5. Singular Value Decomposition (SVD)#

5.1. Overview#

The singular value decomposition (SVD) is a work-horse in applications of least squares projection that form foundations for many statistical and machine learning methods.

After defining the SVD, we’ll describe how it connects to

four fundamental spaces of linear algebra
under-determined and over-determined least squares regressions
principal components analysis (PCA)

Like principal components analysis (PCA), DMD can be thought of as a data-reduction procedure that represents salient patterns by projecting data onto a limited set of factors.

In a sequel to this lecture about Dynamic Mode Decompositions, we’ll describe how SVD’s provide ways rapidly to compute reduced-order approximations to first-order Vector Autoregressions (VARs).

5.2. The Setting#

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

Necessarily, $p \leq min (m, n)$ .

In much of this lecture, we’ll think of $X$ as a matrix of data in which

each column is an individual – a time period or person, depending on the application
each row is a random variable describing an attribute of a time period or a person, depending on the application

We’ll be interested in two situations

A short and fat case in which $m << n$ , so that there are many more columns (individuals) than rows (attributes).
A tall and skinny case in which $m >> n$ , so that there are many more rows (attributes) than columns (individuals).

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

In the $m << n$ case in which there are many more individuals $n$ than attributes $m$ , we can calculate sample moments of a joint distribution by taking averages across observations of functions of the observations.

In this $m << n$ case, we’ll look for patterns by using a singular value decomposition to do a principal components analysis (PCA).

In the $m >> n$ case in which there are many more attributes $m$ than individuals $n$ and when we are in a time-series setting in which $n$ equals the number of time periods covered in the data set $X$ , we’ll proceed in a different way.

We’ll again use a singular value decomposition, but now to construct a dynamic mode decomposition (DMD)

5.3. Singular Value Decomposition#

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

(5.1)#

X = U Σ V^{⊤}

where

\begin{aligned} U U^{⊤} & = I & U^{⊤} U = I \\ V V^{⊤} & = I & V^{⊤} V = I \end{aligned}

and

$U$ is an $m \times m$ orthogonal matrix of left singular vectors of $X$
Columns of $U$ are eigenvectors of $X X^{⊤}$
$V$ is an $n \times n$ orthogonal matrix of right singular vectors of $X$
Columns of $V$ are eigenvectors of $X^{⊤} X$
$Σ$ is an $m \times n$ matrix in which the first $p$ places on its main diagonal are positive numbers $σ_{1}, σ_{2}, \dots, σ_{p}$ called singular values; remaining entries of $Σ$ are all zero
The $p$ singular values are positive square roots of the eigenvalues of the $m \times m$ matrix $X X^{⊤}$ and also of the $n \times n$ matrix $X^{⊤} X$
We adopt a convention that when $U$ is a complex valued matrix, $U^{⊤}$ denotes the conjugate-transpose or Hermitian-transpose of $U$ , meaning that $U_{i j}^{⊤}$ is the complex conjugate of $U_{j i}$ .
Similarly, when $V$ is a complex valued matrix, $V^{⊤}$ denotes the conjugate-transpose or Hermitian-transpose of $V$

The matrices $U, Σ, V$ entail linear transformations that reshape in vectors in the following ways:

multiplying vectors by the unitary matrices $U$ and $V$ rotates them, but leaves angles between vectors and lengths of vectors unchanged.
multiplying vectors by the diagonal matrix $Σ$ leaves angles between vectors unchanged but rescales vectors.

Thus, representation (5.1) asserts that multiplying an $n \times 1$ vector $y$ by the $m \times n$ matrix $X$ amounts to performing the following three multiplications of $y$ sequentially:

rotating $y$ by computing $V^{⊤} y$
rescaling $V^{⊤} y$ by multiplying it by $Σ$
rotating $Σ V^{⊤} y$ by multiplying it by $U$

This structure of the $m \times n$ matrix $X$ opens the door to constructing systems of data encoders and decoders.

Thus,

$V^{⊤} y$ is an encoder
$Σ$ is an operator to be applied to the encoded data
$U$ is a decoder to be applied to the output from applying operator $Σ$ to the encoded data

We’ll apply this circle of ideas later in this lecture when we study Dynamic Mode Decomposition.

Road Ahead

What we have described above is called a full SVD.

In a full SVD, the shapes of $U$ , $Σ$ , and $V$ are $(m, m)$ , $(m, n)$ , $(n, n)$ , respectively.

