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

\documentclass[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=Friday, December 11, 2020 07:49:30}
%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=9.5in
\oddsidemargin=0in
\evensidemargin=0in
\topmargin=-.70in
\setlength{\parskip}{.05in}
\setlength{\parindent}{0in}
\begin{document}

\begin{center}
Math 3013.69033\\[0pt]Solutions to Final Exam\\[0pt]6:00pm - 8:00pm CDT,
December 10, 2020\\[0pt]
\end{center}

1. Give the definitions of the following linear algebraic concepts:

(a) (5 pts)\ a \textbf{subspace} of a vector space $V$.

\begin{itemize}
\item A subspace of vector space $V$ is a subset $W$ of $V$ such that

\begin{itemize}
\item[(i)] whenever $\lambda\in\mathbb{R}$, $\mathbf{w}\in W\ ,\quad
\quad\left(  \lambda\mathbf{w}\right)  \in W$

\item[(ii)] whenever $\mathbf{w}_{1},\mathbf{w}_{2}\in W\ ,\ \quad\left(
\mathbf{w}_{1}+\mathbf{w}_{2}\right)  \in W$
\end{itemize}
\end{itemize}

(b) (5 pts) a set of \textbf{linearly independent} vectors

\begin{itemize}
\item A set of vectors $\left\{  \mathbf{v}_{1},\ldots,\mathbf{v}_{k}\right\}
$ is a linearly independent set if the only solution of%
\[
x_{1}\mathbf{v}_{1}+\cdots+x_{k}\mathbf{v}_{k}=\mathbf{0}%
\]
is%
\[
x_{1}=0,\ldots,x_{k}=0
\]

\end{itemize}

(c) (5 pts) a \textbf{linear transformation} between two vector spaces $V$ and
$W$.

\begin{itemize}
\item A linear transformation between two vector space $V$ and $W$ is a
function $T:V\rightarrow W$ such that

\begin{itemize}
\item[(i)] $T\left(  \lambda\mathbf{v}\right)  =\lambda T\left(
\mathbf{v}\right)  $ for all $\lambda\in\mathbb{R}$ and all $\mathbf{v}\in V$

\item[(ii)] $T\left(  \mathbf{v}_{1}+\mathbf{v}_{2}\right)  =T\left(
\mathbf{v}_{1}\right)  +T\left(  \mathbf{v}_{2}\right)  $ for all
$\mathbf{v}_{1},\mathbf{v}_{2}\in V$
\end{itemize}
\end{itemize}

(d) (5 pts) a \textbf{basis} for a subspace $W$ of a vector space $V$

\begin{itemize}
\item A set of vectors $\left\{  \mathbf{v}_{1},\ldots,\mathbf{v}_{k}\right\}
$ is a basis for a subspace $W$ if

\begin{itemize}
\item[(i)] $W=span\left(  \mathbf{v}_{1},\ldots,\mathbf{v}_{k}\right)  $

\item[(ii)] If $\mathbf{w}=c_{1}\mathbf{v}_{1}+\cdots+c_{k}\mathbf{v}_{k}$,
then coefficients $c_{1},\ldots,c_{k}$ are unique.
\end{itemize}
\end{itemize}

2. For each of the following augmented matrices, describe the solution space
of the corresponding linear system. (Determine if there are solutions; and, if
there are solutions, how many free parameters are needed to describe the
general solution.)

(a) (5 pts) $\left[
\begin{array}
[c]{cccc}%
1 & 0 & 2 & 2\\
0 & 1 & 0 & 1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}
\left\vert
\begin{array}
[c]{c}%
1\\
2\\
0\\
0
\end{array}
\right.  \right]  $

\begin{itemize}
\item The corresponding linear system has infinitely many solutions. \ Since
columns 3 and 4 do not contain pivots, the solution will have $2$ free parameters.
\end{itemize}

(b) (5 pts)$\left[
\begin{array}
[c]{ccc}%
1 & 0 & 4\\
0 & 2 & 1\\
0 & 0 & 1\\
0 & 0 & 0
\end{array}
\left\vert
\begin{array}
[c]{c}%
1\\
2\\
1\\
0
\end{array}
\right.  \right]  $

\begin{itemize}
\item The corresponding linear system has a unique solution (no free parameters)
\end{itemize}

