-
Notifications
You must be signed in to change notification settings - Fork 0
/
section-18.html
223 lines (220 loc) · 16.6 KB
/
section-18.html
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
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
<!DOCTYPE html>
<!--********************************************-->
<!--* Generated from PreTeXt source *-->
<!--* on 2021-08-31T10:06:16-05:00 *-->
<!--* A recent stable commit (2020-08-09): *-->
<!--* 98f21740783f166a773df4dc83cab5293ab63a4a *-->
<!--* *-->
<!--* https://pretextbook.org *-->
<!--* *-->
<!--********************************************-->
<html lang="en-US">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>Equivalent forms of Invertibility and Singularity of Matrices</title>
<meta name="Keywords" content="Authored in PreTeXt">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<script src="https://sagecell.sagemath.org/embedded_sagecell.js"></script><script>window.MathJax = {
tex: {
inlineMath: [['\\(','\\)']],
tags: "none",
useLabelIds: true,
tagSide: "right",
tagIndent: ".8em",
packages: {'[+]': ['base', 'extpfeil', 'ams', 'amscd', 'newcommand', 'knowl']}
},
options: {
ignoreHtmlClass: "tex2jax_ignore",
processHtmlClass: "has_am",
renderActions: {
findScript: [10, function (doc) {
document.querySelectorAll('script[type^="math/tex"]').forEach(function(node) {
var display = !!node.type.match(/; *mode=display/);
var math = new doc.options.MathItem(node.textContent, doc.inputJax[0], display);
var text = document.createTextNode('');
node.parentNode.replaceChild(text, node);
math.start = {node: text, delim: '', n: 0};
math.end = {node: text, delim: '', n: 0};
doc.math.push(math);
});
}, '']
},
},
chtml: {
scale: 0.88,
mtextInheritFont: true
},
loader: {
load: ['input/asciimath', '[tex]/extpfeil', '[tex]/amscd', '[tex]/newcommand', '[pretext]/mathjaxknowl3.js'],
paths: {pretext: "https://pretextbook.org/js/lib"},
},
};
</script><script src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-chtml.js"></script><script xmlns:svg="http://www.w3.org/2000/svg" src="https://pretextbook.org/js/lib/jquery.min.js"></script><script xmlns:svg="http://www.w3.org/2000/svg" src="https://pretextbook.org/js/lib/jquery.sticky.js"></script><script xmlns:svg="http://www.w3.org/2000/svg" src="https://pretextbook.org/js/lib/jquery.espy.min.js"></script><script xmlns:svg="http://www.w3.org/2000/svg" src="https://pretextbook.org/js/0.13/pretext.js"></script><script xmlns:svg="http://www.w3.org/2000/svg" src="https://pretextbook.org/js/0.13/pretext_add_on.js"></script><script xmlns:svg="http://www.w3.org/2000/svg" src="https://pretextbook.org/js/lib/knowl.js"></script><!--knowl.js code controls Sage Cells within knowls--><script xmlns:svg="http://www.w3.org/2000/svg">sagecellEvalName='Evaluate (Sage)';
</script><link xmlns:svg="http://www.w3.org/2000/svg" href="https://fonts.googleapis.com/css?family=Open+Sans:400,400italic,600,600italic" rel="stylesheet" type="text/css">
<link xmlns:svg="http://www.w3.org/2000/svg" href="https://fonts.googleapis.com/css?family=Inconsolata:400,700&subset=latin,latin-ext" rel="stylesheet" type="text/css">
<link xmlns:svg="http://www.w3.org/2000/svg" href="https://pretextbook.org/css/0.31/pretext.css" rel="stylesheet" type="text/css">
<link xmlns:svg="http://www.w3.org/2000/svg" href="https://pretextbook.org/css/0.31/pretext_add_on.css" rel="stylesheet" type="text/css">
<link xmlns:svg="http://www.w3.org/2000/svg" href="https://pretextbook.org/css/0.31/banner_default.css" rel="stylesheet" type="text/css">
