1
2%%% First homework for Lenstra's 54.
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4\documentclass[12pt]{article}
5\usepackage{amsmath}
6\newcommand{\R}{\mathbf{R}}
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8\newcommand{\prob}{\vspace{.2in}\par\noindent {\bf #1}\par}
9
10
11\title{MATH 54\\Linear Algebra and Diff. Eqns.\\
12       Section 1 (Lenstra), assignment \# 5}
13\author{William Stein}
14
15\begin{document}
16\maketitle
17
18\begin{center}
19{\bf {\em Anton}: V.1, \# 6,12; V.2, \# 2bc, 4ace, 7abcd}
20\end{center}
21
22\prob{V.1.6}
23{\em Is the set $W=\{(x,y):x\geq 0, x,y\in\R\}$ with the standard
24operations a vector space?}
25
26No, axiom 5 fails since $-(1,0)=(-1,0)\not\in W$. Axiom 6
27fails since $-2\cdot(2,3)=(-4,-6)\not\in W$.
28
29\prob{V.1.12}
30{\em Is the set of all $2\times 2$ matrices of the form
31$\bigl(\begin{smallmatrix}a&a+b\\a+b&b\end{smallmatrix}\bigr)$ with
32matrix addition and scalar multiplication a vector space?}
33
34Yes, just check all of the axioms.
35
36
37\prob{V.2.2}
38
39{\em (b) Is the set of matrices
40$\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)$ with
41$a+b+c+d=0$ a subspace of $M_{22}$?}
42
43Yes. Apply theorem 5.2.1. First check that the sum of
44two such matrices is again such a matrix. If the sum of
45the entries of $A$ is zero and the sum of the entries
46of $B$ is zero then the sum of the entries of $A+B$ is zero.
47Next check that if the sum of the entries of $A$ is zero
48then the sum of the entries of $kA$ is zero. Do this by
49assuming $A=\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)$
50and $a+b+c+d=0$. Then $kA=\bigl(\begin{smallmatrix}ka&kb\\kc&kd\end{smallmatrix}\bigr)$
51so the sum of the entries of $kA$ is
52$$ka+kb+kc+kd=k(a+b+c+d)=k0=0.$$
53
54{\em (c) Is the set of all $2\times 2$ matrix $A$ such that
55$\det(A)=0$ a subspace?}
56
57No. Although property (b) of theorem 5.2.1 is satisfied, property
58(a) is not. For example, if
59$A=\bigl(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\bigr)$
60and
61$B=\bigl(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\bigr)$
62then $\det(A)=\det(B)=0$ but $\det(A+B)=1$.
63
64\prob{V.2.4}
65{\em (a) Is the set of $f$ such that $f(x)\leq 0$ for all $x$
66a subspace of $F(-\infty,\infty)$?}
67
68No, since property (b) of theorem 5.2.1 fails. For example,
69if $f(x)$ is the function $f(x)=-3$ and $k=-7$ then $k\cdot{}f(x)=21$
70so $kf(x)$ does not lie in the set under consideration.
71
72{\em (c) Is the set of $f$ such that $f(0)=2$ a subspace of
73$F(-\infty,\infty)$?}
74
75No, since property (a) of theorem 5.2.1 fails (in fact, property
76(b) fails as well). We can show that property (a) fails by
77exhibiting one example. Let $f$ be the function $f(x)=2$ and
78let $g$ be the function $g(x)=x+2$. Then $f$ and $g$ lie
79in the subset under consideration, but $(f+g)(0)=f(0)+g(0)=2+2=4$
80so $f+g$ does not lie in the subset.
81
82{\em (e) Is the set
83$$W=\{k_1+k_2\sin x:k_1,k_2\in\R\}$$
84a subspace of $F(-\infty,\infty)$?}
85
86Yes. To see this check properties (a) and (b) in theorem 5.2.1.
87Let $k_1+k_2\sin x$ and $k_3+k_4\sin x$ be elements of $W$.
88Then $$(k_1+k_2\sin x) +(k_3+k_4\sin x)=(k_1+k_3)+(k_2+k_4)\sin x\in W.$$
89If $a\in\R$ and $k_1+k_2\sin x\in W$ then
90$$a(k_1+k_2\sin x)=ak_1+ak_2\sin x\in W.$$
91
92\prob{V.2.7}
93{\em (a) Write $x=(2,2,2)$ as a linear combination of $u=(0,-2,2)$ and
94$v=(1,3,-1)$. }
95
96We just need to find $k_1$ and $k_2$ so that
97$$x=k_1u+k_2v.$$
98This amounts to solving a system of 3 linear equations in
99the two unknown $k_1$ and $k_2$. To see this expand both
100sides and simplify to get
101$$(2,2,2)=(k_2,-2k_1+3k_2,2k_1-k_2).$$
102Then equating corresponding entries gives the system
103\begin{align*}
1042&=k_2\\
1052&=-2k_1+3k_2\\
1062&=2k_1-k_2
107\end{align*}
108Solving gives $k_1=2$ and $k_2=2$. Check that this works.
109Note that if you are working a similiar problem and the system
110turns out to have no solution, then this means that there is
111no way to write $x$ as a linear combination of $u$ and $v$ (or
113
114{\em (b) Write $x=(3,1,5)$ as a linear combination of $u=(0,-2,2)$ and
115$v=(1,3,-1)$.}
116
117Do exactly the same thing as you did for (a) above. You will
118get $k_1=4$ and $k_2=3$.
119
120{\em (d) Show that $x=(0,4,5)$ can not be written
121as a linear combination of $u=(0,-2,2)$ and $v=(1,3,-1)$.}
122
123As in part (a) we obtain a system of equations by equating corresponding
124entries in the equation
125$$(0,4,5)=(k_2,-2k_1+3k_2,2k_1-k_2).$$
126The system is
127\begin{align*}
1280&=k_2\\
1294&=-2k_1+3k_2\\
1305&=2k_1-k_2
131\end{align*}
132We must show that this system has no solutions. If it did, then
133we must have $k_2=0$ from the first equation. Then the third equation implies
134$k_1=5/2$. But this contradicts the second equation which implies
135$k_1=-2$.
136
137\end{document}
138
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