\documentstyle[12pt]{article} \input amssym.def \input amssym \newcounter{probno} \newcommand{\prob}[1]{\addtocounter{probno}{1} \par\vfill\noindent\theprobno . (#1 points) } \newcommand{\by }{\times} \newcommand{\norm}[1]{||#1||} \newcommand{\complex }{\Bbb C} \newcommand{\field }{\Bbb F} \newcommand{\reals }{\Bbb R} \begin{document} \begin{flushright} A. Miller\\ M340 Exam 3\\ April 16, 1990 \end{flushright} {\bf There is a total of 26 points. Put your answers on separate sheets of paper. Prove everything you use. It is OK to quote any theorem that was proved in your book or in class, but you must include a proof of that theorem. } \prob 5 Suppose that $V$ is a vector space over the field $\field$, and $v_j$'s are vectors in $V$ such that $v_1\in span(\{v_2,v_3,\ldots,v_m\})$. Prove that $v_1,v_2,\ldots,v_m$ are linearly dependent. \prob 6 Suppose that the function $L:\reals^{n\by n}\mapsto \reals^{n\by n}$ is defined by $L(A)=A-A^{\prime}$ where $A^{\prime}$ is the transpose of $A$. Prove that $L$ is a linear transformation. \prob 7 Suppose $v\in \complex^{n\by 1}$ and $v$ is not the zero vector and $E$ is the set of all $A\in\complex^{n\by n}$ such that $v$ is an eigenvector of $A$. (This means that $A\in E$ iff there exists $\lambda\in\complex$ such that $Av=\lambda v$.) Prove that $E$ is a subspace of $\complex^{n\by n}$. %\prob 6 Suppose that $V$ is a vector space over the field $\field$, %and $u_i$'s and $v_j$'s are vectors in $V$ such that %$\{v_1,\ldots,v_m\}\subseteq span(\{u_1,u_2,\ldots,u_n\})$. %Prove that %$span(\{v_1,\ldots,v_m\})\subseteq span(\{u_1,u_2,\ldots,u_n\})$. \prob 8 Suppose that $V$ is a vector space over $\field$, $\{u_1,u_2,\ldots,u_n\}$ is a basis for $V$, and $u\in V$ is not the zero vector. Prove that for some $i$ with $1\leq i\leq n$ that $$\{u_1,u_2,\ldots,u_{i-1},u_{i+1},\ldots,u_n, u\}$$ is a basis for $V$. \newpage \begin{flushright} A. Miller\\ M340 Exam 3 answers\\ April 1990 \end{flushright} \bigskip 1. Since $v_1\in span(\{v_2,\ldots,v_m\})$ there exists $c_2,c_3,\ldots,c_m\in \field$ such that $v_1=c_2v_2+\cdots+c_mv_m$. But then ${\bf 0}=(-1)v_1+c_2v_2+\cdots+c_mv_m$ and since the coefficient of $v_1$ is not zero, $v_1,v_2,\ldots,v_m$ are linearly dependent. \bigskip 2. $L(A+B)=(A+B)-(A+B)^\prime= (A+B)-(A^\prime+B^\prime) =A-A^\prime+B-B^\prime$ $=L(A)+L(B)$ $L(cA)=(cA)-(cA)^\prime=cA+cA^\prime=c(A-A^\prime)=cL(A)$ \bigskip 3. Since $[0]_{n\by n}v=0v$ we have that $[0]_{n\by n}\in E$ which is the zero vector of $\complex^{n\by n}$. If $A_0\in E$ and $A_1\in E$, then there exists $\lambda_0,\lambda_1\in\complex$ such that $A_0v=\lambda_0v$ and $A_1v=\lambda_1v$ and so $$(A_0+A_1)v=A_0v+A_1v=\lambda_0v+\lambda_1v=(\lambda_0+\lambda_1)v$$ so $(A_0+A_1)\in E$. If $c\in\complex$ and $A\in E$, then there exists $\lambda\in\complex$ such that $Av=\lambda v$ and $$(cA)v=c(Av)=c(\lambda v)=(c\lambda)v$$ so $cA\in E$. \bigskip 4. Since $\{u_1,u_2,\ldots,u_n\}$ is a basis for $V$ it spans $V$ and therefore there are scalars $c_1,\ldots,c_n$ such that $u=c_1u_1+\cdots+c_nu_n$. Since $u$ is not the zero vector and $u_1,u_2,\ldots,u_n$ are linearly independent some $c_i\not=0$. For this particular $i$ we will show that $\{u_1,u_2,\ldots,u_{i-1},u_{i+1},\ldots,u_n, u\}$ is a basis. \begin{center} Linear independence \end{center} Suppose that $$d_1u_1+d_2u_2+\cdots +d_{i-1}u_{i-1}+d_{i+1}u_{i+1}+\cdots+d_nu_n+d u={\bf 0}.$$ Replacing $u$ by $c_1u_1+\cdots+c_nu_n$ and combining the coefficients we get $$(d_1+dc_1) u_1 +\cdots+ (d_{i-1}+dc_{i-1})u_{i-1}+ (dc_i) u_i+ (d_{i+1}+d{c_{i+1}})u_{i+1}+\cdots $$ $$\cdots+(d_n+d{c_{n}})u_{n} ={\bf 0}$$ Since $u_1,u_2,\ldots,u_n$ are linearly independent it must be that all these coefficients are zero, in particular $dc_i=0$. But $c_i\not=0$ so that it must be that $d=0$. But replacing $d$ with $0$ gives $$d_1u_1+d_2u_2+\cdots +d_{i-1}u_{i-1}+d_{i+1}u_{i+1}+\cdots+d_nu_n={\bf 0}$$ and since $u_1,u_2,\ldots,u_n$ are linearly independent all the $d_j$'s must be zero as well. Hence $u_1,u_2,\ldots,u_{i-1},u_{i+1},\ldots,u_n, u$ are linearly independent. \begin{center} Spanning \end{center} Since $c_i\not=0$ we can solve for $u_i$ $$u_i=-{c_1\over c_i} u_1 +\cdots+ -{c_{i-1}\over c_i}u_{i-1}+ -{c_{i+1}\over c_i}u_{i+1}+\cdots+ -{c_{n}\over c_i}u_{n}+{1\over c_i}u.$$ To simplify the notation write these scalars as $a$'s $$u_i=a_1 u_1 +\cdots+ a_{i-1}u_{i-1}+ a_{i+1}u_{i+1}+\cdots +a_nu_{n}+au.$$ Let $v\in V$ be arbitrary. Since $\{u_1,u_2,\ldots,u_n\}$ is a basis for $V$ there exists scalars $d_1,\ldots,d_n$ such that $v=d_1u_1+\cdots+d_nu_n$. Replace $d_iu_i$ by $d_ia_1 u_1 +\cdots+ d_ia_{i-1}u_{i-1}+ d_ia_{i+1}u_{i+1}+\cdots +d_ia_nu_{n}+d_iau$. and combine coefficients to get $$v=(d_1+d_ia_1) u_1 +\cdots+ (d_{i-1}+d_ia_{i-1})u_{i-1}+ (d_{i+1}+d_ia_{i+1})u_{i+1}+\cdots $$ $$\cdots +(d_n+d_ia_n)u_{n}+(d_ia)u$$ Hence $v\in span(\{u_1,u_2,\ldots,u_{i-1},u_{i+1},\ldots,u_n, u\})$. Since $v$ was arbitrary $V=span(\{u_1,u_2,\ldots,u_{i-1},u_{i+1},\ldots,u_n, u\})$. \end{document} \end{document}