5. Linear Algebra for Machine Learning: Singular Value Decomposition and Principal Component Analysis

Seminar Series on
Linear Algebra for Machine Learning
Part 5: Singular Value Decomposition and
Principal Component Analysis
Dr. Ceni Babaoglu
Data Science Laboratory
Ryerson University
cenibabaoglu.com
Dr. Ceni Babaoglu cenibabaoglu.com
Linear Algebra for Machine Learning: Singular Value Decomposition and Principal Component Analysis

Overview
1 Spectral Decomposition
2 Singular Value Decomposition
3 Principal Component Analysis
4 References

Spectral Decomposition
An n × n symmetric matrix A can be expressed as the matrix
product
A = PDPT
where D is a diagonal matrix and P is an orthogonal matrix.
The diagonal entries of D are the eigenvalues of A,
λ1, λ2, · · · , λn, and the columns of P are associated
orthonormal eigenvectors x1, x2, · · · , xn.

The expression
A = PDPT
is called the spectral decomposition of A. We can write it as
A = x1 x2 · · · xn








λ1 0 · · · · · · 0
0 λ2 0 · · · 0
... 0
...
... 0
...
...
...
... 0
0 0 · · · 0 λn













xT
1
xT
2
...
xT
n






The matrix product DPT gives
DPT
=








λ1 0 · · · · · · 0
0 λ2 0 · · · 0
... 0
...
... 0
...
...
...
... 0
0 0 · · · 0 λn













xT
1
xT
2
...
xT
n





=





λ1xT
1
λ2xT
2
...
λnxT
n





A = x1 x2 · · · xn





λ1xT
1
λ2xT
2
...
λnxT
n






We can express A as a linear combination of the matrices
xj xT
j , and the coeﬃcients are the eigenvalues of A,
A =
n
j=1
λj xj xT
j = λ1x1xT
1 + λ2x2xT
2 + · · · + λnxnxT
n .

Singular Value Decomposition
Singular Value Decomposition is based on a theorem from linear
algebra which says the following:
a rectangular matrix A can be broken down into the product of
three matrices:
an orthogonal matrix U,
a diagonal matrix S,
the transpose of an orthogonal matrix V .

Let A be an m × n real matrix. Then there exist orthogonal
matrices U of size m × m and V of size n × n such that
A = USV T
where S is an m × n matrix with nondiagonal entries all zero and
s11 ≥ s12 ≥ · · · ≥ spp ≥ 0, p = min{m, n}.
the diagonal entries of S are called the singular values of A,
the columns of U are called the left singular vectors of A,
the columns of V are called the right singular vectors of A,
the factorization USV T is called the singular value
decomposition of A.

A = USV T
,
UT
U = I, V T
V = I,
the columns of U are orthonormal eigenvectors of AAT ,
the columns of V are orthonormal eigenvectors of AT A,
S is a diagonal matrix containing the square roots of
eigenvalues from U or V in descending order.

Example
Let’s ﬁnd the singular value decomposition of A=
3 1 1
−1 3 1
.
In order to ﬁnd U, we start with AAT .
AAT
=
3 1 1
−1 3 1


3 −1
1 3
1 1

 =
11 1
1 11
Eigenvalues: λ1 = 12 and λ2 = 10.
Eigenvectors: u1 =
1
1
and u2 =
1
−1
.

Example
Using Gram-Schmidt Process
v1 = u1 w1 =
v1
|v1|
=
1/
√
2
1/
√
2
v2 = u2 −
(u2, v1)
(v1, v1)
v1
v2 =
1
−1
− 0
1
1
=
1
−1
− [0, 0] =
1
−1
w2 =
v2
|v2|
=
1
√
2
,
−1
√
2
U =
1/
√
2 1/
√
2
1/
√
2 −1/
√
2

Example
The calculation of V is similar. V is based on AT A, so we have
AT
A =
3 −1
1 3
1 1
3 1 1
−1 3 1
=
10 0 2
0 10 4
2 4 2
We ﬁnd the eigenvalues of AT A
Eigenvalues: λ1 = 12, λ2 = 10 and λ3 = 0.
Eigenvectors: u1 =
1
2
1
, u2 =
2
−1
0
and u3 =
1
2
−5
.

