# 6. Singular Value Decomposition (SVD)¶

In addition to regular packages contained in Anaconda by default, this notebook also requires:

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import numpy as np
import numpy.linalg as LA
import matplotlib.pyplot as plt
%matplotlib inline
import quandl as ql
import pandas as pd


## 6.1. Overview¶

The singular value decomposition is a work-horse in applications of least squares projection that form the backbone of important parts of modern machine learning methods.

This lecture describes the singular value decomposition and two of its uses:

• principal components analysis (PCA)

• dynamic mode decomposition (DMD)

Each of these can be thought of as data-reduction methods that are designed to capture principal patterns in data by projecting data onto a limited set of factors.

## 6.2. The Setup¶

Let $$X$$ be an $$m \times n$$ matrix of rank $$r$$.

In this notebook, we’ll think of $$X$$ as a matrix of data.

• each column is an individual – a time period or person, depending on the application

• each row is a random variable measuring an attribute of a time period or a person, depending on the application

We’ll be interested in two distinct cases

• The short and fat case in which $$m << n$$, so that there are many more columns than rows.

• The tall and skinny case in which $$m >> n$$, so that there are many more rows than columns.

We’ll apply a singular value decomposition of $$X$$ in both situations.

In the first case in which there are many more observations $$n$$ than there are random variables $$m$$, we learn about the joint distribution of the random variables by taking averages across observations of functions of the observations. Here we’ll look for patterns by using a singular value decomosition to do a principal components analysis (PCA).

In the second case in which there are many more random variables $$m$$ than observations $$n$$, we’ll proceed in a different way. We’ll again use a singular value decomposition, but now to do a dynamic mode decomposition (DMD)

## 6.3. Singular Value Decomposition¶

The singular value decomposition of an $$m \times n$$ matrix $$X$$ of rank $$r \leq \min(m,n)$$ is

$X = U \Sigma V^T$

where

\begin{align*} UU^T & = I & \quad U^T U = I \cr
VV^T & = I & \quad V^T V = I \end{align*}

where

• $$U$$ is an $$m \times m$$ matrix whose columns are eigenvectors of $$X^T X$$

• $$V$$ is an $$n \times n$$ matrix whose columns are eigenvectors of $$X X^T$$

• $$\Sigma$$ is an $$m \times r$$ matrix in which the first $$r$$ places on its main diagonal are positive numbers $$\sigma_1, \sigma_2, \ldots, \sigma_r$$ called singular values; remaining entries of $$\Sigma$$ are all zero

• The $$r$$ singular values are square roots of the eigenvalues of the $$m \times m$$ matrix $$X X^T$$ and the $$n \times n$$ matrix $$X^T X$$

• When $$U$$ is a complex valued matrix, $$U^T$$ denotes the conjugate-transpose or Hermitian-transpose of $$U$$, meaning that $$U_{ij}^T$$ is the complex conjugate of $$U_{ji}$$. Similarly, when $$V$$ is a complex valued matrix, $$V^T$$ denotes the conjugate-transpose or Hermitian-transpose of $$V$$

The shapes of $$U$$, $$\Sigma$$, and $$V$$ are $$\left(m, m\right)$$, $$\left(m, n\right)$$, $$\left(n, n\right)$$, respectively.

Below, we shall assume these shapes.

However, there is an alternative shape convention that we could have used, though we chose not to.

Thus, note that because we assume that $$A$$ has rank $$r$$, there are only $$r$$ nonzero singular values, where $$r=rank(A)\leq\min\left(m, n\right)$$.

Therefore, we could also write $$U$$, $$\Sigma$$, and $$V$$ as matrices with shapes $$\left(m, r\right)$$, $$\left(r, r\right)$$, $$\left(r, n\right)$$.

Sometimes, we will choose the former one to be consistent with what is adopted by numpy.

At other times, we’ll use the latter convention in which $$\Sigma$$ is an $$r \times r$$ diagonal matrix.

Also, when we discuss the dynamic mode decomposition below, we’ll use a special case of the latter convention in which it is understood that $$r$$ is just a pre-specified small number of leading singular values that we think capture the most interesting dynamics.

## 6.4. Digression: the polar decomposition¶

Through the following identities, the singular value decomposition (SVD) is related to the polar decomposition of $$X$$

\begin{align*} X & = SQ \cr
S & = U\Sigma U^T \cr Q & = U V^T \end{align*}

where $$S$$ is evidently a symmetric matrix and $$Q$$ is an orthogonal matrix.

