\documentclass{letter}
\usepackage{latexsym}
\begin{document}

\textbf{First name:} Christian \\
\textbf{Last name:} Osterrieder \\
\textbf{Date:} \today \\
\textbf{Homework number:} 2 \\
\textbf{Homework Title:} Exercise 2.8 \newline

\textbf{Problem description:} \\
Let $A$ and $B$ be any two $n$ $x$ $n$ matrices. \\
(a) Prove that $(AB)^{T} = B^{T}A^{T}$. \\
(b) If $A$ and $B$ are both nonsingular, prove that $(AB)^{-1} = B^{-1}A^{-1}$. \newline

\textbf{Results:} \\
\begin{description}

\item{(a)}
 Let $C = AB$ then 

\begin{eqnarray*}
 c_{i,j} & = & \sum_{k=1}^{n}a_{i,k}b_{k,j}.
\end{eqnarray*}

Thus,
\begin{eqnarray*}
c_{j,i}^{T}  =  c_{i,j} & = & \sum_{k=1}^{n}a_{i,k}b_{k,j}. 
\end{eqnarray*}

Let $D = B^{T}A^{T}$ then 
\begin{eqnarray*}
d_{j,i} = \sum_{k=1}^{n}b_{j,k}^{T}a_{k,i}^{T} & = & \sum_{i=1}^{n}a_{i,k}b_{k,j}. 
\end{eqnarray*}

\item{(b)}
$B^{-1}A^{-1}$ exists because $A$ and $B$ are nonsingular. \newline
First, $B^{-1}A^{-1}$ is a left inverse of $AB$:

\begin{eqnarray*}
(B^{-1}A^{-1})(AB) & = & B^{-1}(A^{-1}A)B \\
& = & B^{-1}IB \\
& = &  B^{-1}B \\
& = & I
\end{eqnarray*}

Next, $B^{-1}A^{-1}$ is a right inverse of $AB$.

\begin{eqnarray*}
(AB)(B^{-1}A^{-1}) & = & A(BB^{-1})A^{-1} \\
& = & AIA^{-1} \\
& = & AA^{-1} \\
& = & I
\end{eqnarray*}

Therefore, $AB$ in nonsingular and its inverse is given by 
\begin{eqnarray*}
(AB)^{-1} & = & B^{-1}A^{-1}.
\end{eqnarray*}

\end{description}

\end{document}
