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\noindent{\bf Math 4303\hfill Worksheet \#3 \hfill\today}\medskip\\
Let $\cal{K}$ denote the set  of $2\times 2$ matrices of the form \ilf{\left(\begin{array}{rr} a&b\\-b&a\end{array}\right)}, where $a,b\in\real$. 
\begin{enumerate}
\item Show: If $A$ and $B$ are elements in $\cal{K}$, then $A+B\in\cal{K}$.%Leave out next time
\item Show: If $A$ and $B$ are elements in $\cal{K}$, then $A\cdot B\in\cal{K}$.
\item Show that $A\cdot B=B\cdot A$ holds for all elements $A$, $B \in\cal{K}$.%Leave out next  time
\item  Note that the identity matrix \ilf{\left(\begin{array}{rr} 1&0\\0&1\end{array}\right)} is an element of $\cal{K}$. Show: If the matrix \ilf{A=\left(\begin{array}{rr} a&b\\-b&a\end{array}\right)} in $\cal{K}$ satisfies $a^2+b^2\neq 0$, then $A$ has a matrix inverse. What is $A^{-1}$? Does $A^{-1}$ always lie in $\cal{K}$? 
\item Show that we can identify $\cal{K}$ with $\cpl$, i.e., find a bijection $f:\cal{K}\to\cpl$ such that $f(A+B)=f(A)+f(B)$ and $f(A\cdot B)=f(A) \cdot f(B)$. (This is easier than it sounds.)
\item Under this identification, find the element $I$ in $\cal{K}$ that corresponds to $i\in\cpl$ (not surprisingly, there are actually two possible choices). Compute $I\cdot I$.
\item Under this identification, what is the ``conjugate'' of an element in $\cal{K}$, what is the ``modulus'' of an element in $\cal{K}$?
\end{enumerate}
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