## 6.5. Principle Componenents Analysis (PCA)¶

Let’s begin with the case in which $$n >> m$$, so that we have many more observations $$n$$ than random variables $$m$$.

The data matrix $$X$$ is short and fat in the $$n >> m$$ case as opposed to a tall and skinny case with $$m > > n$$ to be discussed later in this notebook.

We regard $$X$$ as an $$m \times n$$ matrix of data:

$X = \begin{bmatrix} X_1 \mid X_2 \mid \cdots \mid X_n\end{bmatrix}$

where for $$j = 1, \ldots, n$$ the column vector $$X_j = \begin{bmatrix}X_{1j}\\X_{2j}\\\vdots\\X_{mj}\end{bmatrix}$$ is a vector of observations on variables $$\begin{bmatrix}x_1\\x_2\\\vdots\\x_m\end{bmatrix}$$.

In a time series setting, we would think of columns $$j$$ as indexing different times at which random variables are observed, while rows index different random variables.

In a cross section setting, we would think of columns $$j$$ as indexing different individuals for which random variables are observed, while rows index different random variables.

The number of singular values equals the rank of matrix $$X$$.

Arrange the singular values in decreasing order.

Arrange the positive singular values on the main diagonal of the matrix $$\Sigma$$ of into a vector $$\sigma_R$$.

Set all other entries of $$\Sigma$$ to zero.

## 6.6. Relationship of PCA to SVD¶

To relate a SVD to a PCA (principal component analysis) of data set $$X$$, first construct the SVD of the data matrix $$X$$:

(6.1)$X = U \Sigma V^T = \sigma_1 U_1 V_1^T + \sigma_2 U_2 V_2^T + \cdots + \sigma_r U_r V_r^T$

where

$U=\begin{bmatrix}U_1|U_2|\ldots|U_m\end{bmatrix}$
$\begin{split} V^T = \begin{bmatrix}V_1^T\\V_2^T\\\ldots\\V_n^T\end{bmatrix} \end{split}$

In equation (6.1), each of the $$m \times n$$ matrices $$U_{j}V_{j}^T$$ is evidently of rank $$1$$.

Thus, we have

(6.2)$\begin{split} X = \sigma_1 \begin{pmatrix}U_{11}V_{1}^T\\U_{21}V_{1}^T\\\cdots\\U_{m1}V_{1}^T\\\end{pmatrix} + \sigma_2\begin{pmatrix}U_{12}V_{2}^T\\U_{22}V_{2}^T\\\cdots\\U_{m2}V_{2}^T\\\end{pmatrix}+\ldots + \sigma_r\begin{pmatrix}U_{1r}V_{r}^T\\U_{2r}V_{r}^T\\\cdots\\U_{mr}V_{r}^T\\\end{pmatrix} \end{split}$

Here is how we would interpret the objects in the matrix equation (6.2) in a time series context:

• $$V_{k}^T= \begin{bmatrix}V_{k1} & V_{k2} & \ldots & V_{kn}\end{bmatrix} \quad \textrm{for each} \ k=1, \ldots, n$$ is a time series $$\lbrace V_{kj} \rbrace_{j=1}^n$$ for the $$k$$th principal component

• $$U_j = \begin{bmatrix}U_{1k}\\U_{2k}\\\ldots\\U_{mk}\end{bmatrix} \ k=1, \ldots, m$$ is a vector of loadings of variables $$X_i$$ on the $$k$$th principle component, $$i=1, \ldots, m$$

• $$\sigma_k$$ for each $$k=1, \ldots, r$$ is the strength of $$k$$th principal component

## 6.7. Digression: reduced (or economy) versus full SVD¶

Let’s do a small experiment to see the difference

import numpy as np
X = np.random.rand(5,2)
U, S, V = np.linalg.svd(X,full_matrices=True)  # full SVD
Uhat, Shat, Vhat = np.linalg.svd(X,full_matrices=False) # economy SVD
print('U, S, V ='), U, S, V