Later we’ll also describe an economy or reduced SVD.

Before we study a reduced SVD we’ll say a little more about properties of a full SVD.

5.4. Four Fundamental Subspaces#

Let $C$ denote a column space, $N$ denote a null space, and $R$ denote a row space.

Let’s start by recalling the four fundamental subspaces of an $m \times n$ matrix $X$ of rank $p$ .

The column space of $X$ , denoted $C (X)$ , is the span of the columns of $X$ , i.e., all vectors $y$ that can be written as linear combinations of columns of $X$ . Its dimension is $p$ .
The null space of $X$ , denoted $N (X)$ consists of all vectors $y$ that satisfy $X y = 0$ . Its dimension is $n - p$ .
The row space of $X$ , denoted $R (X)$ is the column space of $X^{⊤}$ . It consists of all vectors $z$ that can be written as linear combinations of rows of $X$ . Its dimension is $p$ .
The left null space of $X$ , denoted $N (X^{⊤})$ , consist of all vectors $z$ such that $X^{⊤} z = 0$ . Its dimension is $m - p$ .

For a full SVD of a matrix $X$ , the matrix $U$ of left singular vectors and the matrix $V$ of right singular vectors contain orthogonal bases for all four subspaces.

They form two pairs of orthogonal subspaces that we’ll describe now.

Let $u_{i}, i = 1, \dots, m$ be the $m$ column vectors of $U$ and let $v_{i}, i = 1, \dots, n$ be the $n$ column vectors of $V$ .

Let’s write the full SVD of X as

(5.2)#

X = [\begin{matrix} U_{L} & U_{R} \end{matrix}] [\begin{matrix} Σ_{p} & 0 \\ 0 & 0 \end{matrix}] {[\begin{matrix} V_{L} & V_{R} \end{matrix}]}^{⊤}

where $Σ_{p}$ is a $p \times p$ diagonal matrix with the $p$ singular values on the diagonal and

\begin{aligned} U_{L} & = [\begin{array}{c} u_{1} & \dots & u_{p} \end{array}], U_{R} = [\begin{array}{c} u_{p + 1} & \dots u_{m} \end{array}] \\ V_{L} & = [\begin{array}{c} v_{1} & \dots & v_{p} \end{array}], U_{R} = [\begin{array}{c} v_{p + 1} & \dots u_{n} \end{array}] \end{aligned}

Representation (5.2) implies that

X [\begin{matrix} V_{L} & V_{R} \end{matrix}] = [\begin{matrix} U_{L} & U_{R} \end{matrix}] [\begin{matrix} Σ_{p} & 0 \\ 0 & 0 \end{matrix}]

(5.3)#

\begin{aligned} X V_{L} & = U_{L} Σ_{p} \\ X V_{R} & = 0 \end{aligned}

(5.4)#

\begin{aligned} X v_{i} & = σ_{i} u_{i}, i = 1, \dots, p \\ X v_{i} & = 0, i = p + 1, \dots, n \end{aligned}

Equations (5.4) tell how the transformation $X$ maps a pair of orthonormal vectors $v_{i}, v_{j}$ for $i$ and $j$ both less than or equal to the rank $p$ of $X$ into a pair of orthonormal vectors $u_{i}, u_{j}$ .

Equations (5.3) assert that

\begin{aligned} C (X) & = C (U_{L}) \\ N (X) & = C (V_{R}) \end{aligned}

Taking transposes on both sides of representation (5.2) implies

X^{⊤} [\begin{matrix} U_{L} & U_{R} \end{matrix}] = [\begin{matrix} V_{L} & V_{R} \end{matrix}] [\begin{matrix} Σ_{p} & 0 \\ 0 & 0 \end{matrix}]

(5.5)#

\begin{aligned} X^{⊤} U_{L} & = V_{L} Σ_{p} \\ X^{⊤} U_{R} & = 0 \end{aligned}

(5.6)#

\begin{aligned} X^{⊤} u_{i} & = σ_{i} v_{i}, i = 1, \dots, p \\ X^{⊤} u_{i} & = 0 i = p + 1, \dots, m \end{aligned}

Notice how equations (5.6) assert that the transformation $X^{⊤}$ maps a pair of distinct orthonormal vectors $u_{i}, u_{j}$ for $i$ and $j$ both less than or equal to the rank $p$ of $X$ into a pair of distinct orthonormal vectors $v_{i}, v_{j}$ .