(c) (5 pts) $\left[
\begin{array}
[c]{ccccc}%
0 & -1 & 0 & 1 & 2\\
0 & 0 & 2 & 0 & 1\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{array}
\left\vert
\begin{array}
[c]{c}%
1\\
0\\
1\\
0
\end{array}
\right.  \right]  $

\begin{itemize}
\item The corresponding linear system has no solution (the third row
corresponds to the equation $0=1$, which is a mathematical contradiction).
\end{itemize}

\newpage

3. (10 pts)\ Compute the inverse of $\mathbf{A}=\left[
\begin{array}
[c]{ccc}%
1 & 2 & 0\\
1 & 3 & 0\\
1 & 2 & 1
\end{array}
\right]  $

\begin{itemize}
\item
\[
\quad\Rightarrow\quad\mathbf{A}^{-1}=\left[
\begin{array}
[c]{ccc}%
3 & -2 & 0\\
-1 & 1 & 0\\
-1 & 0 & 1
\end{array}
\right]
\]

\end{itemize}

4. (10 pts) Let $W=\left\{  \left[  x,y\right]  \in\mathbb{R}^{2}\mid
x+2y=3\in\mathbb{R}\right\}  $. Prove or disprove that $W$ is a subspace of
$\mathbb{R}^{2}$.

\begin{itemize}
\item Let $\mathbf{w}=\left[  1,1\right]  $. Since $\left(  1\right)
+2\left(  1\right)  =3$, this vector is in $W$. \ Now consider the scalar
multiple of $\mathbf{w}$ by $\lambda=0$. \ We have%
\[
\left(  0\right)  \mathbf{w}=\left[  0,0\right]
\]
but%
\[
\left(  0\right)  +\left(  2\right)  \left(  0\right)  =0\neq3
\]
and so for this $\lambda$, $\lambda\mathbf{w}\notin W$. \ This means $W$ is
not closed under scalar multiplication, and so $W$ is not a subspace.
\end{itemize}

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

(a) (10 pts) Find a matrix that represents $T$.

\begin{itemize}
\item
\[
\mathbf{A}_{T}=\left[
\begin{array}
[c]{ccc}%
\uparrow & \uparrow & \uparrow\\
T\left(  \left[  1,0,0\right]  \right)  & T\left(  \left[  0,1,0\right]
\right)  & T\left(  \left[  0,0,1\right]  \right) \\
\downarrow & \downarrow & \downarrow
\end{array}
\right]  =\left[
\begin{array}
[c]{ccc}%
0 & 1 & -1\\
1 & 0 & -1
\end{array}
\right]
\]

\end{itemize}

(b) (10 pts) Find a basis for the kernel of $T$ (i.e. the set of vectors
$\mathbf{x}$ such that $T\left(  \mathbf{x}\right)  =\mathbf{0}$).

\begin{itemize}
\item
\[
\quad\Rightarrow\quad\text{basis for }\ker\left(  T\right)  =\left\{  \left[
\begin{array}
[c]{c}%
1\\
1\\
1
\end{array}
\right]  \right\}
\]

\end{itemize}

\newpage

6. Consider the vectors $\left\{  [1,-2,2,1],[1,-1,3,1],[0,1,1,0],\left[
2,-3,5,2\right]  \right\}  \in\mathbb{R}^{4}$

(a) (10 pts) Determine if these vectors are linearly independent.

\begin{itemize}
\item We form a matrix $\mathbf{A}$ using the given vectors as rows and row
reduce $\mathbf{A}$ to Row Echelon Form%
\[
\left[
\begin{array}
[c]{cccc}%
1 & -2 & 2 & 1\\
1 & -1 & 3 & 1\\
0 & 1 & 1 & 0\\
2 & -3 & 5 & 2
\end{array}
\right]  \quad\underrightarrow{\text{ row reduction }}\quad\left[
\begin{array}
[c]{cccc}%
1 & 0 & 4 & 1\\
0 & 1 & 1 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}
\right]
\]
since we have a zero row (or 2 zero rows), the original set of vectors are not
linearly independent.
\end{itemize}

(b) (5 pts) What is the dimension of the subspace generated by these vectors
(i.e. the subspace spanned by these vectors)?

