38. The Perron-Frobenius Theorem
In addition to what’s in Anaconda, this lecture will need the following libraries:
!pip install quantecon
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In this lecture we will begin with the foundational concepts in spectral theory.
Then we will explore the Perron-Frobenius theorem and connect it to applications in Markov chains and networks.
We will use the following imports:
import numpy as np
from numpy.linalg import eig
import scipy as sp
import quantecon as qe
38.1Nonnegative matrices¶
Often, in economics, the matrix that we are dealing with is nonnegative.
Nonnegative matrices have several special and useful properties.
In this section we will discuss some of them --- in particular, the connection between nonnegativity and eigenvalues.
An matrix is called nonnegative if every element of is nonnegative, i.e., for every .
We denote this as .
38.1.1Irreducible matrices¶
We introduced irreducible matrices in the Markov chain lecture.
Here we generalize this concept:
Let be element of .
An nonnegative matrix is called irreducible if , where indicates that every element in is strictly positive.
In other words, for each with , there exists a such that .
38.1.2Left eigenvectors¶
Recall that we previously discussed eigenvectors in Eigenvalues and Eigenvectors.
In particular, λ is an eigenvalue of and is an eigenvector of if is nonzero and satisfy
In this section we introduce left eigenvectors.
To avoid confusion, what we previously referred to as “eigenvectors” will be called “right eigenvectors”.
Left eigenvectors will play important roles in what follows, including that of stochastic steady states for dynamic models under a Markov assumption.
A vector is called a left eigenvector of if is a right eigenvector of .
In other words, if is a left eigenvector of matrix , then , where λ is the eigenvalue associated with the left eigenvector .
This hints at how to compute left eigenvectors
A = np.array([[3, 2],
[1, 4]])
# Compute eigenvalues and right eigenvectors
λ, v = eig(A)
# Compute eigenvalues and left eigenvectors
λ, w = eig(A.T)
# Keep 5 decimals
np.set_printoptions(precision=5)
print(f"The eigenvalues of A are:\n {λ}\n")
print(f"The corresponding right eigenvectors are: \n {v[:,0]} and {-v[:,1]}\n")
print(f"The corresponding left eigenvectors are: \n {w[:,0]} and {-w[:,1]}\n")
The eigenvalues of A are:
[2. 5.]
The corresponding right eigenvectors are:
[-0.89443 0.44721] and [0.70711 0.70711]
The corresponding left eigenvectors are:
[-0.70711 0.70711] and [0.44721 0.89443]
We can also use scipy.linalg.eig
with argument left=True
to find left eigenvectors directly
eigenvals, ε, e = sp.linalg.eig(A, left=True)
print(f"The eigenvalues of A are:\n {eigenvals.real}\n")
print(f"The corresponding right eigenvectors are: \n {e[:,0]} and {-e[:,1]}\n")
print(f"The corresponding left eigenvectors are: \n {ε[:,0]} and {-ε[:,1]}\n")
The eigenvalues of A are:
[2. 5.]
The corresponding right eigenvectors are:
[-0.89443 0.44721] and [0.70711 0.70711]
The corresponding left eigenvectors are:
[-0.70711 0.70711] and [0.44721 0.89443]
The eigenvalues are the same while the eigenvectors themselves are different.
(Also note that we are taking the nonnegative value of the eigenvector of dominant eigenvalue, this is because eig
automatically normalizes the eigenvectors.)
We can then take transpose to obtain and obtain .
This is a more common expression and where the name left eigenvectors originates.
38.1.3The Perron-Frobenius theorem¶
For a square nonnegative matrix , the behavior of as is controlled by the eigenvalue with the largest absolute value, often called the dominant eigenvalue.
For any such matrix , the Perron-Frobenius theorem characterizes certain properties of the dominant eigenvalue and its corresponding eigenvector.
(This is a relatively simple version of the theorem --- for more details see here).
We will see applications of the theorem below.
Let’s build our intuition for the theorem using a simple example we have seen before.
Now let’s consider examples for each case.
38.1.3.1Example: irreducible matrix¶
Consider the following irreducible matrix :
A = np.array([[0, 1, 0],
[.5, 0, .5],
[0, 1, 0]])
We can compute the dominant eigenvalue and the corresponding eigenvector
eig(A)
EigResult(eigenvalues=array([-1.00000e+00, 2.90566e-17, 1.00000e+00]), eigenvectors=array([[ 5.77350e-01, 7.07107e-01, 5.77350e-01],
[-5.77350e-01, 1.35087e-16, 5.77350e-01],
[ 5.77350e-01, -7.07107e-01, 5.77350e-01]]))
Now we can see the claims of the Perron-Frobenius theorem holds for the irreducible matrix :
- The dominant eigenvalue is real-valued and non-negative.