Equations (5.5) assert that

\begin{aligned} R (X) & \equiv C (X^{⊤}) = C (V_{L}) \\ N (X^{⊤}) & = C (U_{R}) \end{aligned}

Thus, taken together, the systems of equations (5.3) and (5.5) describe the four fundamental subspaces of $X$ in the following ways:

(5.7)#

\begin{array}{r} \begin{matrix} \begin{aligned} C (X) & = C (U_{L}) \\ N (X^{⊤}) & = C (U_{R}) \\ R (X) & \equiv C (X^{⊤}) = C (V_{L}) \\ N (X) & = C (V_{R}) \end{aligned} \end{matrix} \end{array}

Since $U$ and $V$ are both orthonormal matrices, collection (5.7) asserts that

$U_{L}$ is an orthonormal basis for the column space of $X$
$U_{R}$ is an orthonormal basis for the null space of $X^{⊤}$
$V_{L}$ is an orthonormal basis for the row space of $X$
$V_{R}$ is an orthonormal basis for the null space of $X$

We have verified the four claims in (5.7) simply by performing the multiplications called for by the right side of (5.2) and reading them.

The claims in (5.7) and the fact that $U$ and $V$ are both unitary (i.e, orthonormal) matrices imply that

the column space of $X$ is orthogonal to the null space of $X^{⊤}$
the null space of $X$ is orthogonal to the row space of $X$

Sometimes these properties are described with the following two pairs of orthogonal complement subspaces:

$C (X)$ is the orthogonal complement of $N (X^{⊤})$
$R (X)$ is the orthogonal complement $N (X)$

Let’s do an example.

import numpy as np
import numpy.linalg as LA
import matplotlib.pyplot as plt

Having imported these modules, let’s do the example.

np.set_printoptions(precision=2)

# Define the matrix
A = np.array([[1, 2, 3, 4, 5],
              [2, 3, 4, 5, 6],
              [3, 4, 5, 6, 7],
              [4, 5, 6, 7, 8],
              [5, 6, 7, 8, 9]])

# Compute the SVD of the matrix
U, S, V = np.linalg.svd(A,full_matrices=True)

# Compute the rank of the matrix
rank = np.linalg.matrix_rank(A)

# Print the rank of the matrix
print("Rank of matrix:\n", rank)
print("S: \n", S)

# Compute the four fundamental subspaces
row_space = U[:, :rank]
col_space = V[:, :rank]
null_space = V[:, rank:]
left_null_space = U[:, rank:]


print("U:\n", U)
print("Column space:\n", col_space)
print("Left null space:\n", left_null_space)
print("V.T:\n", V.T)
print("Row space:\n", row_space.T)
print("Right null space:\n", null_space.T)

Rank of matrix:
 2
S: 
 [2.69e+01 1.86e+00 1.20e-15 2.24e-16 5.82e-17]
U:
 [[-0.27 -0.73  0.63 -0.06  0.06]
 [-0.35 -0.42 -0.69 -0.45  0.12]
 [-0.43 -0.11 -0.24  0.85  0.12]
 [-0.51  0.19  0.06 -0.1  -0.83]
 [-0.59  0.5   0.25 -0.24  0.53]]
Column space:
 [[-0.27 -0.35]
 [ 0.73  0.42]
 [ 0.32 -0.65]
 [ 0.54 -0.39]
 [-0.06 -0.35]]
Left null space:
 [[ 0.63 -0.06  0.06]
 [-0.69 -0.45  0.12]
 [-0.24  0.85  0.12]
 [ 0.06 -0.1  -0.83]
 [ 0.25 -0.24  0.53]]
V.T:
 [[-0.27  0.73  0.32  0.54 -0.06]
 [-0.35  0.42 -0.65 -0.39 -0.35]
 [-0.43  0.11  0.02 -0.29  0.85]
 [-0.51 -0.19  0.61 -0.41 -0.4 ]
 [-0.59 -0.5  -0.31  0.55 -0.04]]
Row space:
 [[-0.27 -0.35 -0.43 -0.51 -0.59]
 [-0.73 -0.42 -0.11  0.19  0.5 ]]
Right null space:
 [[-0.43  0.11  0.02 -0.29  0.85]
 [-0.51 -0.19  0.61 -0.41 -0.4 ]
 [-0.59 -0.5  -0.31  0.55 -0.04]]

5.5. Eckart-Young Theorem#

Suppose that we want to construct the best rank $r$ approximation of an $m \times n$ matrix $X$ .