\begin{itemize}
\item Since the REF of $\mathbf{A}$ has $2$ non-zero rows, its row space is
$2$-dimensional. \ Since this row space coincides with the subspace generated
by the original list of vectors, this subspace is also $2$-dimensional
\end{itemize}

7. (a) (10 pts) Find the eigenvalues and the eigenvectors of the following
matrix : $\mathbf{A=}\left[
\begin{array}
[c]{ccc}%
2 & 2 & 0\\
0 & 0 & 1\\
0 & 0 & 2
\end{array}
\right]  $

\begin{itemize}
\item
\[
p_{\mathbf{A}}\left(  \lambda\right)  =\left(  2-\lambda\right)  ^{2}\left(
-\lambda\right)  \quad\Rightarrow\quad\text{eigenvalues }=\left\{
2,0\right\}
\]


$2$-eigenspace%
\[
E_{\lambda=2}=span\left(  \left[
\begin{array}
[c]{c}%
1\\
0\\
0
\end{array}
\right]  \ ,\ \left[
\begin{array}
[c]{c}%
0\\
0\\
1
\end{array}
\right]  \right)
\]


$0$-eigenspace%
\[
E_{\lambda=0}=span\left(  \left[
\begin{array}
[c]{c}%
1\\
1\\
0
\end{array}
\right]  \right)
\]

\end{itemize}

(b) (5 pts) What are the algebraic and geometric multiplicities of each
eigenvalue of $\mathbf{A}$.

\begin{itemize}
\item The algebraic multiplicity of an eigenvalue $r$ is the number of factors
of $\left(  \lambda-r\right)  $ in the characteristic polynomial
$p_{\mathbf{A}}\left(  \lambda\right)  $ of $\mathbf{A}$. The geometric
multiplicity of an eigenvalue $r$ is the dimension of the corresponding
eigenspace (the number of basis vectors). \ Thus,
\end{itemize}

(c) (5 pts) Is this matrix diagonalizable?

\begin{itemize}
\item We need $3$ linearly independent eigenvectors to diagonalize a
$3\times3$ matrix. \ Since we found three linearly independent eigenvectors
(two for $\lambda=2$ and one for $\lambda=0$), $\mathbf{A}$ \textbf{is
diagonalizable}
\end{itemize}

\newpage

8. (15 pts) Let $\mathbf{A}$ be the matrix $\left[
\begin{array}
[c]{cc}%
1 & 3\\
-2 & 6
\end{array}
\right]  $ . Find a $2\times2$ matrix $\mathbf{C}$ and a diagonal matrix
$\mathbf{D}$ such that $\mathbf{C}^{-1}\mathbf{AC}=\mathbf{D}$.

\begin{itemize}
\item We first need to find the eigenvalues and eigenvectors of $\mathbf{A}$.%
\[
p_{\mathbf{A}}\left(  \lambda\right)  =\left(  \lambda-3\right)  \left(
\lambda-4\right)  \quad\Rightarrow\quad\text{eigenvalues = }\left\{
3,4\right\}
\]


$3$-eigenspace%
\[
E_{\lambda=3}=span\left(  \left[
\begin{array}
[c]{c}%
3\\
2
\end{array}
\right]  \right)
\]
$4$-eigenspace%
\[
E_{\lambda=4}=span\left(  \left[
\begin{array}
[c]{c}%
1\\
1
\end{array}
\right]  \right)
\]
The diagonal matrix $\mathbf{D}$ is formed from the eigenvalues of
$\mathbf{A}$ and the diagonalizing matrix $\mathbf{C}$ is formed by using the
eigenvectors of $\mathbf{A}$ as columns (following your ordering of
eigenvalues). \ Thus,%
\[
\mathbf{D}=\left[
\begin{array}
[c]{cc}%
3 & 0\\
0 & 4
\end{array}
\right]  \quad,\quad\mathbf{C}=\left[
\begin{array}
[c]{cc}%
3 & 1\\
2 & 1
\end{array}
\right]
\]

\end{itemize}

9. (15 pts) Let $\mathbf{v}=\left[  2,1,2\right]  $ and let $W=span\left(
\left[  0,1,1\right]  \ ,\ \left[  1,1,0\right]  \right)  $. Find the
orthogonal decomposition $\mathbf{v}=\mathbf{v}_{W}+\mathbf{v}_{\perp}$ of
$\mathbf{v}$ with respect to the subspace $W$.

\begin{itemize}
\item Note that the two vectors generating $W$ are linearly independent and so
form a basis $B_{W}=\left\{  \left[  0,1,1\right]  \ ,\ \left[  1,1,0\right]
\right\}  $ for $W$.

