Eigenvalues and Eigenvectors

import numpy as np
import matplotlib.pyplot as plt
import scipy.linalg as la

Definition

Let $A$ be a square matrix. A non-zero vector $\mathbf{v}$ is an eigenvector for $A$ with eigenvalue $\lambda$ if

$$ A\mathbf{v} = \lambda \mathbf{v} $$

Rearranging the equation, we see that $\mathbf{v}$ is a solution of the homogeneous system of equations

$$ \left( A - \lambda I \right) \mathbf{v} = \mathbf{0} $$

where $I$ is the identity matrix of size $n$. Non-trivial solutions exist only if the matrix $A - \lambda I$ is singular which means $\mathrm{det}(A - \lambda I) = 0$. Therefore eigenvalues of $A$ are roots of the characteristic polynomial

$$ p(\lambda) = \mathrm{det}(A - \lambda I) $$

scipy.linalg.eig

The function scipy.linalg.eig computes eigenvalues and eigenvectors of a square matrix $A$.

Let's consider a simple example with a diagonal matrix:

A = np.array([[1,0],[0,-2]])
print(A)
[[ 1  0]
 [ 0 -2]]

The function la.eig returns a tuple (eigvals,eigvecs) where eigvals is a 1D NumPy array of complex numbers giving the eigenvalues of $A$, and eigvecs is a 2D NumPy array with the corresponding eigenvectors in the columns:

results = la.eig(A)

The eigenvalues of $A$ are:

print(results[0])
[ 1.+0.j -2.+0.j]

The corresponding eigenvectors are:

print(results[1])
[[1. 0.]
 [0. 1.]]

We can unpack the tuple:

eigvals, eigvecs = la.eig(A)
print(eigvals)
[ 1.+0.j -2.+0.j]
print(eigvecs)
[[1. 0.]
 [0. 1.]]

If we know that the eigenvalues are real numbers (ie. if $A$ is symmetric), then we can use the NumPy array method .real to convert the array of eigenvalues to real numbers:

eigvals = eigvals.real
print(eigvals)
[ 1. -2.]

Notice that the position of an eigenvalue in the array eigvals correspond to the column in eigvecs with its eigenvector:

lambda1 = eigvals[1]
print(lambda1)
-2.0
v1 = eigvecs[:,1].reshape(2,1)
print(v1)
[[0.]
 [1.]]
A @ v1
array([[ 0.],
       [-2.]])
lambda1 * v1
array([[-0.],
       [-2.]])

Examples

Symmetric Matrices

The eigenvalues of a symmetric matrix are always real and the eigenvectors are always orthogonal! Let's verify these facts with some random matrices:

n = 4
P = np.random.randint(0,10,(n,n))
print(P)
[[7 0 6 2]
 [9 5 1 3]
 [0 2 2 5]
 [6 8 8 6]]

Create the symmetric matrix $S = P P^T$:

S = P @ P.T
print(S)
[[ 89  75  22 102]
 [ 75 116  27 120]
 [ 22  27  33  62]
 [102 120  62 200]]

Let's unpack the eigenvalues and eigenvectors of $S$:

evals, evecs = la.eig(S)
print(evals)
[361.75382302+0.j  42.74593101+0.j  26.33718907+0.j   7.16305691+0.j]

The eigenvalues all have zero imaginary part and so they are indeed real numbers:

evals = evals.real
print(evals)
[361.75382302  42.74593101  26.33718907   7.16305691]

The corresponding eigenvectors of $A$ are:

print(evecs)
[[-0.42552429 -0.42476765  0.76464379 -0.23199439]
 [-0.50507589 -0.54267519 -0.64193252 -0.19576676]
 [-0.20612674  0.54869183 -0.05515612 -0.80833585]
 [-0.72203822  0.4733005   0.01415338  0.50442752]]

Let's check that the eigenvectors are orthogonal to each other:

v1 = evecs[:,0] # First column is the first eigenvector
print(v1)
[-0.42552429 -0.50507589 -0.20612674 -0.72203822]
v2 = evecs[:,1] # Second column is the second eigenvector
print(v2)
[-0.42476765 -0.54267519  0.54869183  0.4733005 ]
v1 @ v2
-1.1102230246251565e-16

The dot product of eigenvectors $\mathbf{v}_1$ and $\mathbf{v}_2$ is zero (the number above is very close to zero and is due to rounding errors in the computations) and so they are orthogonal!

Diagonalization

A square matrix $M$ is diagonalizable if it is similar to a diagonal matrix. In other words, $M$ is diagonalizable if there exists an invertible matrix $P$ such that $D = P^{-1}MP$ is a diagonal matrix.

A beautiful result in linear algebra is that a square matrix $M$ of size $n$ is diagonalizable if and only if $M$ has $n$ independent eigevectors. Furthermore, $M = PDP^{-1}$ where the columns of $P$ are the eigenvectors of $M$ and $D$ has corresponding eigenvalues along the diagonal.

Let's use this to construct a matrix with given eigenvalues $\lambda_1 = 3, \lambda_2 = 1$, and eigenvectors $v_1 = [1,1]^T, v_2 = [1,-1]^T$.

P = np.array([[1,1],[1,-1]])
print(P)
[[ 1  1]
 [ 1 -1]]
D = np.diag((3,1))
print(D)
[[3 0]
 [0 1]]
M = P @ D @ la.inv(P)
print(M)
[[2. 1.]
 [1. 2.]]

Let's verify that the eigenvalues of $M$ are 3 and 1:

evals, evecs = la.eig(M)
print(evals)
[3.+0.j 1.+0.j]

Verify the eigenvectors:

print(evecs)
[[ 0.70710678 -0.70710678]
 [ 0.70710678  0.70710678]]

Matrix Powers

Let $M$ be a square matrix. Computing powers of $M$ by matrix multiplication

$$ M^k = \underbrace{M M \cdots M}_k $$

is computationally expensive. Instead, let's use diagonalization to compute $M^k$ more efficiently

$$ M^k = \left( P D P^{-1} \right)^k = \underbrace{P D P^{-1} P D P^{-1} \cdots P D P^{-1}}_k = P D^k P^{-1} $$

Let's compute $M^{20}$ both ways and compare execution time.

Pinv = la.inv(P)
k = 20
%%timeit
result = M.copy()
for _ in range(1,k):
    result = result @ M
42.1 µs ± 11.4 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Let's use diagonalization to do the same computation.

%%timeit
P @ D**k @ Pinv
6.42 µs ± 1.36 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Diagonalization computes $M^{k}$ much faster!

Exercises

Under construction