By best, we mean a matrix $X_{r}$ of rank $r < p$ that, among all rank $r$ matrices, minimizes

| | X - X_{r} | |

where $| | \cdot | |$ denotes a norm of a matrix $X$ and where $X_{r}$ belongs to the space of all rank $r$ matrices of dimension $m \times n$ .

Three popular matrix norms of an $m \times n$ matrix $X$ can be expressed in terms of the singular values of $X$

the spectral or $l^{2}$ norm $| | X | |_{2} = max_{| | y | | \neq 0} \frac{| | X y | |}{| | y | |} = σ_{1}$
the Frobenius norm $| | X | |_{F} = \sqrt{σ_{1}^{2} + \dots + σ_{p}^{2}}$
the nuclear norm $| | X | |_{N} = σ_{1} + \dots + σ_{p}$

The Eckart-Young theorem states that for each of these three norms, same rank $r$ matrix is best and that it equals

(5.8)#

{\hat{X}}_{r} = σ_{1} U_{1} V_{1}^{⊤} + σ_{2} U_{2} V_{2}^{⊤} + \dots + σ_{r} U_{r} V_{r}^{⊤}

This is a very powerful theorem that says that we can take our $m \times n$ matrix $X$ that in not full rank, and we can best approximate it by a full rank $p \times p$ matrix through the SVD.

Moreover, if some of these $p$ singular values carry more information than others, and if we want to have the most amount of information with the least amount of data, we can take $r$ leading singular values ordered by magnitude.

We’ll say more about this later when we present Principal Component Analysis.

You can read about the Eckart-Young theorem and some of its uses here.

We’ll make use of this theorem when we discuss principal components analysis (PCA) and also dynamic mode decomposition (DMD).

5.6. Full and Reduced SVD’s#

Up to now we have described properties of a full SVD in which shapes of $U$ , $Σ$ , and $V$ are $(m, m)$ , $(m, n)$ , $(n, n)$ , respectively.

There is an alternative bookkeeping convention called an economy or reduced SVD in which the shapes of $U, Σ$ and $V$ are different from what they are in a full SVD.

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

A reduced SVD uses this fact to express $U$ , $Σ$ , and $V$ as matrices with shapes $(m, p)$ , $(p, p)$ , $(n, p)$ .

You can read about reduced and full SVD here https://numpy.org/doc/stable/reference/generated/numpy.linalg.svd.html

For a full SVD,

\begin{aligned} U U^{⊤} & = I & U^{⊤} U = I \\ V V^{⊤} & = I & V^{⊤} V = I \end{aligned}

But not all these properties hold for a reduced SVD.

Which properties hold depend on whether we are in a tall-skinny case or a short-fat case.

In a tall-skinny case in which $m >> n$ , for a reduced SVD

\begin{aligned} U U^{⊤} & \neq I & U^{⊤} U = I \\ V V^{⊤} & = I & V^{⊤} V = I \end{aligned}

In a short-fat case in which $m << n$ , for a reduced SVD

\begin{aligned} U U^{⊤} & = I & U^{⊤} U = I \\ V V^{⊤} & = I & V^{⊤} V \neq I \end{aligned}

When we study Dynamic Mode Decomposition below, we shall want to remember these properties when we use a reduced SVD to compute some DMD representations.

Let’s do an exercise to compare full and reduced SVD’s.

To review,

in a full SVD
- $U$ is $m \times m$
- $Σ$ is $m \times n$
- $V$ is $n \times n$
in a reduced SVD
- $U$ is $m \times p$
- $Σ$ is $p \times p$
- $V$ is $n \times p$

First, let’s study a case in which $m = 5 > n = 2$ .

(This is a small example of the tall-skinny case that will concern us when we study Dynamic Mode Decompositions below.)

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 =

(array([[-0.59,  0.32, -0.68, -0.22,  0.17],
        [-0.47, -0.35,  0.26, -0.51, -0.58],
        [-0.42,  0.57,  0.67,  0.07,  0.24],
        [-0.47, -0.33, -0.04,  0.8 , -0.14],
        [-0.18, -0.59,  0.14, -0.2 ,  0.75]]),
 array([1.79, 0.45]),
 array([[-0.88, -0.47],
        [ 0.47, -0.88]]))

