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%%% First homework for Lenstra's 54.
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%%% AmSLaTeX
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\documentclass[12pt]{article}
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\usepackage{amsmath}
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\newcommand{\R}{\mathbf{R}}
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\newcommand{\prob}[1]{\vspace{.2in}\par\noindent {\bf #1}\par}
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\title{MATH 54\\Linear Algebra and Diff. Eqns.\\
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Section 1 (Lenstra), assignment \# 5}
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\author{William Stein}
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\begin{document}
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\maketitle
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\begin{center}
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{\bf {\em Anton}: V.1, \# 6,12; V.2, \# 2bc, 4ace, 7abcd}
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\end{center}
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\prob{V.1.6}
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{\em Is the set $W=\{(x,y):x\geq 0, x,y\in\R\}$ with the standard
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operations a vector space?}
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No, axiom 5 fails since $-(1,0)=(-1,0)\not\in W$. Axiom 6
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fails since $-2\cdot(2,3)=(-4,-6)\not\in W$.
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\prob{V.1.12}
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{\em Is the set of all $2\times 2$ matrices of the form
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$\bigl(\begin{smallmatrix}a&a+b\\a+b&b\end{smallmatrix}\bigr)$ with
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matrix addition and scalar multiplication a vector space?}
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Yes, just check all of the axioms.
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\prob{V.2.2}
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{\em (b) Is the set of matrices
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$\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)$ with
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$a+b+c+d=0$ a subspace of $M_{22}$?}
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Yes. Apply theorem 5.2.1. First check that the sum of
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two such matrices is again such a matrix. If the sum of
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the entries of $A$ is zero and the sum of the entries
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of $B$ is zero then the sum of the entries of $A+B$ is zero.
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Next check that if the sum of the entries of $A$ is zero
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then the sum of the entries of $kA$ is zero. Do this by
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assuming $A=\bigl(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\bigr)$
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and $a+b+c+d=0$. Then $kA=\bigl(\begin{smallmatrix}ka&kb\\kc&kd\end{smallmatrix}\bigr)$
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so the sum of the entries of $kA$ is
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$$ka+kb+kc+kd=k(a+b+c+d)=k0=0.$$
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{\em (c) Is the set of all $2\times 2$ matrix $A$ such that
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$\det(A)=0$ a subspace?}
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No. Although property (b) of theorem 5.2.1 is satisfied, property
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(a) is not. For example, if
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$A=\bigl(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\bigr)$
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and
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$B=\bigl(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\bigr)$
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then $\det(A)=\det(B)=0$ but $\det(A+B)=1$.
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\prob{V.2.4}
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{\em (a) Is the set of $f$ such that $f(x)\leq 0$ for all $x$
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a subspace of $F(-\infty,\infty)$?}
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No, since property (b) of theorem 5.2.1 fails. For example,
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if $f(x)$ is the function $f(x)=-3$ and $k=-7$ then $k\cdot{}f(x)=21$
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so $kf(x)$ does not lie in the set under consideration.
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{\em (c) Is the set of $f$ such that $f(0)=2$ a subspace of
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$F(-\infty,\infty)$?}
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No, since property (a) of theorem 5.2.1 fails (in fact, property
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(b) fails as well). We can show that property (a) fails by
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exhibiting one example. Let $f$ be the function $f(x)=2$ and
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let $g$ be the function $g(x)=x+2$. Then $f$ and $g$ lie
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in the subset under consideration, but $(f+g)(0)=f(0)+g(0)=2+2=4$
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so $f+g$ does not lie in the subset.
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{\em (e) Is the set
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$$W=\{k_1+k_2\sin x:k_1,k_2\in\R\}$$
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a subspace of $F(-\infty,\infty)$?}
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Yes. To see this check properties (a) and (b) in theorem 5.2.1.
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Let $k_1+k_2\sin x$ and $k_3+k_4\sin x$ be elements of $W$.
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Then $$(k_1+k_2\sin x) +(k_3+k_4\sin x)=(k_1+k_3)+(k_2+k_4)\sin x\in W.$$
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If $a\in\R$ and $k_1+k_2\sin x\in W$ then
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$$a(k_1+k_2\sin x)=ak_1+ak_2\sin x\in W.$$
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\prob{V.2.7}
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{\em (a) Write $x=(2,2,2)$ as a linear combination of $u=(0,-2,2)$ and
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$v=(1,3,-1)$. }
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We just need to find $k_1$ and $k_2$ so that
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$$x=k_1u+k_2v.$$
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This amounts to solving a system of 3 linear equations in
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the two unknown $k_1$ and $k_2$. To see this expand both
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sides and simplify to get
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$$(2,2,2)=(k_2,-2k_1+3k_2,2k_1-k_2).$$
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Then equating corresponding entries gives the system
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\begin{align*}
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2&=k_2\\
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2&=-2k_1+3k_2\\
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2&=2k_1-k_2
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\end{align*}
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Solving gives $k_1=2$ and $k_2=2$. Check that this works.
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Note that if you are working a similiar problem and the system
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turns out to have no solution, then this means that there is
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no way to write $x$ as a linear combination of $u$ and $v$ (or
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you made a mistake somewhere).
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{\em (b) Write $x=(3,1,5)$ as a linear combination of $u=(0,-2,2)$ and
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$v=(1,3,-1)$.}
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Do exactly the same thing as you did for (a) above. You will
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get $k_1=4$ and $k_2=3$.
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{\em (d) Show that $x=(0,4,5)$ can not be written
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as a linear combination of $u=(0,-2,2)$ and $v=(1,3,-1)$.}
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As in part (a) we obtain a system of equations by equating corresponding
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entries in the equation
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$$(0,4,5)=(k_2,-2k_1+3k_2,2k_1-k_2).$$
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The system is
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\begin{align*}
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0&=k_2\\
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4&=-2k_1+3k_2\\
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5&=2k_1-k_2
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\end{align*}
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We must show that this system has no solutions. If it did, then
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we must have $k_2=0$ from the first equation. Then the third equation implies
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$k_1=5/2$. But this contradicts the second equation which implies
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$k_1=-2$.
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\end{document}
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