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%\header{MATH 11}{Exam 1}{Fall 2016}
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\begin{center}\huge MATH 3362 - EXAM 2 - FALL 2017\\
SOLUTION
\end{center}
\begin{flushleft}\vspace{0.5in}
Friday 13 October 2017\\
Instructor: Tom Cuchta
\end{flushleft}
\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}
{\flushleft \underline{Axioms of Vector Spaces}}\\
We say that $(V,\mathbb{F})$, where $V$ is a set and $\mathbb{F}$ is a field, is a vector space provided that for all $\vec{u}, \vec{v}, \vec{w} \in V$ and $\alpha, \beta, \gamma \in \mathbb{F}$,
\begin{enumerate}[1.)]
\item $\vec{u}+\vec{v} \in V$
\item $\vec{u}+\vec{v}=\vec{v}+\vec{u}$
\item $(\vec{u}+\vec{v})+\vec{w}=\vec{u}+(\vec{v}+\vec{w})$
\item there exists $\vec{0} \in V$ such that $\vec{u}+\vec{0}=\vec{u}$
\item there exists $-\vec{u} \in V$ such that $\vec{u}+(-\vec{u})=\vec{0}$
\item $\alpha \vec{u} \in V$
\item $\alpha (\vec{u}+\vec{v}) = \alpha \vec{u} + \alpha \vec{v}$
\item $(\alpha + \beta)\vec{u} = \alpha \vec{u} + \beta \vec{u}$
\item $\alpha (\beta \vec{u}) = (\alpha \beta) \vec{u}$
\item $1 \vec{u} = \vec{u}$
\end{enumerate}
\end{coverpages}

\begin{questions} 
\question[16] Find the inverse of the matrix. If no inverse exists, explain why not.
\begin{parts}
\part[8] $\begin{bmatrix} 1 & 5 \\ 2 & 10 \end{bmatrix}$ \\
\textit{Solution}: Using the simple formula
$$\begin{bmatrix} a&b \\ c&d \end{bmatrix}^{-1} = \dfrac{1}{ad-bc} \begin{bmatrix} d&-b & \\ -c& a \end{bmatrix},$$
we observe that the matrix $\begin{bmatrix} 1&5 \\ 2&10 \end{bmatrix}$ is not invertible because $\mathrm{det} \left( \begin{bmatrix} 1&5 \\ 2&10 \end{bmatrix} \right) = 1(10)-2(5)=0$.
\part[8] $\begin{bmatrix} 2&0&1 \\ -2&1&1 \\ 2&-1&0 \end{bmatrix}$ \\
\textit{Solution}: We will find the inverse using a reduced echelon form: \href{http://www.wolframalpha.com/input/?i=rref+%7B%7B2,0,1,1,0,0%7D,%7B-2,1,1,0,1,0%7D,%7B2,-1,0,0,0,1%7D%7D}{compute}
$$\begin{array}{ll}
\begin{bmatrix} 2&0&1&1&0&0 \\ -2&1&1&0&1&0 \\ 2&-1&0&0&0&1 \end{bmatrix} &\sim \begin{bmatrix} 2&0&1&1&0&0 \\ 0&1&2&1&1&0 \\ 0&-1&-1&-1&0&1 \end{bmatrix} \\
&\sim \begin{bmatrix} 2&0&1&1&0&0 \\ 0&1&2&1&1&0 \\ 0&0&1&0&1&1 \end{bmatrix} \\
&\sim \begin{bmatrix} 2&0&0&1&-1&-1 \\ 0&1&0&1&-1&-2 \\ 0&0&1&0&1&1 \end{bmatrix} \\
&\sim \begin{bmatrix} 1&0&0&\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2} \\ 0&1&0&1&-1&-2 \\ 0&0&1&0&1&1 \end{bmatrix}.
\end{array}$$
From this we \href{http://www.wolframalpha.com/input/?i=inverse+of+%7B%7B2,0,1%7D,%7B-2,1,1%7D,%7B2,-1,0%7D%7D}{observe}
$$\begin{bmatrix} 2&0&1 \\ -2&1&1 \\ 2&-1&0 \end{bmatrix}^{-1}=\begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ 1&-1&-2 \\ 0&1&1 \end{bmatrix}.$$
\end{parts}

\question[1] Is $(W,\mathbb{F})$ a subspace of $(V,\mathbb{F})$ or not? If so, prove it. If not, explain why not (preferably using a counterexample).
\begin{parts}
\part[24] Let $(V,\mathbb{F})=(\mathscr{P}_2,\mathbb{R})$ with the usual operations and let $W=\{t^2+bt \colon a \in \mathbb{R}\}$.  \\
\textit{Solution}: Not a subspace. It is not closed under addition because, e.g. $(t^2+t)+(t^2+t) = 2t^2 + 2t \not\in W$.