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

Uhat, Shat, Vhat =

(array([[-0.59,  0.32],
        [-0.47, -0.35],
        [-0.42,  0.57],
        [-0.47, -0.33],
        [-0.18, -0.59]]),
 array([1.79, 0.45]),
 array([[-0.88, -0.47],
        [ 0.47, -0.88]]))

rr = np.linalg.matrix_rank(X)
print(f'rank of X = {rr}')

rank of X = 2

Properties:

Where $U$ is constructed via a full SVD, $U^{⊤} U = I_{m \times m}$ and $U U^{⊤} = I_{m \times m}$
Where $\hat{U}$ is constructed via a reduced SVD, although ${\hat{U}}^{⊤} \hat{U} = I_{p \times p}$ , it happens that $\hat{U} {\hat{U}}^{⊤} \neq I_{m \times m}$

We illustrate these properties for our example with the following code cells.

UTU = U.T@U
UUT = U@U.T
print('UUT, UTU = ')
UUT, UTU

UUT, UTU =

(array([[ 1.00e+00,  1.88e-16,  2.23e-16,  1.65e-16, -5.60e-18],
        [ 1.88e-16,  1.00e+00,  3.96e-17,  1.61e-16, -3.12e-17],
        [ 2.23e-16,  3.96e-17,  1.00e+00,  3.36e-17, -6.92e-17],
        [ 1.65e-16,  1.61e-16,  3.36e-17,  1.00e+00,  7.54e-17],
        [-5.60e-18, -3.12e-17, -6.92e-17,  7.54e-17,  1.00e+00]]),
 array([[ 1.00e+00, -9.71e-17,  1.62e-16,  7.46e-17, -6.64e-17],
        [-9.71e-17,  1.00e+00, -2.53e-17, -3.18e-17, -1.39e-16],
        [ 1.62e-16, -2.53e-17,  1.00e+00,  5.77e-17,  2.66e-17],
        [ 7.46e-17, -3.18e-17,  5.77e-17,  1.00e+00, -4.94e-17],
        [-6.64e-17, -1.39e-16,  2.66e-17, -4.94e-17,  1.00e+00]]))

UhatUhatT = Uhat@Uhat.T
UhatTUhat = Uhat.T@Uhat
print('UhatUhatT, UhatTUhat= ')
UhatUhatT, UhatTUhat

UhatUhatT, UhatTUhat=

(array([[ 0.45,  0.17,  0.43,  0.17, -0.08],
        [ 0.17,  0.34, -0.  ,  0.34,  0.29],
        [ 0.43, -0.  ,  0.5 ,  0.01, -0.26],
        [ 0.17,  0.34,  0.01,  0.33,  0.28],
        [-0.08,  0.29, -0.26,  0.28,  0.38]]),
 array([[ 1.00e+00, -9.71e-17],
        [-9.71e-17,  1.00e+00]]))

Remarks:

The cells above illustrate the application of the full_matrices=True and full_matrices=False options. Using full_matrices=False returns a reduced singular value decomposition.

The full and reduced SVD’s both accurately decompose an $m \times n$ matrix $X$

When we study Dynamic Mode Decompositions below, it will be important for us to remember the preceding properties of full and reduced SVD’s in such tall-skinny cases.

Now let’s turn to a short-fat case.

To illustrate this case, we’ll set $m = 2 < 5 = n$ and compute both full and reduced SVD’s.

import numpy as np
X = np.random.rand(2,5)
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 =

(array([[-0.61,  0.79],
        [-0.79, -0.61]]),
 array([2.03, 0.58]),
 array([[-0.61, -0.57, -0.37, -0.31, -0.26],
        [ 0.16, -0.09, -0.66,  0.02,  0.73],
        [ 0.14, -0.7 ,  0.58, -0.02,  0.4 ],
        [-0.27, -0.16, -0.07,  0.95, -0.05],
        [-0.72,  0.39,  0.31, -0.09,  0.48]]))

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

Uhat, Shat, Vhat =

(array([[-0.61,  0.79],
        [-0.79, -0.61]]),
 array([2.03, 0.58]),
 array([[-0.61, -0.57, -0.37, -0.31, -0.26],
        [ 0.16, -0.09, -0.66,  0.02,  0.73]]))

Let’s verify that our reduced SVD accurately represents $X$

SShat=np.diag(Shat)
np.allclose(X, Uhat@SShat@Vhat)

True

5.7. Polar Decomposition#

A reduced singular value decomposition (SVD) of $X$ is related to a polar decomposition of $X$

X = S Q

where

\begin{aligned} S & = U Σ U^{⊤} \\ Q & = U V^{⊤} \end{aligned}

Here

$S$ is an $m \times m$ symmetric matrix
$Q$ is an $m \times n$ orthogonal matrix

and in our reduced SVD

$U$ is an $m \times p$ orthonormal matrix
$Σ$ is a $p \times p$ diagonal matrix
$V$ is an $n \times p$ orthonormal

5.8. Application: Principal Components Analysis (PCA)#

Let’s begin with a case in which $n >> m$ , so that we have many more individuals $n$ than attributes $m$ .

