#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}} \end_preamble \language english \inputencoding auto \fontscheme default \graphics default \paperfontsize 12 \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 Sample \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 (25) \series bold 1. \series default In the following you are given that \begin_inset Formula \[ A=[\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{v}_{4},\mathbf{v}_{5}]=\left[\begin{array}{ccccc} 1 & -1 & 5 & 1 & 0\\ 2 & 1 & 4 & 2 & 1\\ 3 & 0 & 9 & 3 & 1\\ -1 & -1 & -1 & -1 & 2\end{array}\right]\text{ has reduced row echelon form }E=\left[\begin{array}{ccccc} 1 & 0 & 3 & 1 & 0\\ 0 & 1 & -2 & 0 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\end{array}\right].\] \end_inset \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (a) What is the rank of \begin_inset Formula $A$ \end_inset and dimensions of \begin_inset Formula ${\mathcal{R}}(A)$ \end_inset , \begin_inset Formula ${\mathcal{C}}(A)$ \end_inset and \begin_inset Formula ${\mathcal{N}}(A)$ \end_inset ? \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (b) Find a basis for 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{0.1in} \end_layout \end_inset (c) Find a basis for 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{0.1in} \end_layout \end_inset (d) Find a basis for 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{0.1in} \end_layout \end_inset (e) Express \begin_inset Formula $\mathbf{v}_{3}$ \end_inset as a linear combination of \begin_inset Formula $\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{4},\mathbf{v}_{5}$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.15in} \end_layout \end_inset (12) \series bold 2. \series default In the following, \begin_inset Formula $\mathbf{u}_{1}=(1,0,1)$ \end_inset and \begin_inset Formula $\mathbf{u}_{2}=(1,-1,1)$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (a) Show that \begin_inset Formula $\mathbf{u}_{1},\mathbf{u}_{2}$ \end_inset form a linearly independent set. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (b) Does \begin_inset Formula $\mathbf{v}=(2,1,2)$ \end_inset belong to \begin_inset Formula $\spann\{\mathbf{u}_{1},\mathbf{u}_{2}\}$ \end_inset ? Give reasons for your answer. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (c) Find a basis of \begin_inset Formula $\mathbb{R}^{3}$ \end_inset which contains \begin_inset Formula $\mathbf{u}_{1}$ \end_inset and \begin_inset Formula $\mathbf{u}_{2}$ \end_inset . Give reasons for your answer. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.15in} \end_layout \end_inset (12) \series bold 3. \series default Use the subspace test to decide if \begin_inset Formula $W$ \end_inset is a subspace of the given (abstract) vector space \begin_inset Formula $V$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (a) \begin_inset Formula $V$ \end_inset is the space of all \begin_inset Formula $2\times2$ \end_inset matrices over the reals with the usual matrix addition and scalar multiplicatio n and \begin_inset Formula $W$ \end_inset is the set of \begin_inset Formula $2\times2$ \end_inset matrices of the form \begin_inset Formula $\{\left[\begin{array}{cc} a & b\\ 0 & b\end{array}\right]|\, a,b$ \end_inset are reals}. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (b) \begin_inset Formula $V$ \end_inset is the space \begin_inset Formula $C[0,1]$ \end_inset of continuuous functions on [0,1] and W is set of \begin_inset Formula $f$ \end_inset in \begin_inset Formula $C[0,1]$ \end_inset such that \begin_inset Formula $f(0)=2$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.15in} \end_layout \end_inset (12) \series bold 4. \series default Let \begin_inset Formula $T:\mathcal{P}_{2}\rightarrow\mathbb{R}^{2}$ \end_inset be the linear operator defined by \begin_inset Formula \[ T\left(a+bx+cx^{2}\right)=\left[\begin{array}{c} a+c\\ a-b\end{array}\right].\] \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 the kernel and range of \begin_inset Formula $T$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (b) Is this operator one-to-one? Onto? An isomorphism? Explain. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.15in} \end_layout \end_inset (24) \series bold 5. \series default Fill in the blanks or answer T/F: \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (a) Every spanning set of a vector space contains a basis of the 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{.1in} \end_layout \end_inset (b) 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{.1in} \end_layout \end_inset (c) The vectors \begin_inset Formula $\mathbf{v}_{1},\mathbf{v}_{2},\ldots,\mathbf{v}_{n}$ \end_inset are linearly dependent means that \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \begin_layout Standard \end_layout \begin_layout Standard \backslash rule{5in}{.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) Every subspace of a finite dimensional space is itself finite dimensional (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{.1in} \end_layout \end_inset (e) The span of the set \begin_inset Formula $\left\{ \mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3}\right\} $ \end_inset in a vector space is the set of all vectors \begin_inset Formula $\mathbf{v}$ \end_inset of the form \begin_inset Formula $\mathbf{v}=$ \end_inset \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{.2in} \end_layout \begin_layout Standard \end_layout \begin_layout Standard \backslash rule{3in}{.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 (f) The set consisting of the zero vector is a linearly independent set (T/F) \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1.3in}{.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) The function \begin_inset Formula $T:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ \end_inset given by \begin_inset Formula $T((x,y))=(x+y,x-2xy)$ \end_inset is linear (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{.1in} \end_layout \end_inset (h) The adjoint of the matrix \begin_inset Formula $\left[\begin{array}{ccc} 1 & 0 & 1\\ 0 & 1 & 0\\ 2 & 0 & 0\end{array}\right]$ \end_inset is \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1.3in}{.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) The cofactor matrix of \begin_inset Formula $A=\left[\begin{array}{cc} a & b\\ c & d\end{array}\right]$ \end_inset is \begin_inset Formula $A_{\mbox{cof}}=$ \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 (j) The inverse of a matrix \begin_inset Formula $A$ \end_inset with nonzero determinant is \begin_inset Formula ${\displaystyle \frac{1}{\det A}A_{\mbox{cof}}}$ \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 (k) The dimension of \begin_inset Formula $\mathbb{C}$ \end_inset as a vector space over \begin_inset Formula $\mathbb{R}$ \end_inset 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 (l) The Dimension Theorem asserts that any two bases of a vector space \begin_inset ERT status collapsed \begin_layout Standard \backslash rule{1.5in}{.01in} \end_layout \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.15in} \end_layout \end_inset (15) \series bold 6. \series default Suppose that the linear system \begin_inset Formula $A\mathbf{x}=\mathbf{b}$ \end_inset is a consistent system of equations, where \begin_inset Formula $A$ \end_inset is an mxn matrix and \begin_inset Formula $\mathbf{x}=[x_{1},...,x_{n}]^{T}$ \end_inset . Prove the following: \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (a) \begin_inset Formula $\mathbf{b}\in{\mathcal{C}}(A)$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (b) If \begin_inset Formula $\mathbf{x}_{0}$ \end_inset is any particular solution to the system, then every vector of the form \begin_inset Formula $\mathbf{x}_{0}+\mathbf{z}$ \end_inset , where \begin_inset Formula $z\in{\mathcal{N}}(A)$ \end_inset , is a solution to the system. \end_layout \begin_layout Standard \begin_inset ERT status collapsed \begin_layout Standard \backslash vspace{0.1in} \end_layout \end_inset (c) If the set of columns of \begin_inset Formula $A$ \end_inset has redundant vectors in it, show the system has more than one solution. \end_layout \end_body \end_document