%% This document created by Scientific Word (R) Version 3.0

\documentclass[oneside,12pt]{amsart}%
\usepackage{float}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{amsfonts}
\usepackage{amssymb}%
\setcounter{MaxMatrixCols}{30}
%TCIDATA{OutputFilter=latex2.dll}
%TCIDATA{Version=5.50.0.2960}
%TCIDATA{CSTFile=amsart.cst}
%TCIDATA{LastRevised=Sunday, March 27, 2022 12:14:41}
%TCIDATA{<META NAME="GraphicsSave" CONTENT="32">}
%TCIDATA{<META NAME="SaveForMode" CONTENT="1">}
%TCIDATA{BibliographyScheme=Manual}
%BeginMSIPreambleData
\providecommand{\U}[1]{\protect\rule{.1in}{.1in}}
%EndMSIPreambleData
\textwidth=6.45in
\textheight=10in
\oddsidemargin=0in
\evensidemargin=0in
\topmargin=-.70in
\setlength{\parskip}{0in}
\setlength{\parindent}{0in}
\begin{document}

\begin{center}
\textbf{Math 3013.21403} \\%[0pt] 
SOLUTIONS TO SECOND EXAM \\%[0pt] 
March 28, 2022
\end{center}

1. Define, precisely, the following notions (where $V$, $W$ are to be regarded
as general vector spaces). space $V$).

(a) (5 pts) a \textbf{subspace} of $V$ is

\begin{itemize}
\item a subset $W$ of $V$ that is
closed under both scalar multiplication and vector addition; i.e.,

\begin{itemize}
\item For all $\lambda\in\mathbb{R}$ and all $\mathbf{v}\in W$, $\lambda
\mathbf{v}\in W$

\item For all $\mathbf{v}_{1},\mathbf{v}_{2}\in W$, $\mathbf{v}_{1}%
+\mathbf{v}_{2}\in W$
\end{itemize}
\end{itemize}

(b) (5 pts) a \textbf{basis} for a vector space $V$ is

\begin{itemize}
\item a set of vectors $\left\{  \mathbf{b}%
_{1},\ldots,\mathbf{b}_{k}\right\}  $ such that every vector $\mathbf{v}\in V$
can be uniquely expressed as%
\[
\mathbf{v}=c_{1}\mathbf{b}_{1}+c_{2}\mathbf{b}_{2}+\cdots+c_{k}\mathbf{b}_{k}%
\]

\end{itemize}

(c) (5 pts) a \textbf{set of linearly independent vectors} in $V$ is

\begin{itemize}
\item a set of vectors $\left\{  \mathbf{v}_{1},\ldots,\mathbf{v}_{k}\right\}
\subset V$ such that the only solution of
\[
c_{1}\mathbf{v}_{1}+c_{2}\mathbf{v}_{2}+\cdots+c_{k}\mathbf{v}_{k}=\mathbf{0}%
\]
is $c_{1}=0$, $c_{2}=0,$ $\ldots,c_{k}=0$
\end{itemize}

(d) (5 pts) a \textbf{linear transformation} from a vector space $V$ to vector
space $W$ is

\begin{itemize}
\item a function $T:V\rightarrow W$ such that

\begin{itemize}
\item $T\left(  \lambda\mathbf{x}\right)  =\lambda T\left(  \mathbf{x}\right)
$ for all $\mathbf{x}\in V$

\item $T\left(  \mathbf{x}_{1}+\mathbf{x}_{2}\right)  =T\left(  \mathbf{x}%
_{1}\right)  +T\left(  \mathbf{x}_{2}\right)  $ for all $\mathbf{x}%
_{1},\mathbf{x}_{2}\in V$
\end{itemize}
\end{itemize}

2. (10 pts) Prove or disprove that the points on the set $S=\left\{  \left[
x,y\right]  \in\mathbb{R}^{2}\mid y =x+1\right\}  $ is a subspace of
$\mathbb{R}^{2}$.

\begin{itemize}
\item A subspace has to be closed under both scalar multiplication and vector addition.

\begin{itemize}
\item \emph{closure under scalar multiplication}. Consider $\mathbf{v}=\left[
0,1\right]  \in S$ and let $\lambda=0$. Then%
\[
\lambda\mathbf{v}=\left[  0\cdot0,0\cdot1\right]  =\left[  0,0\right]
\quad\text{but }\left[  0,0\right]  \notin S\text{ since }0\neq0+1
\]
So $S$ is not closed under scalar multiplication and so \thinspace$S$ is not a subspace.

