#LyX 1.4.3 created this file. For more info see http://www.lyx.org/ \lyxformat 245 \begin_document \begin_header \textclass amsart \begin_preamble \setlength{\topmargin}{-.7in} \setlength{\headheight}{0in} \setlength{\headsep}{0in} \setlength{\footskip}{0in} \setlength{\evensidemargin}{-.5in} \setlength{\oddsidemargin}{-.5in} \setlength{\textheight}{10in} \setlength{\textwidth}{7in} \setlength{\parindent}{0in} \newcommand{\rank}{\operatorname{rank}} \newcommand{\dimn}{\operatorname{dim}} \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{\prj}{\operatorname{proj}} \newcommand{\D}{\mathrm{d}} \newcommand{\E}{\mathrm{e}} \let\eul=\E \newcommand{\I}{{\rm i}} \let\imag=\I \end_preamble \language english \inputencoding auto \fontscheme default \graphics default \paperfontsize 12 \spacing single \papersize default \use_geometry false \use_amsmath 1 \cite_engine basic \use_bibtopic false \paperorientation portrait \topmargin -0.7cm \headheight 0cm \headsep 0cm \footskip 0cm \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 \begin_inset FormulaMacro \newcommand{\realn}[1]{\mathbb{R}^{#1}} \end_inset \end_layout \begin_layout Standard \begin_inset FormulaMacro \newcommand{\cplxn}[1]{\mathbb{C}^{#1}} \end_inset \end_layout \begin_layout Standard \begin_inset FormulaMacro \newcommand{\mapp}[3]{#1:#2\rightarrow#3} \end_inset \end_layout \begin_layout Standard \series bold Final\InsetSpace ~ Exam \hfill Math\InsetSpace ~ 314, Section 006 \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. \begin_inset ERT status collapsed \begin_layout Standard \backslash thispagestyle{empty} \end_layout \end_inset \newline \begin_inset ERT status collapsed \begin_layout Standard \backslash rule[.2in]{7.5in}{.01in} \end_layout \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{-0.1in} \end_layout \end_inset (30) \series bold 1. \series default In the following problem use the fact that \begin_inset Formula $R$ \end_inset is the reduced row-echelon form of the augmented matrix \begin_inset Formula $\widetilde{A}=\left[A\,|\,\mathbf{b}\right]$ \end_inset where \end_layout \begin_layout Standard \begin_inset Formula $\ds\widetilde{A}=\left[\begin{array}{cccccc} 3 & 1 & -2 & 0 & 1 & 1\\ 1 & 1 & 0 & -1 & 1 & 2\\ 3 & 2 & -1 & 1 & 1 & 9\\ 0 & 2 & 2 & -1 & 1 & 8\end{array}\right]=\left[\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v_{\mathbf{3}}},\mathbf{v}_{4},\mathbf{v}_{5},\mathbf{b}\right]\mbox{ and }R=\left[\begin{array}{cccccc} 1 & 0 & -1 & 0 & 0 & -3\\ 0 & 1 & 1 & 0 & 0 & 3\\ 0 & 0 & 0 & 1 & 0 & 5\\ 0 & 0 & 0 & 0 & 1 & 7\end{array}\right].$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (a) Find a basis for the row space of \begin_inset Formula $A$ \end_inset and the dimension of the row space. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.3in} \end_layout \end_inset (b) Find a basis for the column space of \begin_inset Formula $A$ \end_inset and the rank of \begin_inset Formula $A$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.3in} \end_layout \end_inset (c) Find a basis for the null space of \begin_inset Formula $A$ \end_inset and the nullity of \begin_inset Formula $A$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.8in} \end_layout \end_inset (d) Find the general solution to the system \begin_inset Formula $A\mathbf{x}=\mathbf{b}$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.5in} \end_layout \end_inset (e) Fill in the blanks below if possible or give a reason if not possible. (Hint: \begin_inset Formula $A\mathbf{x}$ \end_inset as a l.c.) \end_layout \begin_layout Standard \begin_inset Formula \[ \begin{array}{ccccccc} \rule{.5in}{.01in}\mathbf{v}_{3} & + & \rule{.5in}{.01in}\mathbf{v}_{4} & + & \rule{.5in}{.01in}\mathbf{v}_{5} & = & \mathbf{b}\end{array}\] \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash