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\noindent\textbf{First name:} Christoph\\
\textbf{Last Name:} Oberauer\\
\textbf{Email:} coberau@cosy.sbg.ac.at\\
\textbf{Date:} 15.11.2002\\
\textbf{Homework number:} 2\\
\textbf{Homework Title:} Exercise 2.26


\paragraph{Problem description:\\}
An $n \times n$ matrix $A$ is said to be elementary if it differs from the identity matrix by a matrix of rank one, i.e., if $A=I-uv^T$ for some $n$-vectors $u$ and $v$.
\begin{enumerate}
\item 
If $A$ is elementary, what condition on $u$ and $v$ ensures that $A$ is nonsingular?
\item 
If $A$ is elemantary and nonsingular, prove that $A^{-1}$ is also elementary by showing that $A^{-1}=I-\sigma{}uv^T$ for some scalar $\sigma$. What is the specific value for $\sigma{}$, in terms of $u$ and $v$?
\item 
Is an elementary elemination matrix, as defined in section 2.4.3, elementary? If so, what are $u$, $v$ and $\sigma$ in this case?
\end{enumerate}


\paragraph{Problem solution:}
\begin{enumerate}
\item
Try to find values for $u$ and $v$ so that $uv^T$ creates a matrix where the $k$-th column or row vector equals $e_k$
\item
Use Sherman-Morrison formula:
$$(A-uv^T)^{-1} = A^{-1} + A^{-1}u (1-v^TA^{-1}u)^{-1} v^TA^{-1}$$
\item
Try to find appropriate $u$, $v$ and calculate $\sigma$
\end{enumerate}


\paragraph{Results:}
\begin{enumerate}
\item
$u$($v$)=$e_k$, $v$($u$)=vector with 1 in $k$-th position, other values arbitrary. Calculating $I-uv^T$ results in a matrix with one of the rows(columns) filled with zeros which is singular.
\item
Use $A^{-1}=I-\sigma{}uv^T$ and substitute $A$ by $I - uv^T$
$$(I - uv^T)^{-1} = I-\sigma{}uv^T$$
Apply Sherman-Morrison formula:
$$I^{-1} + I^{-1}u(1-v^TI^{-1}u)^{-1}v^TI^{-1}=I-\sigma{}uv^T$$
$$(1-v^Tu)^{-1}uv^T=-\sigma{}uv^T$$
$$\sigma=-(1-v^Tu)^{-1}$$
\item
Yes, $u = (0,..,m_{k+1},..,m_n)$, $v = (0,...,1,...,0)$ where the index of 1 is $k$. Because of $v^Tu=0$, $\sigma=-(1-v^Tu)^{-1}$ always evaluates to $-1$.
\end{enumerate}


\paragraph{Discussion and Comments:}
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