U, S, V =

(None,
array([[-0.44001183,  0.74675094, -0.40132959,  0.0192523 , -0.29549371],
[-0.19356345, -0.31010806,  0.01996547, -0.73520809, -0.57047054],
[-0.41871908,  0.21027745,  0.8797327 ,  0.07273434, -0.03518258],
[-0.4108698 , -0.49742115, -0.14503514,  0.63000638, -0.40720385],
[-0.65175388, -0.23356288, -0.20873696, -0.23853741,  0.64822376]]),
array([1.87500521, 0.75576908]),
array([[-0.66574138, -0.74618256],
[-0.74618256,  0.66574138]]))

print('Uhat, Shat, Vhat = '), Uhat, Shat, Vhat

Uhat, Shat, Vhat =

(None,
array([[-0.44001183,  0.74675094],
[-0.19356345, -0.31010806],
[-0.41871908,  0.21027745],
[-0.4108698 , -0.49742115],
[-0.65175388, -0.23356288]]),
array([1.87500521, 0.75576908]),
array([[-0.66574138, -0.74618256],
[-0.74618256,  0.66574138]]))

rr = np.linalg.matrix_rank(X)
rr

2


## 6.8. PCA with eigenvalues and eigenvectors¶

We now turn to using the eigen decomposition of a sample covariance matrix to do PCA.

Let $$X_{m \times n}$$ be our $$m \times n$$ data matrix.

Let’s assume that sample means of all variables are zero.

We can make sure that this is true by pre-processing the data by substracting sample means appropriately.

Define the sample covariance matrix $$\Omega$$ as

$\Omega = XX^T$

Then use an eigen decomposition to represent $$\Omega$$ as follows:

$\Omega =P\Lambda P^T$

Here

• $$P$$ is $$m×m$$ matrix of eigenvectors of $$\Omega$$

• $$\Lambda$$ is a diagonal matrix of eigenvalues of $$\Omega$$

We can then represent $$X$$ as

$X=P\epsilon$

where

$\epsilon\epsilon^T=\Lambda$

We can verify that

$XX^T=P\Lambda P^T$

It follows that we can represent the data matrix as

\begin{equation*} X=\begin{bmatrix}X_1|X_2|\ldots|X_m\end{bmatrix} =\begin{bmatrix}P_1|P_2|\ldots|P_m\end{bmatrix} \begin{bmatrix}\epsilon_1\epsilon_2\ldots\epsilon_m\end{bmatrix} = P_1\epsilon_1+P_2\epsilon_2+\ldots+P_m\epsilon_m \end{equation*}

where

$\epsilon\epsilon^T=\Lambda$

To reconcile the preceding representation with the PCA that we obtained through the SVD above, we first note that $$\epsilon_j^2=\lambda_j\equiv\sigma^2_j$$.

Now define $$\tilde{\epsilon_j} = \frac{\epsilon_j}{\sqrt{\lambda_j}}$$ which evidently implies that $$\tilde{\epsilon}_j\tilde{\epsilon}_j^T=1$$.

Therefore

\begin{aligned} X&=\sqrt{\lambda_1}P_1\tilde{\epsilon_1}+\sqrt{\lambda_2}P_2\tilde{\epsilon_2}+\ldots+\sqrt{\lambda_m}P_m\tilde{\epsilon_m}\ &=\sigma_1P_1\tilde{\epsilon_2}+\sigma_2P_2\tilde{\epsilon_2}+\ldots+\sigma_mP_m\tilde{\epsilon_m} \end{aligned}

which evidently agrees with

$X=\sigma_1U_1{V_1}^{T}+\sigma_2 U_2{V_2}^{T}+\ldots+\sigma_{r} U_{r}{V_{r}}^{T}$

provided that we set

• $$U_j=P_j$$ (the loadings of variables on principal components)

• $${V_k}^{T}=\tilde{\epsilon_k}$$ (the principal components)

Since there are several possible ways of computing $$P$$ and $$U$$ for given a data matrix $$X$$, depending on algorithms used, we might have sign differences or different orders between eigenvectors.

We want a way that leads to the same $$U$$ and $$P$$.

In the following, we accomplish this by

1. sorting eigenvalues and singular values in descending order

2. imposing positive diagonals on $$P$$ and $$U$$ and adjusting signs in $$V^T$$ accordingly

## 6.9. Summary of Connections¶

To pull things together, it is useful to assemble and compare some formulas presented above.

First, consider the following SVD of an $$m \times n$$ matrix:

$X = U\Sigma V^T$

Compute:

\begin{align*} XX^T&=U\Sigma V^TV\Sigma^T U^T\cr &\equiv U\Sigma\Sigma^TU^T\cr &\equiv U\Lambda U^T \end{align*}

Thus, $$U$$ in the SVD is the matrix $$P$$ of eigenvectors of $$XX^T$$ and $$\Sigma \Sigma^T$$ is the matrix $$\Lambda$$ of eigenvalues.