clearpage \end_layout \end_inset (18) \series bold 2. \series default Calculate \begin_inset Formula $A^{-1}$ \end_inset and \begin_inset Formula $\det\left(A\right)$ \end_inset , where \begin_inset Formula $A=\left[\begin{array}{rrr} 1 & 2 & 0\\ 0 & -1 & 0\\ 2 & 4 & 2\end{array}\right]$ \end_inset and use \begin_inset Formula $A^{-1}$ \end_inset to solve the equation \begin_inset Formula $A\mathbf{x}=\mathbf{b}$ \end_inset , where \begin_inset Formula $\mathbf{b}=\left(2,0,4\right)$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{4.2in} \end_layout \end_inset (16) \series bold 3. \series default Let \begin_inset Formula $A=\left[\begin{array}{rr} 1 & -2\\ 0 & 1\\ 1 & 1\end{array}\right]$ \end_inset and \begin_inset Formula $B=\left[\begin{array}{rr} 0 & 1\\ 1 & 1\\ 3 & 0\end{array}\right]$ \end_inset , so that \begin_inset Formula $R=\left[\begin{array}{rrcr} 1 & 0 & 2 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 1\end{array}\right]$ \end_inset is the reduced row echelon form of \begin_inset Formula $\left[A,B\right]$ \end_inset (assume this). Find bases and dimensions of the following subspaces of \begin_inset Formula $\realn{3}$ \end_inset : \end_layout \begin_layout Standard (a) \begin_inset Formula $\mathcal{C}\left(A\right)$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.9in} \end_layout \end_inset (b) \begin_inset Formula $\mathcal{C}\left(B\right)$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.9in} \end_layout \end_inset (c) \begin_inset Formula $\mathcal{C}\left(A\right)+\mathcal{C}\left(B\right)$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.9in} \end_layout \end_inset (d) \begin_inset Formula $\mathcal{C}\left(A\right)\cap\mathcal{C}\left(B\right)$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash clearpage \end_layout \end_inset (16) \series bold 4. \series default Let \begin_inset Formula $\mathbf{w}_{1}=\left(1,0,1,0\right)$ \end_inset , \begin_inset Formula $\mathbf{w}_{2}=\left(0,1,0,1\right)$ \end_inset and \begin_inset Formula $\mathbf{w}_{3}=\left(0,1,0,-1\right)$ \end_inset . \end_layout \begin_layout Standard (a) Show these vectors are orthogonal in the inner product space \begin_inset Formula $\realn{4}$ \end_inset with the standard inner product. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.4in} \end_layout \end_inset (b) Why are these vectors linearly independent? \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.0in} \end_layout \end_inset (c) Are these vectors are orthogonal in the inner product space \begin_inset Formula $\realn{4}$ \end_inset with the non-standard inner product \begin_inset Formula $\left\langle \mathbf{x},\mathbf{y}\right\rangle =x_{1}y_{1}+2x_{2}y_{2}+3x_{3}y_{3}+x_{4}y_{4}$ \end_inset ? If not, use Gram-Schmidt to generate an orthogonal set from them. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{3.3in} \end_layout \end_inset \end_layout \begin_layout Standard (16) \series bold 5. \series default Are following subsets \begin_inset Formula $W$ \end_inset of vector space \begin_inset Formula $V$ \end_inset are subspaces of \begin_inset Formula $V$ \end_inset ? Justify your answers. \end_layout \begin_layout Standard (a) \begin_inset Formula $W=\left\{ \mathbf{x}\in\realn{3}\,|\,\mathbf{x}^{T}A\mathbf{x}=-1\right\} \subseteq V=\realn{3}$ \end_inset , where \begin_inset Formula $A=\left[\begin{array}{rrr} 1 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 2\end{array}\right]$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.0in} \end_layout \end_inset (b) \begin_inset Formula $W=\left\{ \mathbf{x}\in\realn{2}\,|\, A\mathbf{x}=0\right\} \subseteq V=\realn{2}$ \end_inset , where \begin_inset Formula $A=\left[\begin{array}{rr} 1 & 2\\ 0 & 4\end{array}\right].