<link xmlns:svg="http://www.w3.org/2000/svg" href="https://pretextbook.org/css/0.31/toc_default.css" rel="stylesheet" type="text/css">
<link xmlns:svg="http://www.w3.org/2000/svg" href="https://pretextbook.org/css/0.31/knowls_default.css" rel="stylesheet" type="text/css">
<link xmlns:svg="http://www.w3.org/2000/svg" href="https://pretextbook.org/css/0.31/style_default.css" rel="stylesheet" type="text/css">
<link xmlns:svg="http://www.w3.org/2000/svg" href="https://pretextbook.org/css/0.31/colors_brown_gold.css" rel="stylesheet" type="text/css">
<link xmlns:svg="http://www.w3.org/2000/svg" href="https://pretextbook.org/css/0.31/setcolors.css" rel="stylesheet" type="text/css">
<!-- 2019-10-12: Temporary - CSS file for experiments with styling --><link xmlns:svg="http://www.w3.org/2000/svg" href="developer.css" rel="stylesheet" type="text/css">
</head>
<body class="pretext-book has-toc has-sidebar-left">
<a class="assistive" href="#content">Skip to main content</a><div xmlns:svg="http://www.w3.org/2000/svg" id="latex-macros" class="hidden-content" style="display:none">\(\def\R{{\mathbb R}}
\def\C{{\mathbb C}}
\def\Q{{\mathbb Q}}
\def\Z{{\mathbb Z}}
\def\N{{\mathbb N}}
\def\vec#1{\mathbf #1}
\newcommand{\adj}{\mathop{\mathrm{adj}}}
\newcommand{\diag}{\mathop{\mathrm{diag}}}
\newcommand{\proj}{\mathop{\mathrm{proj}}}
\newcommand{\Span}{\mathop{\mathrm{span}}}
\newcommand{\sgn}{\mathop{\mathrm{sgn}}}
\newcommand{\tr}{\mathop{\mathrm{tr}}}
\newcommand{\rowint}[2]{R_{#1} \leftrightarrow R_{#2}}
\newcommand{\rowmul}[2]{R_{#1}\gets {#2}R_{#1}}
\newcommand{\rowadd}[3]{R_{#1}\gets R_{#1}+#2R_{#3}}
\newcommand{\rowsub}[3]{R_{#1}\gets R_{#1}-#2R_{#3}}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\)</div>
<header id="masthead" class="smallbuttons"><div class="banner"><div class="container">
<a id="logo-link" href="http://www.umanitoba.ca" target="_blank"><img src="images/umlogo.png" alt="Logo image"></a><div class="title-container">
<h1 class="heading"><a href="mblinalg.html"><span class="title">Manitoba linear algebra</span></a></h1>
<p class="byline">Michael Doob</p>
</div>
</div></div>
<nav xmlns:svg="http://www.w3.org/2000/svg" id="primary-navbar" class="navbar"><div class="container">
<div class="navbar-top-buttons">
<button class="sidebar-left-toggle-button button active" aria-label="Show or hide table of contents sidebar">Contents</button><div class="tree-nav toolbar toolbar-divisor-3"><span class="threebuttons"><a id="previousbutton" class="previous-button toolbar-item button" href="section-17.html" title="Previous">Prev</a><a id="upbutton" class="up-button button toolbar-item" href="MatrixTheoryIntro.html" title="Up">Up</a><a id="nextbutton" class="next-button button toolbar-item" href="Determinants.html" title="Next">Next</a></span></div>
</div>
<div class="navbar-bottom-buttons toolbar toolbar-divisor-4">
<button class="sidebar-left-toggle-button button toolbar-item active">Contents</button><a class="previous-button toolbar-item button" href="section-17.html" title="Previous">Prev</a><a class="up-button button toolbar-item" href="MatrixTheoryIntro.html" title="Up">Up</a><a class="next-button button toolbar-item" href="Determinants.html" title="Next">Next</a>
</div>
</div></nav></header><div class="page">
<div xmlns:svg="http://www.w3.org/2000/svg" id="sidebar-left" class="sidebar" role="navigation"><div class="sidebar-content">
<nav id="toc"><ul>
<li class="link frontmatter"><a href="Frontmatter.html" data-scroll="Frontmatter"><span class="title">Title Page</span></a></li>
