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\noindent\textbf{First name:} Christoph\\
\textbf{Last Name:} Oberauer\\
\textbf{Email:} coberau@cosy.sbg.ac.at\\
\textbf{Date:} 07.12.2002\\
\textbf{Homework number:} 3\\
\textbf{Homework Title:} Exercise 4.1


\paragraph{Problem description:}
\begin{enumerate}
\item
Prove that 5 is an eigenvalue of the matrix 
$$A=\left[\begin{array}{cccc}6&3&3&1\\0&7&4&5\\0&0&5&4\\0&0&0&8\end{array}\right].$$
\item
Exhibit an eigenvector of $A$ corresponding to the eigenvalue 5.
\end{enumerate}


\paragraph{Problem solution:}
\begin{enumerate}
\item
Trivial.
\item
Solving $Ax=\lambda{}x$ yields eigenvector corresponding to $\lambda$.
\end{enumerate}


\paragraph{Results:}
\begin{enumerate}
\item
Since $A$ is a upper-triangular matrix, the diagonal contains all eigenvalues of $A$. Second possibility if $A$ were not triangular: $det(A-\lambda{}I)=0$
\item
Solving linear system
\begin{align*}
6x_1+3x_2+3x_3+ x_4 & = 5x_1\\
     7x_2+4x_3+5x_4 & = 5x_2\\
          5x_3+4x_4 & = 5x_3\\
	       8x_4 & = 5x_4
\end{align*}
using backward substitution: $x_4=0 \Rightarrow x_3=1 \Rightarrow x_2=-2 \Rightarrow x_1=3$
\end{enumerate}


\paragraph{Discussion and Comments:}
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