- All other eigenvalues have absolute values less than or equal to the dominant eigenvalue.
- A non-negative and nonzero eigenvector is associated with the dominant eigenvalue.
- As the matrix is irreducible, the eigenvector associated with the dominant eigenvalue is strictly positive.
- There exists no other positive eigenvector associated with the dominant eigenvalue.
38.1.4Primitive matrices¶
We know that in real world situations it’s hard for a matrix to be everywhere positive (although they have nice properties).
The primitive matrices, however, can still give us helpful properties with looser definitions.
Let be a square nonnegative matrix and let be the power of .
A matrix is called primitive if there exists a such that is everywhere positive.
We can see that if a matrix is primitive, then it implies the matrix is irreducible but not vice versa.
Now let’s step back to the primitive matrices part of the Perron-Frobenius theorem
38.1.4.1Example 1: primitive matrix¶
Consider the following primitive matrix :
B = np.array([[0, 1, 1],
[1, 0, 1],
[1, 1, 0]])
np.linalg.matrix_power(B, 2)
array([[2, 1, 1],
[1, 2, 1],
[1, 1, 2]])
We compute the dominant eigenvalue and the corresponding eigenvector
eig(B)
EigResult(eigenvalues=array([-1., 2., -1.]), eigenvectors=array([[-0.8165 , 0.57735, 0.22646],
[ 0.40825, 0.57735, -0.79259],
[ 0.40825, 0.57735, 0.56614]]))
Now let’s give some examples to see if the claims of the Perron-Frobenius theorem hold for the primitive matrix :
- The dominant eigenvalue is real-valued and non-negative.
- All other eigenvalues have absolute values strictly less than the dominant eigenvalue.
- A non-negative and nonzero eigenvector is associated with the dominant eigenvalue.
- The eigenvector associated with the dominant eigenvalue is strictly positive.
- There exists no other positive eigenvector associated with the dominant eigenvalue.
- The inequality holds for all eigenvalues λ of distinct from the dominant eigenvalue.
Furthermore, we can verify the convergence property (7) of the theorem on the following examples:
def compute_perron_projection(M):
eigval, v = eig(M)
eigval, w = eig(M.T)
r = np.max(eigval)
# Find the index of the dominant (Perron) eigenvalue
i = np.argmax(eigval)
# Get the Perron eigenvectors
v_P = v[:, i].reshape(-1, 1)
w_P = w[:, i].reshape(-1, 1)
# Normalize the left and right eigenvectors
norm_factor = w_P.T @ v_P
v_norm = v_P / norm_factor
# Compute the Perron projection matrix
P = v_norm @ w_P.T
return P, r
def check_convergence(M):
P, r = compute_perron_projection(M)
print("Perron projection:")
print(P)
# Define a list of values for n
n_list = [1, 10, 100, 1000, 10000]
for n in n_list:
# Compute (A/r)^n
M_n = np.linalg.matrix_power(M/r, n)
# Compute the difference between A^n / r^n and the Perron projection
diff = np.abs(M_n - P)
# Calculate the norm of the difference matrix
diff_norm = np.linalg.norm(diff, 'fro')
print(f"n = {n}, error = {diff_norm:.10f}")
A1 = np.array([[1, 2],
[1, 4]])
A2 = np.array([[0, 1, 1],
[1, 0, 1],
[1, 1, 0]])
A3 = np.array([[0.971, 0.029, 0.1, 1],
[0.145, 0.778, 0.077, 0.59],
[0.1, 0.508, 0.492, 1.12],
[0.2, 0.8, 0.71, 0.95]])
for M in A1, A2, A3:
print("Matrix:")
print(M)
check_convergence(M)
print()
print("-"*36)
print()
Matrix:
[[1 2]
[1 4]]
Perron projection:
[[0.1362 0.48507]
[0.24254 0.8638 ]]
n = 1, error = 0.0989045731
n = 10, error = 0.0000000001
n = 100, error = 0.0000000000
n = 1000, error = 0.0000000000
n = 10000, error = 0.0000000000
------------------------------------
Matrix:
[[0 1 1]
[1 0 1]
[1 1 0]]
Perron projection:
[[0.33333 0.33333 0.33333]
[0.33333 0.33333 0.33333]
[0.33333 0.33333 0.33333]]
n = 1, error = 0.7071067812
n = 10, error = 0.0013810679
n = 100, error = 0.0000000000
n = 1000, error = 0.0000000000
n = 10000, error = 0.0000000000
------------------------------------
Matrix:
[[0.971 0.029 0.1 1. ]
[0.145 0.778 0.077 0.59 ]
[0.1 0.508 0.492 1.12 ]
[0.2 0.8 0.71 0.95 ]]
Perron projection:
[[0.12506 0.31949 0.20233 0.43341]
[0.07714 0.19707 0.1248 0.26735]
[0.12158 0.31058 0.19669 0.42133]
[0.13885 0.3547 0.22463 0.48118]]
n = 1, error = 0.5361031549
n = 10, error = 0.0000434043
n = 100, error = 0.0000000000
n = 1000, error = 0.0000000000
n = 10000, error = 0.0000000000
------------------------------------
The convergence is not observed in cases of non-primitive matrices.