<li class="link"><a href="SysLinEq.html" data-scroll="SysLinEq"><span class="codenumber">1</span> <span class="title">Systems of Linear Equations</span></a></li>
<li class="link"><a href="MatrixTheoryIntro.html" data-scroll="MatrixTheoryIntro"><span class="codenumber">2</span> <span class="title">Matrix Theory</span></a></li>
<li class="link"><a href="Determinants.html" data-scroll="Determinants"><span class="codenumber">3</span> <span class="title">The Determinant</span></a></li>
<li class="link"><a href="EuclideanSpace.html" data-scroll="EuclideanSpace"><span class="codenumber">4</span> <span class="title">Vectors in Euclidean \(n\) space</span></a></li>
<li class="link"><a href="chapter-5.html" data-scroll="chapter-5"><span class="codenumber">5</span> <span class="title">Eigenvalues and eigenvectors</span></a></li>
<li class="link"><a href="LinearTransformations.html" data-scroll="LinearTransformations"><span class="codenumber">6</span> <span class="title">Linear transformations</span></a></li>
<li class="link"><a href="ExtraTopics.html" data-scroll="ExtraTopics"><span class="codenumber">7</span> <span class="title">Additional Topics</span></a></li>
</ul></nav><div class="extras"><nav><a class="pretext-link" href="https://pretextbook.org">Authored in PreTeXt</a><a href="https://www.mathjax.org"><img title="Powered by MathJax" src="https://www.mathjax.org/badge/badge.gif" alt="Powered by MathJax"></a></nav></div>
</div></div>
<main class="main"><div id="content" class="pretext-content"><section xmlns:svg="http://www.w3.org/2000/svg" class="section" id="section-18"><h2 class="heading hide-type">
<span class="type">Section</span> <span class="codenumber">2.11</span> <span class="title">Equivalent forms of Invertibility and Singularity of Matrices</span>
</h2>
<p id="p-497">We have already defined singular and nonsingular matrices in <a class="xref" data-knowl="./knowl/SingularMatrixDefinition.html" title="Definition 2.7.5: Matrix singularity">Definition 2.7.5</a> and the inverse of a matrix in <a class="xref" data-knowl="./knowl/InverseMatrixDefinition.html" title="Definition 2.7.1: The inverse of a matrix">Definition 2.7.1</a>. We now have a new goal: to show that a square matrix is invertible if and only if it is nonsingular. In fact we want to give several other conditions that are equivalent to being invertible. By this we mean that if any one of the conditions are true for a matrix \(A\text{,}\) then all them are true for that matrix \(A\text{,}\) and we then call \(A\) <dfn class="terminology">nonsingular</dfn> or <dfn class="terminology">invertible</dfn>. Equivalently, if any one of the conditions are false for a matrix \(A\text{,}\) then all of them are false for that matrix \(A\text{.}\) In this case the matrix \(A\) is called <dfn class="terminology">singular</dfn>.</p>
<p id="p-498">The method for showing statements to be equivalent is a little different. Suppose we have (as will be our case) six statements numbered \((1), (2), \ldots,(6)\) that we wish to show equivalent. We start by showing that if \((1)\) is true, then it follows that \((2)\) is true. We denote this by \((1)\Rightarrow(2) \text{.}\) We continue with \((2)\Rightarrow(3), (3)\Rightarrow(4), (4)\Rightarrow(5),
(5)\Rightarrow(6), (6)\Rightarrow(1)\text{.}\) Once this has been done, we may visualize our result:</p>
<figure class="figure figure-like" id="figure-11"><div class="image-box" style="width: 30%; margin-left: 35%; margin-right: 35%;"><div class="asymptote-box" style="padding-top: 86.0213226655274%"><iframe src="images/image-11.html" class="asymptote"></iframe></div></div>
