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\begin{document}

\leftline{\Large \bf Department of Mathematical Sciences} 
\vspace{12pt}
\leftline{ 21--241 \bf Matrix Algebra \hfill{\bf  Fall 2002}}
\leftline{ \bf Professor A. Frieze and  Professor  S. Ta'asan}


\vspace{1in}
\centerline{\bf Test No. 3}
\vspace{1in}

\leftline{\hspace{3.5in} Name:\hrulefill }
\leftline{\hspace{3.5in} Section:\hrulefill }
\bigskip
\bigskip


\vspace{1.0in}
\hbox{ \hspace{2.0in}
{ \large
\begin{tabular}{||c|c|r||}
\hline
 problem &  points & scores \\ \hline
1 & 25  &\\ \hline
2 & 25 &\\ \hline
3 & 25 &\\ \hline
4 & 25 & \\ \hline
total & 100 & \\
 \hline
\end{tabular}
}
}

\pagebreak

\begin{enumerate}
\item (25 pts) 

(a) Let $\lambda$ be an eigenvalue of an invertible matrix $A$.
Show that $\lambda^{-1}$ is an eigenvalue of $A^{-1}$.

\begin{eqnarray*}
A{\bf x}&=&\lambda {\bf x}\qquad 4\ pts\\
A^{-1}A{\bf x}&=&\lambda A^{-1}{\bf x}\qquad 4\ pts\\
A^{-1}{\bf x}&=&\lambda^{-1}{\bf x}\qquad 4\ pts
\end{eqnarray*}

(b) Show that if $A^2$ is the zero matrix then the only eigenvalue of $A$ is zero.

\begin{eqnarray*}
A{\bf x}&=&\lambda {\bf x}\qquad 3\ pts\\
A^2{\bf x}&=&\lambda A{\bf x}\qquad 3\ pts\\
&=&\lambda^2{\bf x}\qquad 2\ pts\\
\noalign{So}
0&=&\lambda^2{\bf x}\qquad 2\ pts
\end{eqnarray*}
and then $\lambda=0$, since ${\bf x}\neq 0$ 3 pts. 
\newpage
\item (25 points) Let 

$$
A = \left[ \ba{rr} 2 &  3 \\ 
                     4 & 1 \ea \right]  
    $$ 

Diagonalise $A$. 

$$det\ A-\lambda I=(\lambda-5)(\lambda+2)\qquad 5\ pts$$
Eigenvalues are 5 and -2 -- 4 pts.

Eigenvectors are: 5: $[1\ 1]$ (4 pts) and -2: $[3\ -4]$ (4 pts).

$$
P = \left[ \ba{rr} 1 &  3 \\ 
                     1 & -4 \ea \right] \qquad 4\ pts 
    $$ 
$P^{-1}AP$ is diagonal. 4 pts.
\pagebreak


\item (25 points) 
An $n\times n$ matrix is {\em orthonormal} if its columns are an orthonormal set.

(a) Show that the eigenvalues of an $n\times n$ orthonormal matrix $A$ are $\pm 1$.

(Hint: consider the length of vector $A{\bf x}$ when ${\bf x}$ is an eigenvector of $A$.)

\begin{eqnarray*}
A{\bf x}&=&\lambda {\bf x}\qquad 3\ pts\\
{\bf x}^TA^TA{\bf x}&=&\lambda^2 ||{\bf x}||^2\qquad 3\ pts\\
{\bf x}^T{\bf x}&=&\lambda^2 ||{\bf x}||^2\qquad 3\ pts
\end{eqnarray*}
So $\lambda^2=1$ since ${\bf x}\neq 0$ 3 pts.

(b) Let $U,V$ be $n\times n$ orthonormal matrices
Show that $UV$ is also orthonormal.
\begin{eqnarray*}
(UV)^TUV&=&V^TU^TUV\qquad 3\ pts\\
&=&V^TV\qquad 3\ pts\\
&=&I\qquad 3\ pts
\end{eqnarray*}
This implies that $UV$ is orthonormal 4pts.
\vspace{5cm}

\pagebreak 
\item (25 points) 

Let ${\bf y}=\left[\begin{array}{c}2\\6\end{array}\right]$ and
${\bf u}=\left[\begin{array}{c}7\\1\end{array}\right]$  

Write ${\bf y}$ as the sum of a vector in Span$\{{\bf u}\}$ and a vector orthogonal to 
{\bf u}.

\begin{eqnarray*}
{\bf y}&=&\left({\bf y}-\frac{{\bf y}\cdot {\bf u}}{{\bf u}\cdot {\bf u}}{\bf u}\right)
+\frac{{\bf y}\cdot {\bf u}}{{\bf u}\cdot {\bf u}}{\bf u}.\\
&=&\left[\begin{array}{c}-4/5\\28/5\end{array}\right]+\left[\begin{array}{c}14/5\\2/5\end{array}\right].
\end{eqnarray*}
10 pts for correct method and rest for accuracy. Take only a few points off for numerical mistakes.
\end{enumerate}
\end{document}









 









