# Riemannian Newton iteration for Rayleigh quotients on the sphere

In the above figure, each point is in initial value which will be converge to different eigenvectors of an orthogonal 3x3 matrix. The above figure is the equivalent of a Newton fractal, but applied to Rayleigh quotient iteration on a sphere. This post will go through an explanation of the figure, and the numerical method on the sphere, which can be applied to any manifold.

### Rayleigh quotients and Eigenvectors

Given a Hermetian matrix $$M$$ and a vector $$x$$ the Rayleigh quotiend is defined by

$$R(M, x) = \frac{x^TMx}{x^T x}$$

For the sake of this text, let us limit ourselves to looking at $$M = A^T A$$ which are covariance matrices of some data matrix $$A$$. This implies that $$M$$ has unique real values eigenvalues. In addition we know that the eigenvectors of $$M$$ will be orthogonal.

Using these eigenvectors as a basis, let us express an arbitrary vector $$x$$ as a linear combination of eigenvectors.

$$x = \sum_{i = 1}^n \alpha_i v_i$$

Now we have that

$$R(M, x) = \frac{x^T A^T A x}{x^T x} = \frac{x^T A^T A x}{x^T x} =$$

$$= \frac{\left(\sum_{j = 1}^n \alpha_j v_j\right)^T \left( A^T A \right) \left(\sum_{i = 1}^n \alpha_i v_i\right)} {\left(\sum_{j = 1}^n \alpha_j v_j\right)^T \left( \sum_{i = 1}^n \alpha_i v_i \right)} =$$

$$= \frac{\sum_{i = 1}^n \alpha_i^2 \lambda_i}{\sum_{i = 1}^n \alpha_i^2} = \sum_{i = 1}^n \lambda_i \frac{(x^T v_i)^2}{(x^T x)(v_i^T v_I )}$$

since eigenvectors are orthogonal. Additionally, we see that the sum in the last equality will be minimized when $$x$$ is orthogonal to the majority of the $$v_i$$’s. Which is locally the case for any $$x = v_j$$, and globally when the eigenvector $$v_j$$ corresponds to the smallest eigenvalue.

As $$R(M, x)$$ is unaffected by lengths of vectors (due to division by the scalar product) it is defined on the sphere $$S^{n - 1}$$, which is a Riemannian manifold.

$$R(M, \cdot) : S^{n - 1} \rightarrow \mathbb{R}$$

### Newton Iteration

The Newton Method of iteration is defined in $$\mathbb{R}^n$$ by the procedure

1. Solve $$D(\text{grad } \ f)(x)[\eta_x] = -\text{grad } \ f(x)$$ for $$\eta_x$$
2. Set $$x_{i + 1} := x_i + \eta_x$$

In each update of the iteration we solve for the update vector $$\eta_x$$. Optimal points are found by finding the roots of the derivative of the function. Newton iteration is used to find local optima of differentiable functions by in each step making a local second order Taylor approximation of the given function. Since Newton iteration will have information about the local curvature it is normally quicker to converge than gradient descent.

A very classical assignment in courses on numerical methods is to implement the Newton method for root finding to find the complex roots of a polynomial. Usually this entails looking at which root the algorithm will find depending on initial condition. The result is the so called Newton Fractal. (Note that in this case we are looking for the roots of the function in question. For optimizing by finding stationary points we need to look at the derivative of the given function.)

For example, let us find the complex roots of the polynomial $$p(x) = x^3 - 1$$. The iteration becomes

1. Solve $$p(x_k) + p’(x_k) \cdot \Delta x_k = 0$$ for $$\Delta x_k$$
2. Set $$x_{k + 1} := x_k + \Delta x_k$$

Or in other words

$$x_{k + 1} = x_k - \frac{x_k^3 - 1}{2x_k^2}$$

We can simply implement this iteration in the following fashion using Python with NumPy.