\item \emph{closure under vector addition.} Consider $\mathbf{v}_{1}=\left[
0,1\right]  ,\ \mathbf{v}_{2}=\left[  1,2\right]  \in S$. Then%
\[
\mathbf{v}_{1}+\mathbf{v}_{2}=\left[  0,1\right]  +\left[  1,2\right]
=\left[  1,3\right]  \notin S\text{ since }3\neq1+1
\]
So $S$ is not closed under vector addition and so $S$ is not a subspace
\end{itemize}

(Either one of these two arguments is a satisfactory solution.)
\end{itemize}

\newpage

3. (10 pts) Let $W=span\left(  [1,1,1,1],[1,-1,1,0],[2,0,2,1]\right)
\subset\mathbb{R}^{4}$. Find a basis for $W.$

\begin{itemize}
\item
\[
span\left(  W\right)  =RowSp\left(  \left[
\begin{array}
[c]{cccc}%
1 & 1 & 1 & 1\\
1 & 0 & 1 & 0\\
2 & 1 & 2 & 1
\end{array}
\right]  \right)  \quad\underrightarrow{\text{row reduction}}\quad
RowSp\left(  \left[
\begin{array}
[c]{cccc}%
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1\\
0 & 0 & 0 & 0
\end{array}
\right]  \right)
\]
Since the non-zero rows of the matrix in R.E.F. form a basis for the row space
of a matrix
\[
\text{basis for }W=\text{basis for }RowSp=\left\{  \left[  1,0,1,0\right]
,\left[  0,1,0,1\right]  \right\}
\]

\end{itemize}

4. Consider the following matrix: \quad$\displaystyle\mathbf{A}=\left[
\begin{array}
[c]{cccc}%
2 & 0 & 0 & 2\\
2 & 2 & 0 & 4\\
0 & 2 & 0 & 2
\end{array}
\right]  $

(a) (10 pts) Row reduce this matrix to reduced row echelon form

\begin{itemize}
\item
\[
\left[
\begin{array}
[c]{cccc}%
2 & 0 & 0 & 2\\
2 & 2 & 0 & 4\\
0 & 2 & 0 & 2
\end{array}
\right]  \quad\rightarrow\quad\left[
\begin{array}
[c]{cccc}%
1 & 0 & 0 & 1\\
0 & 1 & 0 & 1\\
0 & 0 & 0 & 0
\end{array}
\right]
\]

\end{itemize}

(b) (5 pts) Find a basis for the row space of $\mathbf{A}$.

\begin{itemize}
\item A basis for the row space of the matrix $\mathbf{A}$ is formed by the
non-zero rows of any row echelon form of $\mathbf{A}$. Thus%
\[
\text{basis for }RowSp\left(  \mathbf{A}\right)  =\left\{  \left[
1,0,0,1\right]  ,\left[  0,1,0,1\right]  \right\}
\]

\end{itemize}

(c) (5 pts) Find a basis for the column space of $\mathbf{A}$.

\begin{itemize}
\item A basis for the column space of $\mathbf{A}$ is formed by the columns of
$\mathbf{A}$ corresponding to the columns of a row echelon form of
$\mathbf{A}$ which contain pivots. \ Since the first and second columns of the
RREF of $\mathbf{A}$ are where the pivots of the RREF reside, the first and
second columns of $\mathbf{A}$ will be a basis for the column space of
$\mathbf{A}:$%
\[
\text{basis for }ColSp\left(  \mathbf{A}\right)  =\left\{  \left[
\begin{array}
[c]{c}%
2\\
2\\
0
\end{array}
\right]  ,\left[
\begin{array}
[c]{c}%
0\\
2\\
2
\end{array}
\right]  \right\}
\]

\end{itemize}

(d) (5 pts)\ Find a basis for the null space of $\mathbf{A}$.

\begin{itemize}
\item To find a basis for the null space of $\mathbf{A}$, we must solve
$\mathbf{Ax}=\mathbf{0}$. \ From the reduced row echelon form of $\mathbf{A}$,
we conclude that if $\mathbf{x}=\left[  x_{1},x_{2},x_{3},x_{4}\right]  $ is a
solution of $\mathbf{Ax}=\mathbf{0}$, then%
\[
\left.
\begin{array}
[c]{r}%
x_{1}+0+0+x_{4}=0\\
0+x_{2}+0+x_{4}=0\\
0+0+0+0=0
\end{array}
\right\}  \quad\Rightarrow\quad\left\{
\begin{array}
[c]{c}%
x_{1}=-x_{4}\\
x_{2}=-x_{4}%
\end{array}
\right.
\]
and $x_{3}$ and $x_{4}$ are free parameters: Thus,%
\[
\mathbf{x}=\left[
\begin{array}
[c]{c}%
-x_{4}\\
-x_{4}\\
x_{3}\\
x_{4}%
\end{array}
\right]  =x_{3}\left[
\begin{array}
[c]{c}%
0\\
0\\
1\\
0
\end{array}
\right]  +x_{4}\left[
\begin{array}
[c]{c}%
-1\\
-1\\
0\\
1
\end{array}
\right]
\]
and%
\[
\text{basis for }NullSp\left(  \mathbf{A}\right)  =\left\{  \left[
\begin{array}
[c]{c}%
0\\
0\\
1\\
0
\end{array}
\right]  \ ,\ \left[
\begin{array}
[c]{c}%
-1\\
-1\\
0\\
1
\end{array}
\right]  \right\}
\]

\end{itemize}

(e) (5 pts) What is the rank of $\mathbf{A}$?

\begin{itemize}
\item $rank\left(  A\right)  =\dim\left(  RowSp\left(  \mathbf{A}\right)
\right)  =\dim\left(  ColSp\left(  \mathbf{A}\right)  \right)  =2$

