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\noindent
\parbox{2.25in}{\bf Math 5230}
\hfill {\Large \bf Homework \#5} \hfill
\parbox{2.25in}{\bf \hfill Due: Fri., Oct. $7^\mathrm{th}$, 2022}

\vspace{0.3in}

\noindent
\underline{Notation:} $V$ is a vector space over a field $\mathbb{F}$. Do not assume $V$ is finite dimensional (unless told otherwise). 

\vspace{0.2in}

\begin{description}
\item[\#1 Vector Space Definition] Building up from nothing. I'll provide a few example axiom level proofs.

\noindent
\begin{description}
\item[The zero vector is unique:] Suppose $\B 0$ and $\B 0'$ are zero vectors. Then $\B 0' = \B 0+\B 0' = \B 0$ where the first equality follows from the fact the adding the zero vector $\B 0$ does nothing and the second equality follows from the fact that adding the zero vector $\B 0'$ also does nothing.
\item[Additive inverses are unique:] Let $\B u, \B v, \B w \in V$. Suppose that $\B u$ is a left and $\B w$ is a right additive inverse of $\B v$. Then $\B w = \B 0 + \B w = (\B u+\B v)+\B w = \B u + (\B v+\B w) = \B u + \B 0 = \B u$ where  we used the definition of the zero vector, the definitions of left and right additive inverses, and associativity of vector addition to get our equalities.
\item[The scalar zero yields zero:] Let $\B v \in V$. Then $0\B v = (0+0)\B v = 0\B v+0\B v$ since $0+0=0$ (the scalar zero is the additive identity in $\mathbb{F}$). We also used the fact that we can distribute scalar multiplication. Now add $-(0\B v)$ to both sides: $-(0\B v)+ 0\B v = -(0\B v)+(0\B v+ 0\B v)$. Thus $\B 0 = (-(0\B v)+0\B v)+0\B v = \B 0+0\B v = 0\B v$ where we used the definition of additive inverses, associativity, and additive identity properties. Therefore, $0\B v=\B 0$.
\item[Negative one negates:] Let $\B v \in V$. Then $\B 0 = 0\B v = (1-1)\B v = 1\B v + (-1)\B v = \B v + (-1)\B v$ where we have used our result immediately above, the definition of $-1$, distribution, and the axiom that scaling by $1$ does nothing. Now add $-\B v$ to both side: $-\B v+\B 0 = -\B v+(\B v+(-1)\B v)$. Using the additive identity property of the zero vector and associativity we get: $-\B v = (-\B v+\B v)+(-1)\B v=\B 0+(-1)\B v$. Therefore, $(-1)\B v=-\B v$.
\end{description}


\begin{enumerate}[(a)]
\item In horrifying detail (go through all of the axioms), show $\mathbb{R}^2 = \{ (a,b) \;|\; a,b \in \mathbb{R} \}$ is a vector space over $\mathbb{R}$.
\item {}[Cancellation:] Let $\B u, \B v, \B w \in V$. Prove that $\B u+\B v = \B u+\B w$ implies that $\B v = \B w$.\\ \force \hfill Do this {\it only using axioms}. Cite the axioms being used. 
\item {}[Scaling zero does nothing:] Let $c \in \mathbb{F}$ and $\B 0$ be the zero vector in $V$. Prove $c\,\B 0=\B 0$.\\ \force \hfill Carefully explain each step. 
\end{enumerate}

\item[\#2 Easy Subspace] Use the subspace test.
\begin{enumerate}[(a)]
\item Show that $W = \{ (x,y) \in \mathbb{R}^2 \;|\; x+2y=0 \}$ is a subspace of $\mathbb{R}^2$. 
\item Explain why $B = \{ (x,y) \in \mathbb{R}^2 \;|\; x+2y=3 \}$ is not a subspace of $\mathbb{R}^2$.
\end{enumerate}

\newpage

\item[\#3 Trickier Subspace] The collection of all functions from a field to itself, $V=\mathbb{F}^\mathbb{F} = \{ f \;|\; f:\mathbb{F}\to\mathbb{F}\}$ is a vector space (over $\mathbb{F}$). 