f = np.poly1d([1, 0, 0, -1])  # x^3 - 1
fp = np.polyder(f)

def newton(x, y, max_iters):
c = complex(x, y)
for i in range(max_iters):
c = c - f(c) / fp(c)
if np.abs(f(c)) < 1e-4:
return c

return 0.0


Now let us use this to illustrate the dependence on the initial value of the iteration. We simply define a function which performs the iteration until convergence over a grid of starting values in the complex plane. We color the grid by which root is found.

def create_newton_fractal(min_x, max_x, min_y, max_y, image, iters):
height = image.shape[0]
width = image.shape[1]

pixel_size_x = (max_x - min_x) / width
pixel_size_y = (max_y - min_y) / height

for x in range(width):
real = min_x + x * pixel_size_x
for y in range(height):
imag = min_y + y * pixel_size_y
color = newton(real, imag, iters)
image[y, x] = color.imag + 1


This can be generalized to work on any Riemannian manifold $$\mathcal{M}$$ with an affine connection $$\nabla$$ to optimize real valued functions $$f$$. Define the Hessian by

$$\text{Hess} \ f(x_k):= \nabla_{\eta_k} \text{grad } \ f.$$

Also let $$R : T\mathcal{M} \rightarrow \mathcal{M}$$ be a retraction on $$\mathcal{M}$$, a smooth mapping which satisfies

1. $$R(0_x) = x$$, where $$0_x$$ is the origin of $$T_x\mathcal{M}$$
2. $$\frac{d}{dt} R(t \xi_x) \vert_{t=0} = \xi_x$$

Now we can define the new iteration method by

1. Solve $$\text{Hess} \ f(x_k) \eta_k = -\text{grad } \ f(x_k)$$ for $$\eta_k \in T_{x_k} \mathcal{M}$$
2. Set $$x_{k + 1} := R_{x_k}(\eta_k)$$

### Newton Iteration on Spheres

As a first step towards working on the sphere, we need to look at how the iteration would work for manifolds which are sub-manifolds of $$\mathbb{R}^n$$. In this case we pick $$\nabla$$ to be the Levi-Civita connection. What we need is a way of calculating the gradient in $$\mathcal{M}$$.

Every vector $$z \in T_x \mathbb{R}^n \cong \mathbb{R}^n$$ admits a decomposition $$z = P_x z + P^\perp_x z$$ where $$P_x z \in T_x\mathcal{M}$$ and $$P^\perp_x z \in T^\perp_x\mathcal{M}$$; the orthogonal complement to $$T_x\mathcal{M}$$ in $$\mathbb{R}^n$$. Now say that $$\bar{f} : \mathbb{R}^n \rightarrow \mathbb{R}$$, and the function $$f$$ is defined by the restriction $$f = \bar{f} \vert_\mathcal{M}$$. Then the gradient can be calculated by

$$\text{grad } \ f(x) = P_x \ \text{grad } \ \bar{f}(x)$$

Using this identity, we can now realize the Levi-Civita connection by

$$\nabla_{\eta_k} \text{grad} \ f = P_{x_k} \text{D} (\text{grad} \ f(x_k))[\eta_k] =: \text{Hess} \ f(x_k) \eta_k$$

So now in the step solving for $$\eta_k \in T_{x_k} \mathcal{M}$$ we need to know a partitioning.

Let us now pull back from this abstraction and consider the particular case when $$\mathcal{M} = S^{n - 1}$$ as a sub-manifold of $$\mathbb{R}^n$$. That means in this case we have $$f : S^{n - 1} \rightarrow \mathbb{R}^n$$. We let the partitioning projection $$P_x$$ in to $$T_x S^{n - 1}$$ be defined by

$$P_x := \left( I - xx^T \right).$$

(A projection to the orthogonal complement of the vector projection onto $$x$$)