\item Next we need a basis for $W^{\perp}=\left\{  \mathbf{v}\in\mathbb{R}%
^{3}\mid\mathbf{v}\cdot\mathbf{w}=0\text{ for all }\mathbf{w}\in W\right\}  $.
We have%
\[
W^{\perp}=span\left(  \left[
\begin{array}
[c]{c}%
1\\
-1\\
1
\end{array}
\right]  \right)  \qquad,\qquad B_{W^{\perp}}=\left\{  \left[  1,-1,1\right]
\right\}
\]


\item $B=B_{W}\cup B_{W^{\perp}}=\left\{  \left[  0,1,1\right]  \ ,\ \left[
1,1,0\right]  \ ,\ \left[  1,-1,1\right]  \right\}  $ will be a basis for
$\mathbb{R}^{3}$. \ We now find the coordinates $\left[  c_{1},c_{2}%
,c_{3}\right]  $ of $\mathbf{v}$ with respect to $B$, i.e., we solve%
\begin{align*}
\left[
\begin{array}
[c]{c}%
2\\
1\\
2
\end{array}
\right]    & =c_{1}\left[
\begin{array}
[c]{c}%
0\\
1\\
1
\end{array}
\right]  +c_{2}\left[
\begin{array}
[c]{c}%
1\\
1\\
0
\end{array}
\right]  +c_{3}\left[
\begin{array}
[c]{c}%
1\\
-1\\
1
\end{array}
\right]  \\
\quad & \Rightarrow\quad c_{1}=1\quad,\quad c_{2}=1\quad,\quad c_{3}=1
\end{align*}
The sum of the first two vector terms on the right is $\mathbf{v}_{W}$ and the
last term is $\mathbf{v}_{\perp}$. Thus,%
\[
\mathbf{v}_{W}=\left[
\begin{array}
[c]{c}%
1\\
2\\
1
\end{array}
\right]  \qquad,\qquad\mathbf{v}_{\perp}=\left[
\begin{array}
[c]{c}%
1\\
-1\\
1
\end{array}
\right]
\]

\end{itemize}

\newpage

10. (10 pts) Find an orthonormal basis for the subspace $W$ generated by the
vectors $\mathbf{v}_{1}=\left[  1,1,1\right]  $ and $\mathbf{v}_{2}=\left[
1,0,1\right]  $

\begin{itemize}
\item We'll first construct an orthogonal basis $\left\{  \mathbf{o}%
_{1},\mathbf{o}_{2}\right\}  $ from the basis $\left\{  \mathbf{v}%
_{1},\mathbf{v}_{2}\right\}  $ of $W$. \ Thus, we take%
\[
\mathbf{o}_{1}=\mathbf{v}_{1}=\left[  1,1,1\right]
\]
and then construct a second orthogonal basis vector by removing from
$\mathbf{v}_{2}$ the component that runs parallel to $\mathbf{o}_{1}$%
\begin{align*}
\mathbf{o}_{2} &  =\mathbf{v}_{2}-\frac{\mathbf{o}_{1}\cdot\mathbf{v}_{2}%
}{\mathbf{o}_{1}\cdot\mathbf{o}_{1}}\mathbf{o}_{1}=\left[  1,0,1\right]
-\frac{1+0+1}{1+1+1}\left[  1,1,1\right]  \\
&  =\left[
\begin{array}
[c]{ccc}%
\frac{1}{3} & -\frac{2}{3} & \frac{1}{3}%
\end{array}
\right]
\end{align*}%
\begin{align*}
\mathbf{n}_{1} &  =\frac{1}{\sqrt{3}}\left[  1,1,1\right]  =\left[  \frac
{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right]  \\
\mathbf{n}_{2} &  =\frac{3}{\sqrt{6}}\left[
\begin{array}
[c]{ccc}%
\frac{1}{3} & -\frac{2}{3} & \frac{1}{3}%
\end{array}
\right]  =\left[  \frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}%
}\right]
\end{align*}

\end{itemize}


\end{document}