Let $f \in V$. We say that $f$ is {\bf even} if $f(-x)=f(x)$ for all $x \in \mathbb{F}$. Likewise, we say that $f$ is {\bf odd} if $f(-x)=-f(x)$. For example, $f(x)=x^2$ is even whereas $g(x)=x^3$ is odd. On the other hand, generally $h(x)=x^2+x^3$ is neither.

\begin{enumerate}[(a)]
   \item Let $E = \{ f \in V \;|\; f \text{ is even }\}$. Show $E$ is a subspace of $V$.
   \item Let $O = \{ f \in V \;|\; f \text{ is odd }\}$. Show $O$ is a subspace of $V$.
   \item Suppose that $\mathbb{F}$ is a field such that $\mathrm{char}(\mathbb{F})\not=2$ (i.e., $2^{-1}$ exists).\\ \force \hfill For any $f \in V$, let $f_e(x) = \dfrac{f(x)+f(-x)}{2}$ and $f_o(x)=\dfrac{f(x)-f(-x)}{2}$ define functions $f_e$ and $f_o$ in $V$. 
   $$\hspace*{-0.5in} \text{Notice: } \quad f_e(-x) = \dfrac{f(-x)+f(x)}{2} = f_e(x) \qquad \text{ and } \qquad f_o(-x) = \dfrac{f(-x)-f(x)}{2} = -\dfrac{f(x)-f(-x)}{2} = -f_o(x)$$
   Therefore, $f_e \in E$ and $f_o \in O$. 
   
   Show that $V = E \oplus O$. [Thus every function can be uniquely decomposed into an even plus an odd part.]
   
  {\it Note:} Showing $V$ is a direct sum of $E$ and $O$ requires two things. First, you need to show that every $f \in V$ can be written as $f=g+h$ where $g \in E$ and $h \in O$. Then, you need to show that $E \cap O = \{0\}$ where 0 here is the zero function: $0(x)=0$ for all $x \in \mathbb{F}$. The first part shows that $V=E+O$ and the second part shows that the sum is {\it direct}.
   
   \item Working over $\mathbb{F}=\mathbb{R}$, how does $f(x) = x^3-5x^2+\cos(x)+4\sin(x)-12$ decompose?
\end{enumerate}

{\it Side Notes:} 
\begin{itemize}
\item In characteristic 2, $x=-x$ so $-f(x)=f(x)=f(-x)$. This means $V=E=O$. 
\item Again, working over the reals, the even part of $e^x$ is $\cosh(x) = \dfrac{e^x+e^{-x}}{2}$ and the odd part is $\sinh(x) = \dfrac{e^x-e^{-x}}{2}$ (the hyperbolic sine and cosine functions).
\end{itemize}

\item[\#4 Unions Don't Work] Let $W_1$ and $W_2$ be subspaces of $V$.\\ \force \hfill Show that $W_1 \cup W_2$ is a subspace of $V$ if and only if $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$.

\item[\#5 Linear Independence and Spanning] Suppose $V$ has a basis $\beta = \{\B u, \B v, \B w\}$ (distinct vectors).\\ \force \hfill 
Let $S = \{\B u+\B v, \B v +\B w, \B w +\B u \}$ and $T = \{\B u+\B v, \B u - \B v, \B w \}$.

{\it Additional Assumption:} $\mathrm{char}(\mathbb{F}) \not=2$.

\begin{enumerate}[(a)]
    \item Prove $S$ is linearly independent (using the definition of independence).
    \item Prove $T$ spans $V$ (using the definition of span).
    \item Is $S \cup T$ linearly independent? Does it span $V$? Explain.
    \item Is $S \cap T$ linearly independent? Does it span $V$? Explain.
\end{enumerate}

\end{description}

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