\part[12] $(V,\mathbb{F})$ is any (i.e. an ``arbitrary") vector space and $W=\left\{\vec{0}\right\}$ is the set containing only the zero vector of $(V,\mathbb{F})$. \\
\textit{Solution}: It is a subspace. Here the only element of $W$ is the zero vector and so for any $\alpha, \beta \in \mathbb{F}$ and any $\vec{v}_1, \vec{v}_2 \in W$, $\vec{v}_1=\vec{v}_2=\vec{0}$ and we compute 
$$\alpha \vec{v}_1 + \beta \vec{v}_2 = \alpha \vec{0} + \beta \vec{0} = \vec{0}+\vec{0} = \vec{0},$$
completing the proof.
\end{parts}

\question[36] A structure $(V,\mathbb{F})$ is described. Check whether or not the listed axioms hold in that structure. If it does hold, explain why. If it does not hold, explain why not (preferably with a counterexample).
\begin{parts}
\part[12] A matrix $A \in \mathbb{R}^{2 \times 2}$ is called symmetric provided that it is equal to its own transpose, i.e. $A^T=A$. Let $V$ denote the set of symmetric matrices in $\mathbb{R}^{2 \times 2}$, let $\mathbb{F}=\mathbb{R}$, and consider the usual operations. \vspace{0.15in}

Does $(V,\mathbb{F})$ obey the following axioms of a vector space?: \textbf{Axiom 1}, \textbf{Axiom 5} \\
\textit{Solution}: If $\begin{bmatrix} a&b \\ c&d \end{bmatrix} = \begin{bmatrix} a&b \\ c&d \end{bmatrix}^T = \begin{bmatrix} a&c \\ b&d \end{bmatrix}$, then $b=c$ and so such matrices are of the form $\begin{bmatrix} a&c \\ c&d \end{bmatrix}$. \\
\textbf{Check Axiom 1}: Let us add two such matrices:
$$\begin{bmatrix} a_1&c_1\\c_1&d_1 \end{bmatrix} + \begin{bmatrix} a_2 & c_2 \\ c_2 & d_2 \end{bmatrix} = \begin{bmatrix} a_1+a_2 & c_1+c_2 \\ c_1+c_2 & d_1+d_2 \end{bmatrix},$$
and this final matrix equals its transpose. Therefore Axiom~1 \textbf{holds}. \\
\textbf{Check Axiom 5}: If $A = \begin{bmatrix} a&c \\ c&d \end{bmatrix}$ then I propose $-A = \begin{bmatrix} -a&-c \\ -c & -d \end{bmatrix}$ to be the additive inverse of $A$. First off note that $(-A)^T=-A$ and so what remains to show is 
$$A+(-A) = \begin{bmatrix} a&c \\ c&d \end{bmatrix} + \begin{bmatrix} -a & -c \\ -c&-d\end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}=\vec{0},$$
completing the proof. Therefore Axiom~5 \textbf{holds}.

\part[12] A matrix $A \in \mathbb{R}^{2 \times 2}$ is called nilpotent provided that $A^2=0$. For example, $\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$ is called nilpotent because $\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}=\begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}$. Let $V$ denote the set of nilpotent matrices in $\mathbb{R}^{2 \times 2}$ and $\mathbb{F}=\mathbb{R}$ with the usual operations.

Does $(V,\mathbb{F})$ obey the following axioms of a vector space?: \textbf{Axiom 1},  \textbf{Axiom 4} \\
\textit{Solution}: \textbf{Check Axiom 1}: Note that the matrix $\begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$ is also nilpotent. However,
$$\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} + \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 1&0 \end{bmatrix}$$
is not nilpotent, since $\begin{bmatrix} 1&0 \\ 1&0 \end{bmatrix} \begin{bmatrix} 1&0 \\ 1&0 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 1&0 \end{bmatrix}$ (since the square equals itself, we call it \textit{idempotent}!). Therefore Axiom~1 \textbf{does not hold}. \\
\textbf{Check Axiom 4}: The zero matrix $\begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}$ is nilpotent and so Axiom~4 \textbf{holds}.

