\documentclass[12pt]{article} \pagestyle{empty} \usepackage{amssymb} \usepackage{latexsym} \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}