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\section*{MATH 510, Spring 2024}
\subsection*{Homework 1}
\noindent
Due Friday, Jan~26, at 3:20pm. Hand in hard copy at the beginning of class.
\textbf{You should typeset your solutions in \TeX.}
\begin{itemize}
\item Write the problem statement followed by a proof or solution.
\item List problems in the same order they were given.
\item If you skip a problem, include the problem statement with no solution.
\end{itemize}
\bigskip
\begin{enumerate}[{\rm 1.}]
%\item Let $V$ be a vector space, possibly of infinite dimension, and let $W_1,W_2,\dots,W_n \leq V$ be (possibly infinite-dimensional) subspaces with bases $B_1,\ldots,B_n$, respectively. Prove that $W_1,\ldots,W_n$ are independent if and only if $\cup_{j=1}^n B_j$ is a basis for ${W_1 + \cdots + W_n}$ \textbf{and} $B_i \cap B_j = \emptyset$ for every $i \neq j$.
%\bigskip
\item Let $W_1$ and $W_2$ be subspaces of a vector space $V$ such that the union $W_1\cup W_2$ is a subspace of $V$. Prove that one of the subspaces $W_i$ is contained in the other.
\bigskip
\item
Let $V$ be a finite-dimensional space over a field $F$.
Suppose $W_1$ and $W_2$ are subspaces of $V$ with $\dim W_1 = \dim W_2$.
Prove there is a subspace $U \leq V$ such that $V = W_1 \oplus U = W_2 \oplus U$.
\smallskip
\noindent
\emph{Hint:} In the case $W_1\neq W_2$, use the previous problem to show there is a vector in $V$ which is not in $W_1 \cup W_2$.
%When $\dim V_j \neq \dim V$, show $V_1 \cup V_2 \neq V$.
\bigskip
\item Let $V$ be the vector space of all functions from $\mathbb{R}$ to $\mathbb{R}$ (see [HK, \S 2.1, Example 3]). Let $V_\text{e}$ be the subset of even functions, $f(-x)=f(x)$; let $V_\text{o}$ be the subset of odd functions $f(-x)=-f(x)$.
\begin{enumerate}[{\rm (a)}]
\item Prove that $V_\text{e}$ and $V_\text{o}$ are subspaces of $V$.
\item Prove that $V_\text{e}+V_\text{o}=V$.
\item Prove that $V_\text{e}\cap V_\text{o}=\{0\}$.
\end{enumerate}
\bigskip
\item Let $V=\mathbb{R}$ be the set of all real numbers. Regard $V$ as a vector space over the field of \emph{rational} numbers $\mathbb{Q}$, with the usual operations. Prove that this vector space is not finite-dimensional.
\bigskip
\item Let $W_1, W_2, \ldots, W_n$ be subspaces of a vector space $V$ such that $V=W_1+W_2+\cdots+W_n$. Suppose that $W_i \cap (W_1+W_2+\cdots+\widehat{W_i}+\cdots+W_n)=\{\boldsymbol{0}_V\}$ (here $\widehat{W_i}$ means the term should be \emph{omitted} from the expression, and $\boldsymbol{0}_V$ is the zero vector in $V$) for all $i$. Show that for each vector $v\in V$ there are \emph{unique} vectors $w_i\in W_i$ such that $v=w_1+w_2+\cdots+w_n$.
\end{enumerate}
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