\newpage
\end{itemize}

5. (10 pts) Let $\mathbf{A}$ be an $n\times m$ matrix. Show that the function
$T: \mathbb{R}^{m} \rightarrow\mathbb{R}^{n} : T\left(  \mathbf{x}\right)  =
\mathbf{Ax}$ (via matrix multiplication on the right) is a linear transformation.

\begin{itemize}
\item[(i)] \emph{compatibility with scalar multiplication. }Let $\mathbf{x}%
\in\mathbb{R}^{m}$. Then
\[
T\left(  \lambda\mathbf{x}\right)  =\mathbf{A}\left(  \lambda\mathbf{x}%
\right)  =\lambda\mathbf{Ax}=\lambda T\mathbf{x}%
\]
and $\lambda\mathbf{y}$ is also a solution

\item[(ii)] \emph{compatibility with vector addition.} Let $\mathbf{x}%
_{1},\mathbf{x}_{2}\in\mathbb{R}^{m}$. Then%
\[
T\left(  \mathbf{x}_{1}+\mathbf{x}_{2}\right)  =\mathbf{A}\left(
\mathbf{x}_{1}+\mathbf{x}_{2}\right)  =\mathbf{Ax}_{1}+\mathbf{Ax}%
_{2}=T\left(  \mathbf{x}_{1}\right)  +T\left(  \mathbf{x}_{2}\right)
\]

\end{itemize}

Since $T$ is function between two vector spaces that is compatible with both
scalar multiplication and vector addition, $T$ is a linear transformation.

6. Consider the linear transformation $T:\mathbb{R}^{3}\rightarrow
\mathbb{R}^{2}:T\left(  \left[  x_{1},x_{2},x_{3}\right]  \right)  =\left[
x_{1}-x_{3},x_{2}+x_{3}\right]  $.

(a) (10 pts) Find the matrix $\mathbf{A}_{T}$ corresponding to $T$:%
\[
\mathbf{A}_{T}=\left[
\begin{array}
[c]{ccc}%
\uparrow & \uparrow & \uparrow\\
T\left(  \left[  1,0,0\right]  \right)   & T\left(  \left[  0,1,0\right]
\right)   & T\left(  \left[  0,0,1\right]  \right)  \\
\downarrow & \downarrow & \downarrow
\end{array}
\right]  =\left[
\begin{array}
[c]{ccc}%
1 & 0 & -1\\
0 & 1 & 1
\end{array}
\right]
\]
Note that $\mathbf{A}_{T}$ is already in R.R.E.F.

(b) (5 pts)\ Find a basis for $Range(T)\equiv\left\{  \mathbf{y}\in
\mathbb{R}^{2}\mid\mathbf{y}=T(\mathbf{x})\text{ for some }\mathbf{x}%
\in\mathbb{R}^{4}\right\}  $.

\begin{itemize}
\item
\begin{align*}
\text{basis for }Range\left(  T\right)    & =\text{basis for }ColSp\left(
\mathbf{A}_{T}\right)  \\
& =\text{columns of }\mathbf{A}_{T}\text{ that contain pivots (since
}\mathbf{A}_{T}\text{ is already in R.R.E.F.)}\\
& =\left\{  \left[
\begin{array}
[c]{c}%
1\\
0
\end{array}
\right]  ,\left[
\begin{array}
[c]{c}%
0\\
1
\end{array}
\right]  \right\}  \text{ }%
\end{align*}

\end{itemize}

(c) (5 pts) Find a basis for $ker(T)\equiv\left\{  \mathbf{x}\in\mathbb{R}%
^{3}\mid T(\mathbf{x})=\mathbf{0}\right\}  $

\begin{itemize}
\item We have $\ker\left(  T\right)  =NullSp\left(  \mathbf{A}_{T}\right)  $.
\ From the R.R.E.F. of $\mathbf{A}_{T}$ we see that the solutions
$\mathbf{x}=\left[
\begin{array}
[c]{c}%
x_{1}\\
x_{2}\\
x_{3}%
\end{array}
\right]  $of $\mathbf{A}_{T}\mathbf{x}=\mathbf{0}$ must be satisfy%
\[
\left.
\begin{array}
[c]{r}%
x_{1}+0-x_{3}=0\\
x_{2}+x_{3}=0
\end{array}
\right\}  \quad\Rightarrow\quad\left\{
\begin{array}
[c]{c}%
x_{1}=x_{3}\\
x_{2}=-x_{3}%
\end{array}
\right.  \quad\Rightarrow\quad\mathbf{x}=\left[
\begin{array}
[c]{c}%
x_{3}\\
-x_{3}\\
x_{3}%
\end{array}
\right]  =x_{3}\left[
\begin{array}
[c]{c}%
1\\
-1\\
1
\end{array}
\right]
\]
Therefore,%
\[
\text{basis for }\ker\left(  T\right)  =\text{basis for }NullSp\left(
\mathbf{A}_{T}\right)  =\left\{  \left[
\begin{array}
[c]{c}%
1\\
-1\\
1
\end{array}
\right]  \right\}
\]

\end{itemize}


\end{document}
