Eigenvectors are both important auxiliary tools and also play important role in applications.
To start with all our microworld is governed by the Schrodinger equation which is an eigenvalue problem.
$$
H \psi = E \psi,
$$
where $E$ is the ground state energy, $\psi$ is called wavefunction and $H$ is the Hamiltonian.
More than 50% of the computer power is spent on solving this type of problems for computational material / drug design.
One of the most famous eigenvectors computation is the Google PageRank. It is not actively used by Google nowdays, but it was of the main features in its early stages. The question is how do we rank webpages, which one is important, and which one is not.
All we know about the web is which page referrs to which. PageRank is defined by a recursive definition. Denote by $p_i$ the importance of the $i$-th page. Then we define this importance as an average value of all importances of all pages that refer to the current page. It gives us a linear system
$$
p_i = \sum_{j \in N(i)} \frac{p_j}{L(j)},
$$
where $L(j)$ is the number of outgoing links on the $j$-th page, $N(i)$ are all the neighbours. It can be rewritten as
$$
p = G p,
$$
or as an eigenvalue problem
i.e. the eigenvalue $1$ is already known.
We will use a simple example of Zachary karate club network. This data was collected in 1977, and is a classical social network dataset.
In [2]:
import numpy as np
%matplotlib inline
import matplotlib.pyplot as plt
plt.xkcd()
import networkx as nx
kn = nx.read_gml('karate.gml')
nx.draw_networkx(kn) #Draw the graph
Now we can actually compute the PageRank using the NetworkX built-in function. We also plot the size of the nodes larger if its PageRank is larger.
In [3]:
pr = nx.algorithms.link_analysis.pagerank(kn)
pr_vector = [pr[i+1] for i in xrange(len(pr))]
pr_vector = np.array(pr_vector) * 5000
nx.draw_networkx(kn, node_size=pr_vector, labels=None)
plt.tight_layout()
plt.title('PageRank nodes')
Out[3]:
The eigenvalue problem has the form
$$
Ax = \lambda x,
$$
or
$$
(A - \lambda I) x = 0,
$$
therefore matrix $A - \lambda I$ has non-trivial nullspace and should be singular. That means, that the determinant
$$
p(\lambda) = \det(A - \lambda I) = 0.
$$
The equation is is called characteristic equations and is a polynomial of order $n$.
The $n$-degree polynomial has $n$ complex roots!
The characteristic equation can be used to compute the eigenvalues, which leads to naive algorithm:
Is this a good idea?
We can do a short demo of this
In [15]:
import numpy as np
n = 100
a = [[1.0 / (i - j + 0.5) for i in xrange(n)] for j in xrange(n)]
a = np.array(a)
ev = np.linalg.eigvals(a)
#print 'Eigenvalues using Lapack function:', ev
#There is a special numpy function for chacteristic polynomial
cf = np.poly(a)
ev_roots = np.roots(cf)
#print 'Coefficients of the polynomial:', cf
#print 'Polynomial roots:', ev_roots
plt.scatter(ev.real, ev.imag, marker='x', label='Lapack')
b = a + 1e-1 * np.random.randn(n, n)
ev_b = np.linalg.eigvals(b)
plt.scatter(ev_b.real, ev_b.imag, marker='o')
#plt.scatter(ev_roots.real, ev_roots.imag, marker='o', label='Brute force')
plt.legend(loc='best')
plt.xlabel('Real part')
plt.ylabel('Imaginary part')
Out[15]:
Morale:
There is a very interesting theorem that often helps to localize the eigenvalues. It is called Gershgorin theorem.
It states that all the eigenvalues $\lambda_i, \quad i = 1, \ldots, n$ are located inside the union of Gershgorin circles $C_i$, where $C_i$ is a disk on the complex plane with center $a_{ii}$ and radius
$$r_i = \sum_{j \ne i} |a_{ij}|.$$
Moreover, if the circles do not intersect, that contain only one eigenvalues. The proof is instructive, since it uses the concepts we looked at the previous lectures.
We separate the diagonal part and off-diagonal part, and get
$$
A = D + S,
$$
and $\Vert D^{-1} S\Vert_1 < 1$. Therefore, by using the Neumann series, the matrix $D + S$ is invertible.