<figcaption><span class="type">Figure</span><span class="space"> </span><span class="codenumber">2.11.1<span class="period">.</span></span><span class="space"> </span>A cycle of implications</figcaption></figure><p id="p-499">and so once one statement becomes true, all of them are true. This also means that once one statement is false then all of them are false (think about the logic!).</p>
<article class="theorem theorem-like" id="InvertibilityEquivalence"><h6 class="heading">
<span class="type">Theorem</span><span class="space"> </span><span class="codenumber">2.11.2</span><span class="period">.</span><span class="space"> </span><span class="title">Equivalent Forms of Invertibility.</span>
</h6>
<p id="p-500">Suppose that \(A\) is an \(n\times n\) square matrix. Then the following statements are equivalent:</p>
<ol class="decimal">
<li id="li-218"><p id="p-501">\(A\) is invertible</p></li>
<li id="li-219"><p id="p-502">\(A\vec x=0\) if and only if \(\vec x=0\)</p></li>
<li id="li-220"><p id="p-503">The reduced row echelon form of \(A\) is \(I_n\)</p></li>
<li id="li-221"><p id="p-504">\(A\) is a product of elementary matrices</p></li>
<li id="li-222"><p id="p-505">\(A\vec x=\vec b\) is consistent for any \(\vec b\)</p></li>
<li id="li-223"><p id="p-506">\(A\vec x=\vec b\) has exactly one solution for any \(\vec b\)</p></li>
</ol></article><article class="hiddenproof" id="proof-32"><a data-knowl="" class="id-ref proof-knowl original" data-refid="hk-proof-32"><h6 class="heading"><span class="type">Proof<span class="period">.</span></span></h6></a></article><div class="hidden-content tex2jax_ignore" id="hk-proof-32"><article class="hiddenproof"><p id="p-507">The logic of the proof is to show that</p>
<p id="p-508">(1) true implies (2) true (which we write as (1) \(\Rightarrow \)(2)),</p>
<p id="p-509">(2) true implies (3) true (which we write as (2) \(\Rightarrow\) (3)),</p>
<p id="p-510">(3) true implies (4) true (which we write as (3) \(\Rightarrow\) (4)),</p>
<p id="p-511">(4) true implies (5) true (which we write as (4) \(\Rightarrow\) (5)),</p>
<p id="p-512">(5) true implies (6) true (which we write as (5) \(\Rightarrow\) (6)), and</p>
<p id="p-513">(6) true implies (1) true (which we write as (6) \(\Rightarrow\) (1))</p>
<p id="p-514">We now proceed to prove the implications one at a time.</p>
<p id="p-515">(1) \(\Rightarrow\) (2): Assuming (1) means that \(A\) is invertible. If \(x=\vec0\text{,}\) then \(A\vec x=A\vec0=\vec0\text{.}\) Now suppose that \(A\vec x=\vec0.\) Then, since \(A^{-1}\) exists, we may evaluate \(A^{-1}A\vec x\) in two ways:</p>
<div class="displaymath">
\begin{equation*}
A^{-1}(A\vec x) = A^{-1}\vec0=\vec0\\ (A^{-1}A)\vec x
=I\vec x=\vec x
\end{equation*}
</div>
<p class="continuation">Equating the two evaluations, we get \(\vec x=\vec0.\)</p>
<p id="p-516">(2) \(\Rightarrow\) (3): Consider the augmented matrix for the system of linear equations \(A\vec x=\vec0.\) If the reduced row echelon from of \(A\) is not \(I_n,\) then the form for augmented matrix has a free variable; this may be assigned a nonzero value, resulting in a nonzero solution to \(A\vec x=\vec0.\)</p>
<p id="p-517">(3) \(\Rightarrow\) (4): Suppose \(A\) is reduced to \(I_n\) by a series of elementary row operations whose corresponding elementary matrices are \(E_1,E_2,\dots,E_k\text{.}\) Then \(E_kE_{k-1}\cdots E_2E_1A=I_n\text{,}\) and</p>
<div class="displaymath">
\begin{equation*}
A=E_1^{-1}E_2^{-1}\cdots E_{k-1}^{-1}E_k^{-1}=F_1F_2\cdots
F_{k-1}F_k.
