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