Now the proof follows by contradiction: if any of the eigenvalues lies outside all of the circles, the matrix $(A - \lambda I)$ is strictly diagonally dominant, and thus is invertible. That means, that $(A - \lambda I) x = 0$ means $x = 0$.
In [20]:
import numpy as np
%matplotlib inline
n = 30
fig, ax = plt.subplots(1, 1)
a = [[1.0 / (i - j + 0.5) for i in xrange(n)] for j in xrange(n)]
#a = np.array(a)
#a = np.diag(np.arange(n))
#a = a + 0.1*np.random.randn(n, n)
u = np.random.randn(n, n)
a = np.linalg.inv(u).dot(a).dot(u)
xg = np.diag(a).real
yg = np.diag(a).imag
rg = np.zeros(n)
ev = np.linalg.eigvals(a)
for i in xrange(n):
rg[i] = np.sum(np.abs(a[i, :])) - np.abs(a[i, i])
crc = plt.Circle((xg[i], yg[i]), radius=rg[i], fill=False)
ax.add_patch(crc)
plt.scatter(ev.real, ev.imag, label='x', color='r')
plt.axis('equal')
ax.set_title('Eigenvalues and Gershgorin circles')
fig.tight_layout()
Note: There are more complicated figures, like Cassini ovals, that include the spectra.
We are often interested in the computation of the part of the spectrum, like the largest eigenvalues, smallest eigenvalues. Also it is interesting to note that for the Hermitian matrices $(A = A^*)$ the eigenvalues are always real (prove it!).
A power method is the simplest method for the computation of the largest eigenvalue in modulus. It is also our first example of the iterative method and Krylov method.
Power method has the form $$ x_{k+1} = A x_k, \quad x_{k+1} := \frac{x_{k+1}}{\Vert x_{k+1} \Vert}. $$ The convergence is geometric, but the convergence ratio is $q^k$, where $q = |\frac{\lambda_{2}}{\lambda_{1}}| \leq 1$ and $k$ is the number of iterations. It means, the convergence can be artitrary small. To prove it, it is sufficient to consider a $2 \times 2$ diagonal matrix.
There is one class of matrices when eigenvalues can be found easily: triangular matrices
$$ A = \begin{pmatrix} \lambda_1 & * & * \\ 0 & \lambda_2 & * \\ 0 & 0 & \lambda_3 \\ \end{pmatrix}. $$The eigenvalues of $A$ are $\lambda_1, \lambda_2, \lambda_3$. Why?
Because the determinant is
$$
\det(A - \lambda I) = (\lambda - \lambda_1) (\lambda - \lambda_2) (\lambda - \lambda_3).
$$
Thus, computing the eigenvalues of triangular matrices is easy. Now, the unitary matrices come to help. Let $U$ be a unitary matrix, i.e. $U^* U = I$. Then
$$
\det(A - \lambda I) = \det(U (U^* A U - \lambda I) U^*) = \det(UU^*) \det(U^* A U - \lambda I) = \det(U^* A U - \lambda I),
$$
where we have used the famous multiplicativity property of the determinant, $\det(AB) = \det(A) \det(B)$. It means,
that the matrices $U^* A U$ and $A$ have the same characteristic polynomials, and the same eigenvalues.
If we manage to make $U^* A U = T$, where $T$ is upper triangular then we are done. Multplying from the left and the right by $U$ and $U^*$ respectively, we get the desired decomposition:
$$
A = U T U^*.
$$
This is the celebrated Schur decomposition. Recall that unitary matrices mean stability, thus the eigenvalues are computed very accuratedly. The Schur decomposition shows why we need matrix decompositions: it represents a matrix into a product of three matrices with structure.
Schur theorem
Every $n \times n$ matrix can be represented in the Schur form.
Sketch of the proof.
Theorem: $A$ is a normal matrix, iff $A = U \Lambda U^*$.
Sketch of the proof: One way is obvious (if the decomposition holds, the matrix is normal). The other is more complicated. Consider the Schur form of the matrix $A$. Then $AA^* = A^*A$ means $TT^* = T^* T$. By looking at the elements we immediately see,
that the only uppertriangular matrix $T$ that satisfies $TT^* = T^* T$ is a diagonal matrix!
In [137]:
from IPython.core.display import HTML
def css_styling():
styles = open("./styles/custom.css", "r").read()
return HTML(styles)
css_styling()
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