\part[12] Recall that a real number $x \in \mathbb{R}$ is called rational provided it can be written as the quotient of two integers $m$ and $n$ with $n \neq 0$: $x = \dfrac{m}{n}$ (e.g. $\frac{1}{2}, -\frac{3}{17}$, etc). A number that is not rational is called irrational (e.g. $\pi$, $\sqrt{2}$, etc). Let $V$ denote the set of irrational numbers in $\mathbb{R}$ and let $\mathbb{F}=\mathbb{R}$ with the usual operations. \vspace{0.15in}

Does $(V,\mathbb{F})$ obey the following axioms of a vector space?: \textbf{Axiom 1}, \textbf{Axiom 6} \\
\textit{Solution}: \textbf{Check Axiom 1}: Note that $\sqrt{2}$ and $-\sqrt{2}$ are irrational while $0$ is rational. Hence $\sqrt{2}+(-\sqrt{2})=0$ shows that Axiom~1 \textbf{does not hold}.  \\
\textbf{Check Axiom 6}: Note that $\sqrt{2}$ is irrational and $2$ is rational so $\sqrt{2}(\sqrt{2}) = \sqrt{2}^2=2$ shows that Axiom~6 \textbf{does not hold}.
%\part[1] Recall that a sequence $a_n$ is a function whose domain is the integers; we typically write ``$a_n$" to denote $a(n)$. Example: if $a_n=n^2$ then $a_1=1, a_2=4, a_3=9, \ldots$. Assume that the value of sequences are real numbers, i.e. that $a_n \in \mathbb{R}$ for all $n$.

%Consider the following set of sequences $V=\{ a_n \colon a_n \neq -1\}$, i.e. the set of all (real-valued) sequences which are never equal to $-1$. Define an operation $\oplus$ on $V$ (note: this is ``$+_V$") by 
%$$a_n \oplus b_n = a_n + b_n + a_nb_n.$$
%Let $\alpha \in \mathbb{R}$ and define a scalar multiplication operation $\odot$ (note: this is ``$\cdot_V$") by


\end{parts}



%\question[1] 
%\begin{parts}
%\part[1] Consider the vector space $(\mathbb{R}^{3 \times 1},\mathbb{R})$. Does the vector $\vec{v}=\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ lie in the span of the following set of vectors?: $\left\{ \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\}$
%\part[1] Consider the vector space $(\mathscr{P}_2,\mathbb{R})$. Does
%$$\mathscr{P}_2 = \mathrm{span} \{1, x-x^2 ,x+x^2 \}?$$
%\end{parts}