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

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

X = [\begin{matrix} X_{1} ∣ X_{2} ∣ \dots ∣ X_{n} \end{matrix}]

where for $j = 1, \dots, n$ the column vector $X_{j} = [\begin{matrix} x_{1 j} \\ x_{2 j} \\ ⋮ \\ x_{m j} \end{matrix}]$ is a vector of observations on variables $[\begin{matrix} X_{1} \\ X_{2} \\ ⋮ \\ X_{m} \end{matrix}]$ .

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 attributes.

As we have seen before, the SVD is a way to decompose a matrix into useful components, just like polar decomposition, eigendecomposition, and many others.

PCA, on the other hand, is a method that builds on the SVD to analyze data. The goal is to apply certain steps, to help better visualize patterns in data, using statistical tools to capture the most important patterns in data.

Step 1: Standardize the data:

Because our data matrix may hold variables of different units and scales, we first need to standardize the data.

First by computing the average of each row of $X$ .

\bar{X_{i}} = \frac{1}{n} \sum_{j = 1}^{n} x_{i j}

We then create an average matrix out of these means:

\begin{array}{r} \bar{X} = [\begin{array}{c} \bar{X_{1}} \\ \bar{X_{2}} \\ \dots \\ \bar{X_{m}} \end{array}] [\begin{array}{c} 1 ∣ 1 ∣ \dots ∣ 1 \end{array}] \end{array}

And subtract out of the original matrix to create a mean centered matrix:

B = X - \bar{X}

Step 2: Compute the covariance matrix:

Then because we want to extract the relationships between variables rather than just their magnitude, in other words, we want to know how they can explain each other, we compute the covariance matrix of $B$ .

C = \frac{1}{n} B B^{⊤}

Step 3: Decompose the covariance matrix and arrange the singular values:

Since the matrix $C$ is positive definite, we can eigendecompose it, find its eigenvalues, and rearrange the eigenvalue and eigenvector matrices in a decreasing order.

The eigendecomposition of $C$ can be found by decomposing $B$ instead. Since $B$ is not a square matrix, we obtain an SVD of $B$ :

\begin{array}{r} \begin{aligned} B B^{⊤} & = U Σ V^{⊤} (U Σ V^{⊤})^{⊤} \\ = U Σ V^{⊤} V Σ^{⊤} U^{⊤} \\ = U Σ Σ^{⊤} U^{⊤} \end{aligned} \end{array}

C = \frac{1}{n} U Σ Σ^{⊤} U^{⊤}

We can then rearrange the columns in the matrices $U$ and $Σ$ so that the singular values are in decreasing order.

Step 4: Select singular values, (optional) truncate the rest:

We can now decide how many singular values to pick, based on how much variance you want to retain. (e.g., retaining 95% of the total variance).

We can obtain the percentage by calculating the variance contained in the leading $r$ factors divided by the variance in total:

\frac{\sum_{i = 1}^{r} σ_{i}^{2}}{\sum_{i = 1}^{p} σ_{i}^{2}}

Step 5: Create the Score Matrix:

\begin{aligned} T & = B V \\ = U Σ V^{⊤} V \\ = U Σ \end{aligned}

5.9. Relationship of PCA to SVD#

To relate an SVD to a PCA of data set $X$ , first construct the SVD of the data matrix $X$ :

Let’s assume that sample means of all variables are zero, so we don’t need to standardize our matrix.

(5.9)#

X = U Σ V^{⊤} = σ_{1} U_{1} V_{1}^{⊤} + σ_{2} U_{2} V_{2}^{⊤} + \dots + σ_{p} U_{p} V_{p}^{⊤}

where

U = [\begin{matrix} U_{1} | U_{2} | \dots | U_{m} \end{matrix}]

\begin{array}{r} V^{⊤} = [\begin{array}{c} V_{1}^{⊤} \\ V_{2}^{⊤} \\ \dots \\ V_{n}^{⊤} \end{array}] \end{array}

In equation (5.9), each of the $m \times n$ matrices $U_{j} V_{j}^{⊤}$ is evidently of rank $1$ .