Let’s go through an example
B = np.array([[0, 1, 1],
[1, 0, 0],
[1, 0, 0]])
# This shows that the matrix is not primitive
print("Matrix:")
print(B)
print("100th power of matrix B:")
print(np.linalg.matrix_power(B, 100))
check_convergence(B)
Matrix:
[[0 1 1]
[1 0 0]
[1 0 0]]
100th power of matrix B:
[[1125899906842624 0 0]
[ 0 562949953421312 562949953421312]
[ 0 562949953421312 562949953421312]]
Perron projection:
[[0.5 0.35355 0.35355]
[0.35355 0.25 0.25 ]
[0.35355 0.25 0.25 ]]
n = 1, error = 1.0000000000
n = 10, error = 1.0000000000
n = 100, error = 1.0000000000
n = 1000, error = 1.0000000000
n = 10000, error = 1.0000000000
The result shows that the matrix is not primitive as it is not everywhere positive.
These examples show how the Perron-Frobenius theorem relates to the eigenvalues and eigenvectors of positive matrices and the convergence of the power of matrices.
In fact we have already seen the theorem in action before in the Markov chain lecture.
38.1.4.2Example 2: connection to Markov chains¶
We are now prepared to bridge the languages spoken in the two lectures.
A primitive matrix is both irreducible and aperiodic.
So Perron-Frobenius theorem explains why both Imam and Temple matrix and Hamilton matrix converge to a stationary distribution, which is the Perron projection of the two matrices
P = np.array([[0.68, 0.12, 0.20],
[0.50, 0.24, 0.26],
[0.36, 0.18, 0.46]])
print(compute_perron_projection(P)[0])
[[0.56146 0.15565 0.28289]
[0.56146 0.15565 0.28289]
[0.56146 0.15565 0.28289]]
mc = qe.MarkovChain(P)
ψ_star = mc.stationary_distributions[0]
ψ_star
array([0.56146, 0.15565, 0.28289])
P_hamilton = np.array([[0.971, 0.029, 0.000],
[0.145, 0.778, 0.077],
[0.000, 0.508, 0.492]])
print(compute_perron_projection(P_hamilton)[0])
[[0.8128 0.16256 0.02464]
[0.8128 0.16256 0.02464]
[0.8128 0.16256 0.02464]]
mc = qe.MarkovChain(P_hamilton)
ψ_star = mc.stationary_distributions[0]
ψ_star
array([0.8128 , 0.16256, 0.02464])
We can also verify other properties hinted by Perron-Frobenius in these stochastic matrices.
Another example is the relationship between convergence gap and convergence rate.
In the exercise, we stated that the convergence rate is determined by the spectral gap, the difference between the largest and the second largest eigenvalue.
This can be proven using what we have learned here.
Please note that we use for a vector of ones in this lecture.
With Markov model with state space and transition matrix , we can write as
This is proven in Sargent & Stachurski (2023) and a nice discussion can be found here.
In this formula is an eigenvalue of with corresponding right and left eigenvectors and .
Premultiplying by arbitrary and rearranging now gives
Recall that eigenvalues are ordered from smallest to largest from .
As we have seen, the largest eigenvalue for a primitive stochastic matrix is one.
This can be proven using Gershgorin Circle Theorem, but it is out of the scope of this lecture.
So by the statement (6) of Perron-Frobenius theorem, for all , and when is primitive.
Hence, after taking the Euclidean norm deviation, we obtain
Thus, the rate of convergence is governed by the modulus of the second largest eigenvalue.
38.2Exercises¶
Solution to Exercise 1
A = np.array([[0.3, 0.2, 0.3],
[0.2, 0.4, 0.3],
[0.2, 0.5, 0.1]])
evals, evecs = eig(A)
r = max(abs(λ) for λ in evals) #dominant eigenvalue/spectral radius
print(r)
0.8444086477164554
Since we have we can thus find the solution using the Neumann Series Lemma.
I = np.identity(3)
B = I - A
d = np.array([4, 5, 12])
d.shape = (3,1)
B_inv = np.linalg.inv(B)
x_star = B_inv @ d
print(x_star)
[[38.30189]
[44.33962]
[46.47799]]
- Sargent, T. J., & Stachurski, J. (2023). Economic Networks: Theory and Computation. arXiv Preprint arXiv:2203.11972.

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