\question[24] Determine whether or not the set $\mathscr{B}$ is a basis for the vector space $(V,\mathbb{F})$. 
\begin{parts}
\part[12] Let $(V,\mathbb{F})=(\mathbb{R}^{3 \times 1},\mathbb{R})$ and let $\mathscr{B}=\left\{ \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\}$. \\
\textit{Solution}: First check if $\mathscr{B}$ spans $\mathbb{R}^{3\times 1}$, i.e. let $\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} \in \mathbb{R}^{3 \times 1}$ and solve the following vector equation:
$$c_1 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} + c_3 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix}.$$
This leads to a matrix we can put into reduced echelon form: \href{http://www.wolframalpha.com/input/?i=rref+%7B%7B0,1,0,alpha%7D,%7B1,0,0,beta%7D,%7B1,-1,1,gamma%7D%7D}{compute}
$$\begin{array}{ll}
\begin{bmatrix} 0&1&0&\alpha \\ 1&0&0&\beta \\ 1&-1&1&\gamma \end{bmatrix} &\sim \begin{bmatrix} 1&0&0&\beta \\ 0&1&0&\alpha \\ 1&-1&1&\gamma \end{bmatrix} \\
&\sim \begin{bmatrix} 1&0&0&\beta \\ 0&1&0&\alpha \\ 0&-1&1&\gamma-\beta \end{bmatrix} \\
&\sim \begin{bmatrix} 1&0&0&\beta \\ 0&1&0&\alpha \\ 0&0&1&\alpha+\gamma-\beta, \end{bmatrix}
\end{array}$$
proving that $\mathscr{B}$ spans $\mathbb{R}^{3 \times 1}$. To check that $\mathscr{B}$ is independent, just let $\alpha=\beta=\gamma=0$ in the previous calculation and we see that $\mathscr{B}$ is independent. Hence $\mathscr{B}$ is a basis for $\mathbb{R}^{3 \times 1}$.
\part[12] Let $(V,\mathbb{F})=(\mathscr{P}_2,\mathbb{R})$ and let $\mathscr{B}=\{1, x-x^2 ,x+x^2 \}$.  \\
\textit{Solution}: First check if $\mathscr{B}$ spans $\mathscr{P}_2$. Let $\alpha x^2 + \beta x + \gamma \in \mathscr{P}_2$ and solve the following vector equation:
$$c_1(1) + c_2(x-x^2) + c_3 (x+x^2) = \alpha x^2 + \beta x + \gamma.$$
Combining like-terms on the left-hand side yields
$$(-c_2+c_3)x^2+(c_2+c_3)x+(c_1) = \alpha x^2 + \beta x + \gamma.$$
Hence we may solve using an augmented matrix: \href{http://www.wolframalpha.com/input/?i=rref+%7B%7B0,-1,1,alpha%7D,%7B0,1,1,beta%7D,%7B1,0,0,gamma%7D%7D}{compute}
$$\begin{bmatrix} 0&-1&1&\alpha \\ 0&1&1&\beta \\ 1&0&0&\gamma  \end{bmatrix} \sim \begin{bmatrix} 1&0&0& \gamma \\ 0&1&0&\frac{\beta-\alpha}{2} \\ 0&0&1& \frac{\alpha+\beta}{2} \end{bmatrix},$$
from this we see that $\mathscr{B}$ is a basis.
\end{parts}

\question[4] (Bonus) Recall the properties of integrals, namely that for numbers $\alpha$ and $\beta$ and functions $f(x)$ and $g(x)$,
$$\displaystyle\int_a^b \alpha f(x) + \beta g(x) \mathrm{d}x = \alpha \displaystyle\int_a^b f(x) \mathrm{d}x + \beta \displaystyle\int_a^b g(x) \mathrm{d}x.$$
Notably, if $\alpha$ is a consant, then $\displaystyle\int_0^1 \alpha \mathrm{d}x = \alpha$, and if $f(x)=x$, then $\displaystyle\int_0^1 f(x) \mathrm{d}x = \displaystyle\int_0^1 x \mathrm{d}x = \dfrac{1}{2}$. Recall that a sequence is a function whose domain is the integers. Typically, if $a$ is a sequence, then $a(n)$ is denoted by $a_n$. For example the sequence $a_n=n^2$ has values $a_{-2}=(-2)^2=4$, $a_1=(-1)^2=1$, $a_0=0^2=0$, $a_5=5^2=25$, etc. 

Let $V$ denote the set of real-valued sequences with the property that for all integers $n$, $1+a_n >0$ and let $\mathbb{F}=\mathbb{R}$. Define the addition operation of $(V,\mathbb{F})$ to be denoted by $\oplus$ and be defined by $a_n \oplus b_n = a_b+b_n+a_nb_n$. Let the scalar multiplication of $(V,\mathbb{F})$ be denoted by $\oplus$ be defined by $\alpha \oplus a_n = \alpha a_n \displaystyle\int_0^1 (1+a_nx)^{\alpha-1} \mathrm{d}x$. This structure forms a vector space called the ``reegressive vector space on the time scale of integers". 

Consider the sequence $a_n=n^2 \in V$ and $b_n=n$. Compute $a_n \oplus b_n$ and $2 \odot a_n$. \\
\textit{Solution}: Compute
$$a_n \oplus b_n = a_n+b_n+a_nb_n=n^2+n+(n^2)(n) = n^3+n^2+n.$$
Compute
$$2 \odot a_n = 2n^2 \displaystyle\int_0^1 1+n^2 x dx =2n^2\left(1+ \dfrac{n^2}{2} \right).$$


\end{questions}

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