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\title[252 Worksheet \#12]{Math 252: Abstract Algebra II \\ Worksheet, Day \#12}
\date{Monday, 11 February 2008}
\newcommand{\shortblank}{\underline{\hspace{1in}}\ \ }
\newcommand{\longblank}{\underline{\hspace{2in}}\ \ }
\newcommand{\Longblank}{\underline{\hspace{2.5in}}\ \ }
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\newcommand{\fulllongblank}{\underline{\hspace{\textwidth}}}
\maketitle
Fill in the blanks.
Let $F$ be a field. An $F$-module $V$ is also known as a \underline{\textsf{\ \ $F$-vector space\ \ }} over $F$, and an $F$-module homomorphism $\phi:V \to W$ is called a \longblank.
Let $V$ be a vector space over $F$. Let $v_1,\dots,v_n \in V$. A \emph{linear combination} of $v_1,\dots,v_n$ is \\ \fulllongblank \\ The set of all vectors $w$ which are linear combinations of $v_1,\dots,v_n$ forms a \longblank $W \subset V$, and we say that $W$ is \longblank by $v_1,\dots,v_n$.
A \emph{linear relation} among vectors $v_1,\dots,v_n$ is a linear combination which is equal to zero, i.e., \\ \fulllongblank \\ The vectors $v_1,\dots,v_n$ are called \longblank if there is no nonzero linear relation among the vectors, i.e., if $c_1v_1 + \dots + c_nv_n=0$ then \twolongblank; otherwise $v_1,\dots,v_n$ are called \longblank. By convention, the empty set is considered to be \longblank, and the span of the empty set is \longblank. Two vectors $v_1,v_2$ have no nonzero linear relation if and only if either \\ \longblank or \Longblank.
An ordered set $B=\{v_1,\dots,v_n\}$ of vectors which is linearly independent and spans $V$ is called a \longblank of $V$; for example, for $F^n=\{(a_1,\dots,a_n):a_i \in F\}$ we may take \\ \fulllongblank
\begin{lem}
The set $B$ is a basis for $V$ if and only if every $w \in V$ can be written uniquely as a \\ \fulllongblank
\end{lem}
\begin{prop}
Let $L=\{v_1,\dots,v_n\} \subset V$ be a linearly independent ordered set, and let $v \in V$. Then the ordered set $\{v_1,\dots,v_n,v\}$ is linearly independent if and only if \\ \fulllongblank
\end{prop}
\begin{prop}
For any finite set $S$ which spans $V$, there exists a subset $B \subset S$ which is a basis for $V$.
\end{prop}
\begin{proof}
Suppose that $S=\{v_1,\dots,v_n\}$ and that $S$ is not linearly independent. Then \\
\fulllongblank \\ \fulllongblank \\ \fulllongblank \\ \prooflongblank
\end{proof}
\begin{lem}
Let $V$ be a vector space with a finite basis. Then any spanning set of $V$ contains a basis, and any
\Longblank set $L$ can be extended by adding elements of $V$ to get a basis.
\end{lem}
\begin{cor}
Suppose $V$ has a finite basis $B$ with $\#B=n$. Then any set of linearly independent vectors has at most \shortblank elements, and any spanning set has \longblank elements.
\end{cor}
\begin{proof}
Let $L$ be a linearly independent set of vectors. By the lemma, \\ \fulllongblank \\ \fulllongblank \\ \prooflongblank
\end{proof}
\begin{cor}
If $V$ has a finite basis then any two bases of $V$ have the same cardinality.
\end{cor}
\begin{proof} \prooflongblank \fulllongblank \\ \fulllongblank \\ \prooflongblank
\end{proof}
Let $V$ be a vector space with a finite basis. Then the \underline{\textsf{\ \ dimension\ \ }} of $V$ is defined to be \twolongblank and is denoted $\dim_F V$, and $V$ is said to be \longblank over $F$.
If $F$ is a finite field with $\#F=q$, then a vector space of dimension $n$ over $F$ has \shortblank elements.
\begin{thm}
Let $V$ be a vector space of dimension $n$. Then $V \cong F^n$. In particular, any two vector spaces of the same finite dimension are isomorphic.
\end{thm}
\begin{proof}
Let $v_1,\dots,v_n$ be a basis for $V$. Define the map
\begin{align*}
\phi:F^n &\to V \\
\phi(a_1,\dots,a_n) &= a_1v_1 + \dots + a_nv_n.
\end{align*}
\fulllongblank \\ \fulllongblank \\ \prooflongblank
\end{proof}
\begin{thm}
Let $V$ be a finite-dimensional vector space over $F$ and let $W$ be a subspace of $V$. Then the quotient $V/W$ is a vector space with
\[ \dim (V/W) = \text{\twolongblank}, \]
\end{thm}
\begin{proof}
Since $V$ is finite-dimensional, so is $W$ because \\ \twolongblank. So let $W$ have dimension $m$ and let $w_1,\dots,w_m$ be a basis for $W$. We extend this basis to a basis $w_1,\dots,w_m,v_{m+1},\dots,v_n$
of $V$. Then the projection map $V \to V/W$ maps each $w_i$ to \shortblank and therefore has image spanned by $v_{m+1}+W,\dots,v_n+W$; these vectors are linearly independent because \prooflongblank. So $\dim(V/W)=\text{\longblank}$.
\end{proof}
\begin{cor}
Let $\phi:V \to W$ be a linear transformation. Then
\[ \dim V = \dim \ker\phi + \dim \img\phi. \]
\end{cor}
We also say that $\ker\phi$ is the \longblank of $\phi$ and $\dim\ker\phi$ is the \longblank. The dimension of $\img \phi=\phi(V)$ is called the \longblank.
\begin{cor}
Let $\phi:V \to W$ be a linear transformation of vector spaces of the same finite dimension $n$. Then $\phi$ is an isomorphism if and only if $\phi$ is injective if and only if $\phi$ is surjective.
\end{cor}
\begin{proof} \prooflongblank \fulllongblank \\ \prooflongblank
\end{proof}
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