$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash clearpage \end_layout \end_inset (16) \series bold 6. \series default The matrix \begin_inset Formula $A=\left[\begin{array}{cc} 2 & -1\\ -1 & 2\end{array}\right]$ \end_inset is symmetric with an eigenvector \begin_inset Formula $\mathbf{v}_{1}=\left(1,1\right)$ \end_inset . \end_layout \begin_layout Standard (a) Find an orthonormal basis of \begin_inset Formula $\realn{2}$ \end_inset consisting of eigenvectors of \begin_inset Formula $A$ \end_inset and find the eigenvalues of \begin_inset Formula $A$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{2.1in} \end_layout \end_inset (b) Use (a) to find a matrix formula (with no matrix products) for \begin_inset Formula $A^{k}$ \end_inset , \begin_inset Formula $k>0$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{2.1in} \end_layout \end_inset (20) \series bold 7. \series default For the matrix \begin_inset Formula $\begin{array}{ccc} A & = & \left[\begin{array}{ccc} 1 & 8 & 8\\ 0 & 5 & 4\\ 0 & 0 & 1\end{array}\right]\end{array}$ \end_inset find a matrix \begin_inset Formula $P$ \end_inset and a diagonal matrix \begin_inset Formula $D$ \end_inset such that \begin_inset Formula $P^{-1}AP=D.$ \end_inset (You do \shape italic not \shape default have to find \begin_inset Formula $P^{-1}$ \end_inset .) \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash clearpage \end_layout \end_inset (16) \series bold 8. \series default Find a least squares solution to the system \begin_inset Formula \[ \left[\begin{array}{cc} 1 & 1\\ 1 & 1\\ 1 & 2\end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2}\end{array}\right]=\left[\begin{array}{c} 1\\ 2\\ 3\end{array}\right].\] \end_inset Is the least squares solution a genuine solution? \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{3.5in} \end_layout \end_inset (22) \series bold 9. \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{.1in} \end_layout \end_inset (a) If \begin_inset Formula $A,B$ \end_inset are \begin_inset Formula $2\times2$ \end_inset matrices, then \begin_inset Formula $\left(AB\right)^{2}=A^{2}B^{2}$ \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{.1in} \end_layout \end_inset (b) Every orthonormal set of vectors is linearly independent (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{.1in} \end_layout \end_inset (c) If \begin_inset Formula $A$ \end_inset is real symmetric, then the eigenvalues of \begin_inset Formula $A$ \end_inset are real (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{.1in} \end_layout \end_inset (d) If the linear system \begin_inset Formula $A\mathbf{x}=\mathbf{0}$ \end_inset has infinitely many solutions and \begin_inset Formula $A$ \end_inset is an \begin_inset Formula $m\times n$ \end_inset matrix, then \begin_inset Formula $n>m$ \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{.1in} \end_layout \end_inset (e) If \begin_inset Formula $V=\mbox{span}\left\{ \mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3}\right\} $ \end_inset , then \begin_inset Formula $\dim V=3$ \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 (f) The CBS inequality for an inner product space \begin_inset Formula $V$ \end_inset says that for all vectors \begin_inset Formula $\mathbf{u},\mathbf{v}\in V$ \end_inset , \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{.1in} \end_layout \end_inset (g) If \begin_inset Formula $x$ \end_inset solves the normal equations for \begin_inset Formula $A\mathbf{x}=\mathbf{b}$ \end_inset then \begin_inset Formula $A\mathbf{x}$ \end_inset is the projection