#LyX 1.4.3 created this file. For more info see http://www.lyx.org/ \lyxformat 245 \begin_document \begin_header \textclass article \begin_preamble %% \usepackage{amsfonts} % \setlength{\topmargin}{-.8in} \setlength{\headheight}{0in} \setlength{\headsep}{0in} \setlength{\footskip}{0in} \setlength{\evensidemargin}{-.5in} \setlength{\oddsidemargin}{-.5in} \setlength{\textheight}{9.3in} \setlength{\textwidth}{7in} \setlength{\parindent}{0in} \newcommand{\rank}{\operatorname{rank}} \newcommand{\spann}{\operatorname{span}} \newcommand{\ds}{\displaystyle} \newcommand{\spn}{\operatorname{span}} \newcommand{\rnk}{\operatorname{rank}} \newcommand{\cond}{\operatorname{cond}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\diag}{\operatorname{diag}} \newcommand{\id}{\operatorname{id}} \newcommand{\mybox}{\tag*{$\square$}} \newcommand{\prj}{\operatorname{proj}} \newcommand{\limlim}[1]{@>>#1>} \end_preamble \language english \inputencoding auto \fontscheme default \graphics default \paperfontsize 12 \spacing single \papersize default \use_geometry false \use_amsmath 2 \cite_engine basic \use_bibtopic false \paperorientation portrait \secnumdepth 3 \tocdepth 3 \paragraph_separation indent \defskip medskip \quotes_language english \papercolumns 1 \papersides 1 \paperpagestyle empty \tracking_changes false \output_changes true \end_header \begin_body \begin_layout Standard \series bold Exam\InsetSpace ~ 2 \hfill Math\InsetSpace ~ 314 \hfill Fall 2006 \newline Name: \series default \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{2.4in}{.01in} \end_layout \end_inset \hfill \series bold Score: \series default \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1in}{.01in} \end_layout \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash hspace{.25in} \end_layout \end_inset \shape italic Instructions: \begin_inset ERT status collapsed \begin_layout Standard \backslash /{} \end_layout \end_inset \shape default Show your work in the spaces provided below for full credit. Use the reverse side for additional space, \shape italic but clearly so indicate. \begin_inset ERT status collapsed \begin_layout Standard \backslash /{} \end_layout \end_inset \shape default You must clearly identify answers and show supporting work to receive any credit. Exact answers (e.g., \begin_inset Formula $\pi$ \end_inset ) are preferred to inexact (e.g., \begin_inset Formula $3.14$ \end_inset ). Point values of problems are given in parentheses. Notes or text in \emph on any \emph default form not allowed. Calculator is allowed. \newline \begin_inset ERT status collapsed \begin_layout Standard \backslash rule[.2in]{7in}{.01in} \end_layout \end_inset \end_layout \begin_layout Standard (30) \series bold 1. \series default Let \begin_inset Formula $A=[\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4}]=\left[\begin{array}{cccc} 1 & 4 & 1 & -1\\ 2 & 4 & 1 & -1\\ 4 & 8 & 2 & -2\end{array}\right]$ \end_inset with reduced row echelon form \begin_inset Formula $R=\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & \frac{1}{4} & 0\\ 0 & 0 & 0 & 0\end{array}\right]$ \end_inset . \end_layout \begin_layout Standard (a) Find a basis for \begin_inset Formula $\mathcal{R}(A)$ \end_inset , the row space of \begin_inset Formula $A$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1in} \end_layout \end_inset (b) Find a basis for \begin_inset Formula $\mathcal{C}(A)$ \end_inset , the column space of \begin_inset Formula $A$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.2in} \end_layout \end_inset (c) Find a basis for \begin_inset Formula $\mathcal{N}(A)$ \end_inset , the null space of \begin_inset Formula $A$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.4in} \end_layout \end_inset (d) Find all possible linear combinations of the vectors \begin_inset Formula $\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4}$ \end_inset that sum to \begin_inset Formula $\mathbf{0}.$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.3in} \end_layout \end_inset (e) Which \begin_inset Formula $\mathbf{v}_{j}$ \end_inset 's are redundant in the list of vectors \begin_inset Formula $\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4}?$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.2in} \end_layout \end_inset (f) Find a basis of \begin_inset Formula $\mathbb{R}^{3}$ \end_inset containing a basis of \begin_inset Formula $\mathcal{C}(A)$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash clearpage \end_layout \end_inset \end_layout \begin_layout Standard (16) \series bold 2. \series default Use the Subspace Test to decide if \begin_inset Formula $W$ \end_inset is a subspace of the vector space \begin_inset Formula $V$ \end_inset , where \end_layout \begin_layout Standard (a) \begin_inset Formula $V=\mathbb{R}^{3}$ \end_inset and \begin_inset Formula $W=\left\{ \left(a,b,a-b+1\right)\,|\, a,b\in\mathbb{R}\right\} $ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{3.0in} \end_layout \end_inset (b) \begin_inset Formula $V=C[0,1]$ \end_inset , the continuous functions on \begin_inset Formula $[0,1]$ \end_inset and \begin_inset Formula $W=\left\{ f\left(x\right)\,|\, f\left(x\right)\in C\left[0,1\right]\mbox{ and }f\left(1\right)=0\right\} $ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{3.0in} \end_layout \end_inset (10) \series bold 3. \series default Assume that \begin_inset Formula $1+x$ \end_inset , \begin_inset Formula $x+x^{2}$ \end_inset , \begin_inset Formula $1-x$ \end_inset is a basis of \begin_inset Formula $\mathcal{P}_{2}$ \end_inset , the space of polynomials of degree at most two, and find the coordinates of \begin_inset Formula $2+x^{2}$ \end_inset relative to this basis. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash clearpage \end_layout \end_inset \end_layout \begin_layout Standard (8) \series bold 4. \series default Let \begin_inset Formula $A=\left[\begin{array}{ccc} 1 & 0 & 3\\ 0 & 1 & 0\\ 1 & 0 & -1\end{array}\right]$ \end_inset . Find the adjoint matrix \begin_inset Formula $\operatorname{adj}\left(A\right)$ \end_inset of \begin_inset Formula $A$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{4.0in} \end_layout \end_inset \end_layout \begin_layout Standard (12) \series bold 5. \series default You are given that \begin_inset Formula $\mathbf{w}_{1}=(0,1,0)$ \end_inset , \begin_inset Formula $\mathbf{w}_{2}=(1,1,1)$ \end_inset is a linearly independent set in \begin_inset Formula $V=\mathbb{R}^{3}$ \end_inset and \begin_inset Formula $\mathbf{v}_{1}=(1,3,1)$ \end_inset , \begin_inset Formula $\mathbf{v}_{2}=(2,-1,1)$ \end_inset , \begin_inset Formula $\mathbf{v}_{3}=(1,0,1)$ \end_inset is a basis of \begin_inset Formula $V$ \end_inset . Steinitz substitution says that \begin_inset Formula $\mathbf{w}_{1}$ \end_inset , \begin_inset Formula $\mathbf{w}_{2}$ \end_inset can be substituted into the basis in place of certain \begin_inset Formula $\mathbf{v}_{i}$ \end_inset 's. Which substitutions work? \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash clearpage \end_layout \end_inset \end_layout \begin_layout Standard (16) \series bold 6. \series default Fill in the blanks or answer True/False (T/F). \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.12in} \end_layout \end_inset \end_layout \begin_layout Standard (a) Every vector space is finite dimensional (T/F) \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1in}{.01in} \end_layout \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (b) Elementary row operations on a matrix do not change the column space (T/F) \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1in}{.01in} \end_layout \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (c) If \begin_inset Formula $\mathbf{x}=\mathbf{x}_{0}$ \end_inset and \begin_inset Formula $\mathbf{x}=\mathbf{x}_{1}$ \end_inset are both vector solutions to the linear system \begin_inset Formula $A\mathbf{x}=\mathbf{b},$ \end_inset then \begin_inset Formula $\mathbf{x}_{1}-\mathbf{x}_{0}$ \end_inset is in the null space of \begin_inset Formula $A.$ \end_inset (T/F) \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1in}{.01in} \end_layout \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (d) The function \begin_inset Formula $T:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ \end_inset given by \begin_inset Formula $T((x,y))=(x+y,x-2y)$ \end_inset is linear (T/F) \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1.0in}{.01in} \end_layout \end_inset and one-to-one (T/F) \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1.0in}{.01in} \end_layout \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (e) The Basis Theorem asserts that every finite dimensional vector space \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{2.0in}{.01in} \end_layout \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (f) The Dimension Theorem asserts that \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{3.5in}{.01in} \end_layout \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \end_inset (g) A basis of a vector space is by definition \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{3.5in}{.01in} \end_layout \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.25in} \end_layout \end_inset (10) \series bold 7. \series default (a) Show that the columns of the matrix \begin_inset Formula $\left[\begin{array}{ccc} 1 & 0 & 1\\ 1 & 0 & 1\\ 3 & 0 & 1\end{array}\right]$ \end_inset form a linearly dependent set. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{3.0in} \end_layout \end_inset \end_layout \begin_layout Standard (b) ( \emph on Honors students only \emph default ) Prove that any set of vectors \begin_inset Formula $\mathbf{v}_{1},\mathbf{v}_{2},\ldots,\mathbf{v}_{n}$ \end_inset in a vector space \begin_inset Formula $V$ \end_inset that contains the zero vector is a linearly dependent set. \end_layout \end_body \end_document