\end{equation*}
</div>
<p id="p-518">(4) \(\Rightarrow\) (5): Suppose we want to solve \(A\vec x=\vec b\) where \(\vec b\) is any vector and \(A=F_1F_2\cdots F_{k-1}F_k.\) Then we have</p>
<div class="displaymath">
\begin{equation*}
F_1F_2\cdots F_{k-1}F_k \vec x=\vec b,
\end{equation*}
</div>
<p class="continuation">and</p>
<div class="displaymath">
\begin{equation*}
\vec x=F_k^{-1}F_{k-1}^{-1}\cdots
F_2^{-1}F_1^{-1} \vec b
\end{equation*}
</div>
<p class="continuation">which then satisfies</p>
<div class="displaymath">
\begin{equation*}
\begin{split} A\vec x \amp =(F_1F_2\cdots
F_{k-1}F_k)(F_k^{-1}F_{k-1}^{-1}\cdots F_2^{-1}F_1^{-1}\vec
b)\\ \amp=(F_1F_2\cdots F_{k-1}F_k)(F_k^{-1}F_{k-1}^{-1}\cdots
F_2^{-1}F_1^{-1})\vec b \\ \amp=\vec b \end{split}
\end{equation*}
</div>
<p class="continuation">and so gives a solution to \(A\vec x=\vec b.\) Hence \(A\vec x=\vec b\) is consistent.</p>
<p id="p-519">(5) \(\Rightarrow\) (6): We take the following values for \(\vec b_1,\vec b_2,\dots,\vec b_n\text{:}\)</p>
<div class="displaymath">
\begin{equation*}
\vec b_1=
\begin{bmatrix}
1 \\ 0 \\ 0 \\ \vdots \\ 0
\end{bmatrix},
\quad
\vec b_2=
\begin{bmatrix}
0 \\ 1 \\ 0 \\ \vdots \\ 0
\end{bmatrix},
\quad
\vec b_3=
\begin{bmatrix} 0 \\ 0 \\ 1 \\ \vdots \\ 0
\end{bmatrix},
\ldots,
\quad
\vec b_n=
\begin{bmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ 1
\end{bmatrix} \text{.}
\end{equation*}
</div>
<p class="continuation">Then \(A\vec x=\vec b_1\) has a solution \(\vec c_1\text{,}\) \(A\vec x=\vec b_2\) has a solution \(\vec c_2\text{,}\) \(A\vec x=\vec b_3\) has a solution \(\vec c_3, \dots,\) \(A\vec x=\vec b_n\) has a solution \(\vec c_n\text{.}\)</p>
<p id="p-520">Now define the matrix \(C\) as the one with \(\vec c_1,\vec c_2,\dots,\vec c_n\) as columns, that is,</p>
<div class="displaymath">
\begin{equation*}
C=
\begin{bmatrix}
\vec c_1 \amp\vec c_2 \amp\vec c_3 \amp\cdots \amp\vec c_n
\end{bmatrix}.
\end{equation*}
</div>
<p class="continuation">Then \(AC=I\text{,}\) and so \(C=A^{-1}\) and \(CA=I.\) Now if \(A\vec x=\vec b\) is consistent, then \(\vec
x=I\vec x=CA\vec x=C\vec b\) and so \(\vec
x=C\vec b\) is the one and only solution to \(A\vec
x=\vec b.\)</p>
<p id="p-521">(6) \(\Rightarrow\) (1): Use the vectors \(\vec b_1,
\vec b_2,\dots,\vec b_n\) given just above. In each case there is exactly one vector \(\vec c_k\) which is a solution to the equation \(A\vec x=\vec b_k\text{.}\) Let \(C\) be the matrix with \(\vec c_1,\dots,\vec
c_n\) as columns. This matrix satisfies \(AC=I\) which makes \(C=A^{-1}\) and so \(A\) is invertible.</p></article></div></section></div></main>
</div>
</body>
</html>