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