of \begin_inset Formula $\mathbf{b}$ \end_inset into \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 (h) The cosine of the angle between \begin_inset Formula $\left(1,1\right)$ \end_inset and \begin_inset Formula $\left(1,-3\right)$ \end_inset in \begin_inset Formula $\realn{2}$ \end_inset with the standard inner product is \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{.1in} \end_layout \end_inset (i) \begin_inset Formula $\left[\begin{array}{cc} 1 & 2\\ 0 & 1+i\end{array}\right]$ \end_inset \begin_inset Formula $\left[\begin{array}{ccc} 1 & 2 & 0\\ 0 & i & 1\end{array}\right]=$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.1in} \end_layout \end_inset (j) \begin_inset Formula $\left[\begin{array}{ccc} 1 & 2 & 0\\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc} 1 & 2 & 0\\ 0 & 1 & 1\end{array}\right]^{T}=$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.1in} \end_layout \end_inset (k) \begin_inset Formula $T\left(\left(x,y\right)\right)=\left(x+y,2x,4y-x\right)$ \end_inset is a matrix multiplication operator \begin_inset Formula $T_{A}\left(\left(x,y\right)\right),$ \end_inset where \begin_inset Formula $A=$ \end_inset \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 clearpage \end_layout \end_inset \end_layout \begin_layout Standard (30) \series bold 10. \series default Give brief answers to the following. Do only one (honors students, two) of (d), (e), or (f). \end_layout \begin_layout Standard (a) Let \begin_inset Formula $f\left(x\right)=x$ \end_inset and \begin_inset Formula $g\left(x\right)=x^{2}$ \end_inset in the inner product space \begin_inset Formula $C\left[0,1\right]$ \end_inset with the standard function space inner product ( \begin_inset Formula $\left\langle f,g\right\rangle =\int_{0}^{1}f\left(x\right)g\left(x\right)dx$ \end_inset . Find the projection of \begin_inset Formula $g\left(x\right)$ \end_inset along \begin_inset Formula $f\left(x\right)$ \end_inset in this space. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.3in} \end_layout \end_inset (b) Use determinants to determine for what \begin_inset Formula $x$ \end_inset the matrix \begin_inset Formula $A=\left[\begin{array}{ccc} 1 & 1 & 1\\ 1 & 2 & 3\\ 0 & 4 & x\end{array}\right]$ \end_inset is singular. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.3in} \end_layout \end_inset (c) Find \begin_inset Formula $\Vert2(\mathbf{u}-\mathbf{v})\Vert_{p}$ \end_inset , \begin_inset Formula $p=1,2,\infty$ \end_inset , where \begin_inset Formula $\mathbf{u}=\left(-2,1,1\right)$ \end_inset and \begin_inset Formula $\mathbf{v}=\left(0,3,-3\right)$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.3in} \end_layout \end_inset (d) Show \shape italic from definition \shape default that if \begin_inset Formula $\lambda$ \end_inset is an eigenvalue of invertible \begin_inset Formula $A$ \end_inset , then \begin_inset Formula $1/\lambda$ \end_inset is an eigenvalue of \begin_inset Formula $A^{-1}$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.3in} \end_layout \end_inset (e) Show \shape italic from definition \shape default that if \begin_inset Formula $\mathbf{v}_{1}=\mathbf{0}$ \end_inset then the set of vectors \begin_inset Formula $\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3}$ \end_inset in the vector space \begin_inset Formula $V$ \end_inset is linearly dependent. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{1.3in} \end_layout \end_inset (f) Show that if \begin_inset Formula $\mathbf{v}\in\realn{n}$ \end_inset is a nonzero vector, then the matrix \begin_inset Formula $\ds H=I_{n}-2\frac{\mathbf{v}\mathbf{v}^{T}}{\mathbf{v}^{T}\mathbf{v}}$ \end_inset is orthogonal. \end_layout \end_body \end_document