Thus, we have

(5.10)#

\begin{array}{r} X = σ_{1} (\begin{array}{c} U_{11} V_{1}^{⊤} \\ U_{21} V_{1}^{⊤} \\ \dots \\ U_{m 1} V_{1}^{⊤} \end{array}) + σ_{2} (\begin{array}{c} U_{12} V_{2}^{⊤} \\ U_{22} V_{2}^{⊤} \\ \dots \\ U_{m 2} V_{2}^{⊤} \end{array}) + \dots + σ_{p} (\begin{array}{c} U_{1 p} V_{p}^{⊤} \\ U_{2 p} V_{p}^{⊤} \\ \dots \\ U_{m p} V_{p}^{⊤} \end{array}) \end{array}

Here is how we would interpret the objects in the matrix equation (5.10) in a time series context:

$for each k = 1, \dots, n$ , the object ${V_{k j}}_{j = 1}^{n}$ is a time series for the $k$ th principal component
$U_{j} = [\begin{matrix} U_{1 k} \\ U_{2 k} \\ \dots \\ U_{m k} \end{matrix}] k = 1, \dots, m$ is a vector of loadings of variables $X_{i}$ on the $k$ th principal component, $i = 1, \dots, m$
$σ_{k}$ for each $k = 1, \dots, p$ is the strength of $k$ th principal component, where strength means contribution to the overall covariance of $X$ .

5.10. PCA with Eigenvalues and Eigenvectors#

We now use an 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 assure this by pre-processing the data by subtracting sample means.

Define a sample covariance matrix $Ω$ as

Ω = X X^{⊤}

Then use an eigen decomposition to represent $Ω$ as follows:

Ω = P Λ P^{⊤}

Here

$P$ is $m \times m$ matrix of eigenvectors of $Ω$
$Λ$ is a diagonal matrix of eigenvalues of $Ω$

We can then represent $X$ as

X = P ϵ

where

ϵ = P^{- 1} X

and

ϵ ϵ^{⊤} = Λ .

We can verify that

(5.11)#

X X^{⊤} = P Λ P^{⊤} .

It follows that we can represent the data matrix $X$ as

X = [\begin{matrix} X_{1} | X_{2} | \dots | X_{m} \end{matrix}] = [\begin{matrix} P_{1} | P_{2} | \dots | P_{m} \end{matrix}] [\begin{matrix} ϵ_{1} \\ ϵ_{2} \\ \dots \\ ϵ_{m} \end{matrix}] = P_{1} ϵ_{1} + P_{2} ϵ_{2} + \dots + P_{m} ϵ_{m}

To reconcile the preceding representation with the PCA that we had obtained earlier through the SVD, we first note that $ϵ_{j}^{2} = λ_{j} \equiv σ_{j}^{2}$ .

Now define $\tilde{ϵ_{j}} = \frac{ϵ_{j}}{\sqrt{λ_{j}}}$ , which implies that ${\tilde{ϵ}}_{j} {\tilde{ϵ}}_{j}^{⊤} = 1$ .

Therefore

\begin{array}{r} \begin{aligned} X & = \sqrt{λ_{1}} P_{1} \tilde{ϵ_{1}} + \sqrt{λ_{2}} P_{2} \tilde{ϵ_{2}} + \dots + \sqrt{λ_{m}} P_{m} \tilde{ϵ_{m}} \\ = σ_{1} P_{1} \tilde{ϵ_{2}} + σ_{2} P_{2} \tilde{ϵ_{2}} + \dots + σ_{m} P_{m} \tilde{ϵ_{m}}, \end{aligned} \end{array}

which agrees with

X = σ_{1} U_{1} {V_{1}}^{T} + σ_{2} U_{2} {V_{2}}^{T} + \dots + σ_{r} U_{r} {V_{r}}^{T}

provided that we set

$U_{j} = P_{j}$ (a vector of loadings of variables on principal component $j$ )
${V_{k}}^{T} = \tilde{ϵ_{k}}$ (the $k$ th principal component)

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

We can resolve such ambiguities about $U$ and $P$ by

sorting eigenvalues and singular values in descending order
imposing positive diagonals on $P$ and $U$ and adjusting signs in $V^{⊤}$ accordingly

5.11. Connections#

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

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

X = U Σ V^{⊤}

Compute:

(5.12)#

\begin{aligned} X X^{⊤} & = U Σ V^{⊤} V Σ^{⊤} U^{⊤} \\ \equiv U Σ Σ^{⊤} U^{⊤} \\ \equiv U Λ U^{⊤} \end{aligned}

Compare representation (5.12) with equation (5.11) above.