Example
v1 = u1, w1 =
v1
|v1|
=
1
√
6
,
2
√
6
,
1
√
6
v2 = u2 −
(u2, v1)
(v1, v1)
v1 = [2, −1, 0]
w2 =
v2
|v2|
=
2
√
5
,
−1
√
5
, 0

Example
v3 = u3 −
(u3, v1)
(v1, v1)
v1 −
(u3, v2)
(v2, v2)
v2 =
−2
3
,
−4
3
,
10
3
w3 =
v3
|v3|
=
1
√
30
,
2
√
30
,
−5
√
30
V =



1√
6
2√
5
1√
30
2√
6
−1√
5
2√
30
1√
6
0 −5√
30


 , V T
=



1√
6
2√
6
1√
6
2√
5
−1√
5
0
1√
30
2√
30
−5√
30




Example
A = USV T
A =
1√
2
1√
2
1√
2
−1√
2
√
12 0 0
0
√
10 0


1√
6
2√
6
1√
6
2√
5
−1√
5
0
1√
30
2√
30
−5√
30

 =
3 1 1
−1 3 1

Let λ1, λ2, · · · , λn be the eigenvalues of A and x1, x2, · · · , xn be a
set of associated orthonormal eigenvectors.
Then the spectral decomposition of A is given by
A = λ1x1xT
1 + λ2x2xT
2 + · · · + λnxnxT
n .
If A is a real n × n matrix with real eigenvalues λ1, λ2, · · · , λn,
then an eigenvalue of largest magnitude is called a dominant
eigenvalue of A.

Let X be the multivariate data matrix
X =








x11 x12 · · · x1k · · · x1p
x21 x22 · · · x2k · · · x2p
...
...
...
...
xj1 xj2 · · · xjk · · · xjp
...
...
...
...
xn1 xn2 · · · xnk · · · xnp








.
The measure of association between the ith and kth variables in
the multivariate data matrix is given by the sample covariance
sik =
1
n
n
j=1
(xji − xi ) (xjk − xk) , i = 1, 2, . . . , p, k = 1, 2, . . . , p.

Let Sn be the p × p covariance matrix associated with the
multivariate data matrix X.
Sn =





s11 s12 · · · s1p
s21 s22 · · · s2p
...
...
...
...
sp1 sp2 · · · spp





Let the eigenvalues of Sn be λj , j = 1, 2, . . . , p, where
λ1 ≥ λ2 ≥ · · · ≥ λp ≥ 0 and let the associated orthonormal
eigenvectors be uj , j = 1, 2, . . . , p.
The ith principal component yi is given by the linear
combination of the columns of X, where the coeﬃcients are
the entries of the eigenvector ui ,
yi = Xui

The variance of yi is λi ,
The covariance of yi and yk, i = k, is zero.
If some of the eigenvalues are repeated, then the choices of
the associated eigenvectors are not unique; hence the principal
components are not unique.




Proportion of the
total variance due
to the k th principal
component



 =
λk
λ1 + λ2 + · · · + λp
, k = 1, 2, . . . , p

Example
Let’s compute the first and second principal components for the
multivariate data matrix X given by
X =








39 21
59 28
18 10
21 13
14 13
22 10








.
We first find the sample means x1 ≈ 28.8 and x2 ≈ 15.8.
Then we take the matrix of sample means as
x =
28.8
15.8

Example
The variances are
s11 ≈ 243.1 and s22 ≈ 43.1
while the covariances are
s12 = s21 ≈ 97.8.
We take the covariance matrix as
Sn =
243.1 97.8
97.8 43.1

Example
Eigenvalues: λ1 = 282.9744 and λ2 = 3.2256,
Eigenvectors: u1 =
0.9260
0.3775
and u2 =
0.3775
−0.9260
.
We ﬁnd the ﬁrst principal component as
y1 = 0.9260 col1(X) + 0.3775 col2(X).
It follows that y1 accounts for the proportion
λ1
λ1 + λ2
( about 98.9%)
of the total variance of X.

Example
We ﬁnd the second principal component as
y2 = 0.3775 col1(X) − 0.9260 col2(X).
It follows that y2 accounts for the proportion
λ2
λ1 + λ2
( about 0.011%)
of the total variance of X.

References
Linear Algebra With Applications, 7th Edition
by Steven J. Leon.
Elementary Linear Algebra with Applications, 9th Edition
by Bernard Kolman and David Hill.

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