Second, let’s compute

\begin{align*} X^TX &=V\Sigma^T U^TU\Sigma V^T\ &=V\Sigma^T{\Sigma}V^T \end{align*}

Thus, the matrix $$V$$ in the SVD is the matrix of eigenvectors of $$X^TX$$

Summarizing and fitting things together, we have the eigen decomposition of the sample covariance matrix

$X X^T = P \Lambda P^T$

where $$P$$ is an orthogonal matrix.

Further, from the SVD of $$X$$, we know that

$X X^T = U \Sigma \Sigma^T U^T$

where $$U$$ is an orthonal matrix.

Thus, $$P = U$$ and we have the representation of $$X$$

$X = P \epsilon = U \Sigma V^T$

It follows that

$U^T X = \Sigma V^T = \epsilon$

Note that the preceding implies that

$\epsilon \epsilon^T = \Sigma V^T V \Sigma^T = \Sigma \Sigma^T = \Lambda ,$

so that everything fits together.

Below we define a class DecomAnalysis that wraps PCA and SVD for a given a data matrix X.

class DecomAnalysis:
"""
A class for conducting PCA and SVD.
"""

def __init__(self, X, n_component=None):

self.X = X

self.Ω = (X @ X.T)

self.m, self.n = X.shape
self.r = LA.matrix_rank(X)

if n_component:
self.n_component = n_component
else:
self.n_component = self.m

def pca(self):

𝜆, P = LA.eigh(self.Ω)    # columns of P are eigenvectors

ind = sorted(range(𝜆.size), key=lambda x: 𝜆[x], reverse=True)

# sort by eigenvalues
self.𝜆 = 𝜆[ind]
P = P[:, ind]
self.P = P @ diag_sign(P)

self.Λ = np.diag(self.𝜆)

self.explained_ratio_pca = np.cumsum(self.𝜆) / self.𝜆.sum()

# compute the N by T matrix of principal components
self.𝜖 = self.P.T @ self.X

P = self.P[:, :self.n_component]
𝜖 = self.𝜖[:self.n_component, :]

# transform data
self.X_pca = P @ 𝜖

def svd(self):

U, 𝜎, VT = LA.svd(self.X)

ind = sorted(range(𝜎.size), key=lambda x: 𝜎[x], reverse=True)

# sort by eigenvalues
d = min(self.m, self.n)

self.𝜎 = 𝜎[ind]
U = U[:, ind]
D = diag_sign(U)
self.U = U @ D
VT[:d, :] = D @ VT[ind, :]
self.VT = VT

self.Σ = np.zeros((self.m, self.n))
self.Σ[:d, :d] = np.diag(self.𝜎)

𝜎_sq = self.𝜎 ** 2
self.explained_ratio_svd = np.cumsum(𝜎_sq) / 𝜎_sq.sum()

# slicing matrices by the number of components to use
U = self.U[:, :self.n_component]
Σ = self.Σ[:self.n_component, :self.n_component]
VT = self.VT[:self.n_component, :]

# transform data
self.X_svd = U @ Σ @ VT

def fit(self, n_component):

# pca
P = self.P[:, :n_component]
𝜖 = self.𝜖[:n_component, :]

# transform data
self.X_pca = P @ 𝜖

# svd
U = self.U[:, :n_component]
Σ = self.Σ[:n_component, :n_component]
VT = self.VT[:n_component, :]

# transform data
self.X_svd = U @ Σ @ VT

def diag_sign(A):
"Compute the signs of the diagonal of matrix A"

D = np.diag(np.sign(np.diag(A)))

return D


We also define a function that prints out information so that we can compare decompositions obtained by different algorithms.

def compare_pca_svd(da):
"""
Compare the outcomes of PCA and SVD.
"""

da.pca()
da.svd()

print('Eigenvalues and Singular values\n')
print(f'λ = {da.λ}\n')
print(f'σ^2 = {da.σ**2}\n')
print('\n')

fig, axs = plt.subplots(1, 2, figsize=(14, 5))
axs[0].plot(da.P.T)
axs[0].set_title('P')
axs[0].set_xlabel('m')
axs[1].plot(da.U.T)
axs[1].set_title('U')
axs[1].set_xlabel('m')
plt.show()

# principal components
fig, axs = plt.subplots(1, 2, figsize=(14, 5))
plt.suptitle('principal components')
axs[0].plot(da.ε.T)
axs[0].set_title('ε')
axs[0].set_xlabel('n')
axs[1].plot(da.VT[:da.r, :].T * np.sqrt(da.λ))
axs[1].set_title('$V^T*\sqrt{\lambda}$')
axs[1].set_xlabel('n')
plt.show()


## 6.10. Dynamic Mode Decomposition (DMD)¶

We now turn to the case in which $$m >>n$$ so that there are many more random variables $$m$$ than observations $$n$$.