Evidently, $U$ in the SVD is the matrix $P$ of eigenvectors of $X X^{⊤}$ and $Σ Σ^{⊤}$ is the matrix $Λ$ of eigenvalues.

Second, let’s compute

\begin{array}{r} \begin{aligned} X^{⊤} X & = V Σ^{⊤} U^{⊤} U Σ V^{⊤} \\ = V Σ^{⊤} Σ V^{⊤} \end{aligned} \end{array}

Thus, the matrix $V$ in the SVD is the matrix of eigenvectors of $X^{⊤} X$

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

X X^{⊤} = P Λ P^{⊤}

where $P$ is an orthogonal matrix.

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

X X^{⊤} = U Σ Σ^{⊤} U^{⊤}

where $U$ is an orthogonal matrix.

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

X = P ϵ = U Σ V^{⊤}

It follows that

U^{⊤} X = Σ V^{⊤} = ϵ

Note that the preceding implies that

ϵ ϵ^{⊤} = Σ V^{⊤} V Σ^{⊤} = Σ Σ^{⊤} = Λ,

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.
    X: data matrix
    r_component: chosen rank for best approximation
    """

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

        self.X = X

        self.Ω = (X @ X.T)

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

        if r_component:
            self.r_component = r_component
        else:
            self.r_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.r_component]
        𝜖 = self.𝜖[:self.r_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.r_component]
        Σ = self.Σ[:self.r_component, :self.r_component]
        VT = self.VT[:self.r_component, :]

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

    def fit(self, r_component):

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

        # transform data
        self.X_pca = P @ 𝜖

        # svd
        U = self.U[:, :r_component]
        Σ = self.Σ[:r_component, :r_component]
        VT = self.VT[:r_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')

    # loading matrices
    fig, axs = plt.subplots(1, 2, figsize=(14, 5))
    plt.suptitle('loadings')
    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('$V^\top *\sqrt{\lambda}$')
    axs[1].set_xlabel('n')
    plt.show()

<>:32: SyntaxWarning: invalid escape sequence '\s'
<>:32: SyntaxWarning: invalid escape sequence '\s'
/tmp/ipykernel_33443/4223624786.py:32: SyntaxWarning: invalid escape sequence '\s'
  axs[1].set_title('$V^\top *\sqrt{\lambda}$')

5.12. Exercises#

Exercise 5.1

In Ordinary Least Squares (OLS), we learn to compute $\hat{β} = (X^{⊤} X)^{- 1} X^{⊤} y$ , but there are cases such as when we have colinearity or an underdetermined system: short fat matrix.

In these cases, the $(X^{⊤} X)$ matrix is not not invertible (its determinant is zero) or ill-conditioned (its determinant is very close to zero).

What we can do instead is to create what is called a pseudoinverse, a full rank approximation of the inverted matrix so we can compute $\hat{β}$ with it.

Thinking in terms of the Eckart-Young theorem, build the pseudoinverse matrix $X^{+}$ and use it to compute $\hat{β}$ .

Solution to Exercise 5.1

We can use SVD to compute the pseudoinverse:

X = U Σ V^{⊤}

inverting $X$ , we have:

X^{+} = V Σ^{+} U^{⊤}

where:

\begin{array}{r} Σ^{+} = [\begin{array}{c} \frac{1}{σ_{1}} & 0 & \dots & 0 & 0 \\ 0 & \frac{1}{σ_{2}} & \dots & 0 & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ & ⋮ \\ 0 & 0 & \dots & \frac{1}{σ_{p}} & 0 \\ 0 & 0 & \dots & 0 & 0 \end{array}] \end{array}

and finally:

\hat{β} = X^{+} y = V Σ^{+} U^{⊤} y

For an example PCA applied to analyzing the structure of intelligence tests see this lecture Multivariable Normal Distribution.

Look at parts of that lecture that describe and illustrate the classic factor analysis model.

As mentioned earlier, in a sequel to this lecture about Dynamic Mode Decompositions, we’ll describe how SVD’s provide ways rapidly to compute reduced-order approximations to first-order Vector Autoregressions (VARs).