d matrices. Assume the positive parts of the natural logarithms of the operator norms of A and its inverse are integrable. These are sufficient hypotheses for the two-sided finite-dimensional theorem used here. The cocycle product applies the earliest matrix first, on the right. Reversing the order can change a product when matrices do not commute.

The theorem gives finitely many finite real exponents with multiplicities and a measurable direct-sum decomposition of the vector space at almost every base point. Applying A sends each subspace to its corresponding subspace over the next point. Every nonzero vector in one subspace has its assigned exponential growth rate. In the ergodic case the exponents and dimensions are almost surely constant; without ergodicity they can be invariant measurable functions. Almost-sure statements do not automatically cover all initial base points or supply numerical convergence rates.

ln⁡+∥A∥, ln⁡+∥A−1∥∈L1(μ),Rd=⨁iEi(x),A(x)Ei(x)=Ei(Tx),lim⁡n→∞1nln⁡∥A(n)(x)v∥=λi(0≠v∈Ei(x))\ln^+\|A\|,\ \ln^+\|A^{-1}\|\in L^1(\mu),\qquad \mathbb R^d=\bigoplus_i E_i(x),\quad A(x)E_i(x)=E_i(Tx),\quad \lim_{n\to\infty}\frac1n\ln\|A^{(n)}(x)v\|=\lambda_i\quad(0\ne v\in E_i(x))
This displayed conclusion uses the invertible ergodic probability base and invertible finite-dimensional cocycle described in the text. Natural logarithms give exponents per iterate.↔ Scroll wide equations horizontally.

Growth directions differ from finite-time singular vectors

A vector with components in several Oseledets subspaces has the largest exponent among its nonzero components. Singular values of the full n-step product also have normalized logarithms converging to the exponents, repeated with multiplicity. However, singular vectors of a particular finite product are optimized for that finite interval and need not equal the equivariant splitting directions. This distinction is already visible for a constant nonnormal matrix, where a coordinate direction can feed an expanding eigenvector despite having a contracting diagonal entry.

With only forward integrability, a more general theorem provides a forward filtration, and some exponents may equal minus infinity. One should not claim the full invertible splitting stated above after dropping inverse integrability or invertibility of the base. Different formulations address noninvertible bases, but require their own conclusions and hypotheses. Here the explicit assumptions keep forward and backward growth compatible and make every exponent finite. A small observed singular value at one time is not by itself evidence of a negative limiting exponent.

lim⁡n→∞1nln⁡σj(A(n)(x))=λj with multiplicity,∑i(dim⁡Ei)λi=∫Xln⁡∣det⁡A(x)∣ dμ(x)\lim_{n\to\infty}\frac1n\ln\sigma_j(A^{(n)}(x))=\lambda_j\text{ with multiplicity},\qquad \sum_i(\dim E_i)\lambda_i=\int_X\ln|\det A(x)|\,d\mu(x)
The singular values are ordered from largest to smallest. Under the stated forward and inverse integrability, the logarithmic determinant is integrable; its time average gives the sum of exponents.↔ Scroll wide equations horizontally.

Exterior powers account for the spectrum

The largest stretch of an exterior power measures growth of oriented volume in the corresponding dimension. Applying the subadditive theorem to these exterior-power cocycles identifies sums of the largest exponents; differences of consecutive sums recover individual values. The top exterior power is determinant, whose logarithm is additive along products. Birkhoff then gives the displayed sum rule. This explains why determinant one forces the sum of exponents to vanish but does not force each exponent to vanish. Expansion and contraction can compensate.

The matrix lab's family has bottom-right entry one, so it does not directly represent every matrix in the worked example. It can nevertheless expose the difference between one-step stretching and asymptotic growth for its own selected constant matrix. Its playback interpolates geometry rather than composing a cocycle. Compute powers and their normalized logarithms separately, and treat the readouts as checks on an individual matrix. The theorem establishes the long-time conclusion only after its measure, measurability, and integrability assumptions have been verified.

A contracting eigenvector hidden by an expanding coordinate

On a one-point probability base, take A=(2101/2)A=\begin{pmatrix}2&1\\0&1/2\end{pmatrix}. Find both exponents and their invariant directions, then determine the rate of the second coordinate vector.

  1. The expanding eigenvector is e1=(1,0)e_1=(1,0), with eigenvalue two. The vector v=(−2,3)v=(-2,3) satisfies Av=(−1,3/2)=v/2Av=(-1,3/2)=v/2. Hence the two exponents are ln⁡2\ln2 and −ln⁡2-\ln2, with the corresponding one-dimensional eigenspaces.

  2. Multiplication or a geometric sum gives An=(2n(2/3)(2n−2−n)02−n)A^n=\begin{pmatrix}2^n&(2/3)(2^n-2^{-n})\\0&2^{-n}\end{pmatrix}. Thus Ane2=((2/3)(2n−2−n),2−n)A^ne_2=((2/3)(2^n-2^{-n}),2^{-n}), whose norm grows at rate ln⁡2\ln2.

  3. The decomposition e2=(2/3)e1+(1/3)ve_2=(2/3)e_1+(1/3)v has a nonzero expanding component. The determinant is one, so the sum of the two exponents is zero, matching ln⁡2−ln⁡2\ln2-\ln2. All integrability conditions hold automatically on this finite base.

The result

The second coordinate vector expands asymptotically even though the bottom diagonal entry contracts. The contracting direction is an oblique eigenspace, and determinant one balances the two rates.

Explore the model

Interactive lab

Interactive lab · matrix

One-step stretching can overstate the long-time rate

Pause and Reset; set both a=0.5,b=1a=0.5,b=1. Predict determinant one half and largest singular value (9+65)/8>1\sqrt{(9+\sqrt{65})/8}>1, although powers of this selected matrix have top exponent zero. Reset and set both a=2,b=1a=2,b=1; predict determinant two and stretch 3+5\sqrt{3+\sqrt5}, with top constant-matrix exponent ln⁡2\ln2. These are the lab's matrices, not the worked matrix. Playback interpolates geometry, not powers; edits retain time, so Reset and set both controls for fresh comparisons.

One-step stretching can overstate the long-time rateThe 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 forA one-step singular stretch and an asymptotic Lyapunov exponent answer different questions.

Check your understanding

Under the stated invertible Oseledets hypotheses, which statement is correct?

Try it yourself

Let the base alternate between two points with uniform probability. Set A(0)=diag⁡(3,1/3)A(0)=\operatorname{diag}(3,1/3) and A(1)=diag⁡(1/2,2)A(1)=\operatorname{diag}(1/2,2). Find the two Lyapunov exponents per single iterate, state the product order from point zero, and verify their sum.

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

Give me a hint

Compute a two-step product, then divide its logarithmic rates by two because one full base cycle takes two iterates.

Compare with the solution

From zero the product is A(1)A(0)=diag⁡(3/2,2/3)A(1)A(0)=\operatorname{diag}(3/2,2/3). Repeating full cycles gives exponents 12ln⁡(3/2)\tfrac12\ln(3/2) and 12ln⁡(2/3)=−12ln⁡(3/2)\tfrac12\ln(2/3)=-\tfrac12\ln(3/2) per iterate; the bounded leftover factor does not alter them. Here the diagonal matrices commute, so starting from the other point gives the same full-cycle product. Both determinants equal one, and the exponent sum is zero, as the determinant formula requires.

Take it with you

  • A finite two-sided Lyapunov splitting uses explicit forward and inverse integrability hypotheses.
  • Invariant growth directions and finite-time singular vectors need not coincide.
  • Exterior powers and determinant connect individual exponents to volume growth.

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