This is the tall and skinny case associated with Dynamic Mode Decomposition.

Starting with an $$m \times n$$ matrix of data $$X$$, we form two matrices

$\tilde X = \begin{bmatrix} X_1 \mid X_2 \mid \cdots \mid X_{n-1}\end{bmatrix}$

and

$\tilde X' = \begin{bmatrix} X_2 \mid X_3 \mid \cdots \mid X_n\end{bmatrix}$

In forming $$\tilde X$$ and $$\tilde X'$$, we have in each case dropped a column from $$X$$.

Evidently, $$\tilde X$$ and $$\tilde X'$$ are both $$m \times \tilde n$$ matrices where $$\tilde n = n - 1$$.

We start with a system consisting of $$m$$ least squares regressions of everything on everything:

$\tilde X' = A \tilde X + \epsilon$

where

$A = \tilde X' \tilde X^{+}$

and where the (huge) $$m \times m$$ matrix $$X^{+}$$ is the Moore-Penrose generalize inverse of $$X$$ that we could compute as

$X^{+} = V \Sigma^{-1} U^T$

where the matrix $$\Sigma^{-1}$$ is constructed by replacing each non-zero element of $$\Sigma$$ with $$\sigma_j^{-1}$$.

The idea behind dynamic mode decomposition is to construct an approximation that

• sidesteps computing $$X^{+}$$

• retains only the largest $$\tilde r< < r$$ eigenvalues and associated eigenvectors of $$U$$ and $$V^T$$

• constructs an $$m \times \tilde r$$ matrix $$\Phi$$ that captures effects of $$r$$ dynamic modes on all $$m$$ variables

• uses $$\Phi$$ and the $$\tilde r$$ leading singular values to forecast future $$X_t$$’s

The magic of dynamic mode decomposition is that we accomplish this without ever computing the regression coefficients $$A = X' X^{+}$$.

To accomplish a DMD, we deploy the following steps:

• Compute the singular value decomposition

$X = U \Sigma V^T$

where $$U$$ is $$m \times r$$, $$\Sigma$$ is an $$r \times r$$ diagonal matrix, and $$V^T$$ is an $$r \times \tilde n$$ matrix.

• Notice that (though it would be costly), we could compute $$A$$ by solving

$A = X' V \Sigma^{-1} U^T$

But we won’t do that.

Note that since, $$X' = A U \Sigma V^T$$, we know that

$A U = X' V \Sigma^{-1}$

so that

$U^T X' V \Sigma^{-1} = U^T A U \equiv \tilde A$
• At this point, in constructing $$\tilde A$$ according to the above formula, we take only the columns of $$U$$ corresponding to the $$\tilde r$$ largest singular values.

Tu et al. verify that eigenvalues and eigenvectors of $$\tilde A$$ equal the leading eigenvalues and associated eigenvectors of $$A$$.

• Construct an eigencomposition of $$\tilde A$$ that satisfies

$\tilde A W = W \Lambda$

where $$\Lambda$$ is a $$\tilde r \times \tilde r$$ diagonal matrix of eigenvalues and the columns of $$W$$ are corresponding eigenvectors of $$\tilde A$$. Both $$\Lambda$$ and $$W$$ are $$\tilde r \times \tilde r$$ matrices.

• Construct the $$m \times \tilde r$$ matrix

$\Phi = X' V \Sigma^{-1} W$

Let $$\Phi^{+}$$ be a generalized inverse of $$\Phi$$; $$\Phi^{+}$$ is an $$\tilde r \times m$$ matrix.

• Define an initial vector $$b$$ of dominant modes by

$b= \Phi^{+} X_1$

where evidently $$b$$ is an $$\tilde r \times 1$$ vector.

With $$\Lambda, \Phi$$ in hand, our least-squares fitted dynamics fitted to the $$r$$ dominant modes are governed by

$X_{t+1} = \Phi \Lambda \Phi^{+} X_t$

Conditional on $$X_t$$, forecasts $$\check X_{t+j}$$ of $$X_{t+j}, j = 1, 2, \ldots,$$ are evidently given by

$\check X_{t+j} = \Phi \Lambda^j \Phi^{+} X_t$

## 6.11. Source for some Python code¶

You can find a Python implementation of DMD here:

https://mathlab.github.io/PyDMD/