We now have the iteration defined by

1: Solve

$$P_{x_k} \text{D}(\text{grad} \ f)(x_k)[\eta_k] = -\text{grad} f(x_k)$$

for $$\eta_k \in T_x S^{n - 1} \cong \mathbb{R}^n$$

2: Step and retract back to the sphere by setting

$$x_{k + 1} := \frac{x_k + \eta_k}{\Vert x_k + \eta_k \Vert}$$

### Newton Iteration of Rayleigh quotient

We now have all the necessary tools to solve the eigenvector problem by minimizing the Rayleigh quotient. For gradient evaluation, as $$\frac{\partial}{\partial x_j}(x^T x) = 2x_j$$ and $$\frac{\partial}{\partial x_j}(x^T M x) = 2(Mx)_j$$ it follows that

$$\text{grad} \ R(M, x) = \frac{2}{x^T x} \left( Mx - R(M, x) \cdot x \right)$$

This leads to the Newton equation

$$P_{x_k} M P_{x_k} \eta_k - \eta_k x_k^T M x_k = -P_{x_k} M x_k$$

We know have a concrete iteration method which we can implement and run using Python

I = np.eye(3)

# Retraction at x
def R(x, eta):
return (x + eta) / linalg.norm(x + eta)

# Small helper function
def f(A, x):
return dot(x.T, dot(A, x))

# Projection P
def P(xk):
return I - outer(xk, xk)

# Make a random orthogonal matrix
A = random.randint(0, 10, (3, 3))
A = dot(A, A.T)

# Pick starting point
x0 = np.array([0,-1,0])
xk = x0

# Iterate and store path
path = [xk]
for i in range(10):
Pxk = P(xk)
M = (Pxk.dot(A).dot(Pxk) - f(A, xk) * I)
y = -Pxk.dot(A).dot(xk)
eta_k = linalg.solve(M, y)
path.append(xk + eta_k)
xk = R(xk, eta_k)
path.append(xk)


Let us plot the path of the iteration over the sphere, in the following plot blue lines mean $$x_k + \eta_k$$ (updates in the tangent space) and orange lines $$R_{x_k}(\eta_k)$$ (retraction to the sphere).

In this case the iteration has already converged after five steps, the following five steps are also plotted, but are stable and can’t be seen. The iteration converges when the Rayleigh quotient is minimized, therefore, we have now found an eigenvector to the randomly chosen matrix A.

Now we have everything we need to generate the figure above. First we define a couple of functions.

def RNmS(A, x0):
# Iteration
N = 5
xk = x0
for k in range(0, N + 1):
Pxk = P(xk)
M = (Pxk.dot(A).dot(Pxk) - f(A, xk) * I)
y = -Pxk.dot(A).dot(xk)
eta_k = linalg.solve(M, y)
xk = R(xk, eta_k)

X = A.dot(xk) / xk
return X.mean()

def find_nearest_index(array, value):
index = ((array.real - value.real) ** 2 + \
(array.imag - value.imag) ** 2).argmin(0)

return index


The we pick a matrix and calculate the eigenvalues with the standard algorithm for comparison.

# Choice of 3x3 symmetric matrix for which we want to find the eigenvectors/-values.
# Note that different matrices produce very different images.
A = random.randint(0, 10, (3, 3))
A = dot(A, A.T)

EW, EV = linalg.eig(A)
print "Eigenvalues: ",
print EW


Finally we make a spherical grid, perform the iteration for every point in the grid, and plot the result.

# Resolution
n = 512
nj = n * 1j

# u and v are parametric variables.
u = r_[0:2*pi:nj]
v = r_[0:pi:nj]

x = outer(cos(u), sin(v))
y = outer(sin(u), sin(v))
z = outer(ones(size(u)), cos(v))

# Figure out which eigenvectors/-values the algorithm converges to
# for all the spherical initial vectors.
s = zeros((n, n))
for i in range(0, n):
for j in range(0, n):
try:
X = RNmS(A, array([x[i, j], y[i, j], z[i, j]]))
s[i, j] = find_nearest_index(EW, X)
except:
raise

print X

print "Indexes: ",
print set(s.flatten())

bmap = brewer2mpl.get_map('RdBu', 'Diverging', 3)
cmap = bmap.get_mpl_colormap(N=3, gamma=1.0)

figsize(50, 50)
fig = plt.figure()