1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
|
#set document(
title: [Lecture 2 - Matrix Norms: Low rank Approximation],
author: "Gidon Rosalki",
date: auto,
)
#set heading(numbering: "1.")
#set text(font: "New Computer Modern")
#set page(margin: (
bottom: 0.5cm,
right: 1.5cm,
left: 1.5cm,
top: 1.5cm,
))
#set list(indent: 10pt)
#set enum(indent: 10pt)
#show title: set align(center)
#show link: set text(rgb("005eff"))
#show link: underline
#show math.equation.where(block: true): set block(breakable: true)
#title()
#align(center)[
Gidon Rosalki \
2026-10-20
]
#let colourmaths(x, color) = text(fill: color)[$#x$]
// author: eric1102 - https://sitandr.github.io/typst-examples-book/book/typstonomicon/math_display.html
// may slightly interfere with maths blocks
#show math.equation: it => {
if it.body.fields().at("size", default: none) != "display" {
return math.display(it)
}
it
}
#let Mat = [Mat]
= Reminder
*Definition* (SVD): Let there be $M in Mat_(n times m) (RR)$. The SVD of this matrix is:
$
M = U dot Sigma dot V^T
$
Where $U, V$ are orthogonal. The columns $u_1, dots, u_n$ of $U$ are orthonormal basis for $RR^n$, and the columns $v_1,
dots, v_m$ of $V$ are an orthonormal basis for $RR^m$. \
$Sigma$ is a diagonal matrix where $forall i != j, Sigma_(i j) = 0$. Note that $Sigma$ is not necessarily square. We
denote the values of the diagonal $sigma_1 >= sigma_2 >= dots >= sigma_k : k = min(m, n)$.
*Definition* trace:
$
tr(A) & = sum_(i = 1)^(n)A_(i, i) \
tr(A B) & = tr(B A)
$
= Matrix Norms
*Definition* (norm): A norm on a (real) vector space $V$ is a function $norm(dot.c) : V -> [0, inf)$ such that:
+ $norm(x) = 0 <==> x = 0 quad forall x in V$
+ Triangle inequality: $norm(x + y) <= norm(x) + norm(y) quad forall x, y in V$.
+ $norm(alpha x) = abs(alpha) norm(x) quad forall x in V, alpha in RR$
== Examples
For $V = RR^n$
$
norm(x) = & norm(x)_2 && = sqrt(sum_(i = 1)^(n)x_i^2) \
1 <= p < infinity : & norm(x)_p && = (sum_(i = 1)^(n) abs(x_i)^(p))^(1 / p) \
& norm(x)_infinity && = max_(i in {1, dots, n}) abs(x_i)
$
== Matrix Norms
This is a fancy name where we take $V$ to be the vector space of matrices: $V = Mat_(n times m) (RR) tilde.equiv R^(n dot
m)$. For example, for $M in V$:
+ The operator norm / spectral norm is a measure of the size of a matrix, how much it can increase a vector $v$:
$
norm(M)_"op" = max {norm(M v) : v in RR^m, norm(v) = 1}
$
Recall that $norm(M)_"op" = sigma_1$, the largest singular value of $M$
+ Frobenius norm:
$
norm(M)_F = (sum_(i = 1)^(n) sum_(j = 1)^(m)Mat_(i j)^(2))^(1 / 2)
$
#pagebreak()
*Claim*: $norm(M)_F^2 = tr(M dot M^T)$ \
_Proof_:
$
(M dot M^T)_(i i) & = sum_(j = 1)^(m) Mat_(i j) dot Mat_(j i)^(T) \
& = sum_(j = 1)^(m)Mat_(i j)^(2)
$
*Claim*: $norm(M)_F = (sum_(i = 1)^(k)sigma_i^2)^(1 / 2)$\
_Proof_:
$
norm(M)_F^2 & = tr(M M^T) \
& = tr(underbrace(U Sigma V^T, M) dot underbrace(V Sigma^T U^T, M^T)) \
& = tr(underbrace(U Sigma Sigma^T, A) underbrace(U^T, B)) \
& = tr(U^T U Sigma Sigma^T) \
& = tr(Sigma Sigma^T) \
& = norm(Sigma)_F^2 \
& = sum_(i = 1)^(k)sigma_i^2
$
= Low Rank Approximation
#let rk = [rk]
#let rank = [rank]
#let Im = [Im]
*Definition* $rk(M) = rank(M) = dim(Im(M))$
*Claim*: Let $M in Mat_(n times m) (RR)$. The rank of $M$ is the minimum number $r$ such that
$
exists A in Mat_(n times r) (RR), B in Mat_(r times m) (RR) : M = A dot B
$
It is clear that $rk(A) <= r and rk(A dot B) <= rk(A) <= r$. Therefore, if $M = A dot B => rk(M) <= r$.
*Claim*: Suppose that the singular values of $M$ are $sigma_1 >= sigma_2 >= dots >= sigma_k >= 0$. Then
$rk(M)$ is the number of non-zero singular values.
$
rk(M) = max{i : i < k and sigma_i > 0}
$
_Proof_:
$
M & = U Sigma V^T \
M v_i & = cases(
sigma_i u_i && "if" 1 <= i <= k,
0 && "if" i > k
)
$
Therefore, $Im(M)$ is spanned by $u_i : sigma_i > 0$.
|