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\title{MA20008 Algebra 1, 2004, Sheet 6}
\author{Geoff Smith, {\tt http://www.bath.ac.uk/$\sim$masgcs}}
\date{}
\begin{document}
\maketitle
\begin{enumerate}
\item Let $V$ be a vector space of dimension
$n$. Suppose that $\alpha: V \rightarrow V$ is a linear
map. Show that the following are equivalent.
\begin{enumerate}
\item $\alpha$ is injective.
\item $\alpha$ is bijective.
\item $\alpha$ is surjective.
\item There is a basis ${\bf v_1}, \ldots, {\bf v_n}$ of $V$ such that
$\alpha({\bf v_1}), \ldots, \alpha({\bf v_n})$ is a basis of $V$.
\item For every basis ${\bf v_1}, \ldots, {\bf v_n}$ of $V$, the vectors
$\alpha({\bf v_1}), \ldots, \alpha({\bf v_n})$ also form a basis of $V$.
\end{enumerate}
{\em Hint: the rank-nullity theorem may be useful in places.}
\item Suppose that 
\[ X = \left( \begin{array}{cc} A & B\\
0 & C\end{array}\right)\]
is a $2r$ by $2r$ matrix built from the four $r$ by $r$ matrices
$A, B, C$ and the zero matrix $0$. Suppose that $X$ has an inverse matrix.
Describe that matrix in terms of $A, B, C$ and $0$.
\item The matrix 
\[ F = \left( \begin{array}{cc} 1 & 1\\
1 & 0\end{array}\right) \] has entries in the Field $F_7$, the integers
modulo $7$. Calculate 
\begin{enumerate}
\item $F^2$.
\item $F^5$.
\item $F^{1000}$.
\item \[ \sum_{i= 0}^{999} F^i\]
where $F^0$ denotes the identity matrix.
\end{enumerate}
\item The matrix
\[ F = \left( \begin{array}{cc} 1 & 1\\
1 & 0\end{array}\right) \] has entries in $\mathbb Q$,
Let $I$ denote the 2 by 2 identity matrix.
Show that $I$ and $F$ are linearly independent but
that $I, F$ and $F^2$ are linearly dependent elements
of the vector space of $2$ by $2$ matrices with rational entries
(with scalars in $\mathbb Q$).
\item Suppose that $\alpha, \beta: V \longrightarrow V$ are a 
pair of commuting linear maps. 
\begin{enumerate}
\item Prove that both $\mbox{Im }\alpha$
and $\mbox{Ker }\alpha$ are $\beta$-invariant spaces. 
\item Prove that $\mbox{Im }\alpha + \mbox{Im }\beta$ is
both $\alpha$-invariant and $\beta$-invariant.
\item Prove that $\mbox{Im }\alpha \cap \mbox{Im }\beta$ is
both $\alpha$-invariant and $\beta$-invariant.
\end{enumerate}
\end{enumerate} 
\end{document}
