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\usepackage{amsmath}
\begin{document}

{\bf Question}

In question 3 of exercise sheet 7 you found the eigenvalues and
eigenvectors for the matrices associated to Allyl radical,
Cyclopropenyl, and Trimethlene methane.  Use your results to
diagonalise these matrices.

{\bf Answer} \vspace{.2in}

\begin{description}
\item[(i)]
Allyl radical $\left( \begin{array}{ccc} 0 &amp; 1 &amp; 0 \\ 1 &amp; 0 &amp; 1 \\
0 &amp; 1 &amp; 0 \\ \end{array} \right)$ Eigenvalues:$0$, $\sqrt2$,
$-\sqrt2$

Corresponding eigenvalues $\left( \begin{array}{c} \alpha \\ 0 \\
-\alpha \end{array} \right)$, $\left( \begin{array}{c} \beta \\
\sqrt2 \beta \\ \beta \end{array} \right)$ and $\left(
\begin{array}{c} \gamma \\ -\sqrt2 \gamma \\ \gamma \end{array}
\right)$ which normalise to $\left( \begin{array}{c}
\frac{1}{\sqrt2} \\ 0 \\ \frac{1}{\sqrt2} \end{array} \right)$ ,
$\left( \begin{array}{c} \frac{1}{2} \\ \frac{1}{\sqrt2} \\
\frac{1}{2} \end{array} \right)$ and $\left( \begin{array}{c}
\frac{1}{2} \\ -\frac{1}{\sqrt2} \\ \frac{1}{2} \end{array}
\right)$ $\left[ {\rm Note:\ } \frac{\sqrt2}{2} =
\frac{\sqrt2}{\sqrt2 \sqrt2} = \frac{1}{\sqrt2}\right]$

Take the orthogonal matrix $R = \left(
\begin{array}{ccc} \frac{1}{\sqrt2} &amp; \frac{1}{\sqrt2} &amp; \frac{1}{\sqrt2} \\
 0 &amp; \frac{1}{\sqrt2}  &amp; \frac{-1}{\sqrt2} \\ \frac{-1}{\sqrt2} &amp; \frac{1}{\sqrt2}
  &amp; \frac{1}{\sqrt2} \\ \end{array} \right)$

The original matrix diagonalises to $R^TAR =
\left(\begin{array}{ccc} 0 &amp; 0 &amp; 0 \\
 0 &amp; {\sqrt2}  &amp; 0 \\ 0 &amp; 0  &amp; {-\sqrt2} \\ \end{array} \right)$


\item[(ii)]
Cyclopropenyl $\left( \begin{array}{ccc} 0 &amp; 1 &amp; 1 \\ 1 &amp; 0 &amp; 1
\\ 1 &amp; 1 &amp; 0 \\ \end{array} \right)$ Eigenvalues:$2$, $-1$,
$-1$

Corresponding eigenvalues $\left( \begin{array}{c} \alpha \\
\alpha \\ \alpha \end{array} \right)$, $\left( \begin{array}{c}
\beta \\ -(\beta +\gamma) \\ \gamma \end{array} \right)$.  Using
this last eigenvector we can choose two orthognal eigenvectors.
For example if $\beta = \gamma = 1$ then we have $\left(
\begin{array}{c} 1 \\ -2 \\ 1 \end{array} \right)$ and if $\beta =
1, \gamma = -1$ we have $\left( \begin{array}{c} 1 \\ 0
\\ -1 \end{array} \right)$ so the eigenvalues 2, -1, -1 yield three mutually orthogonal,
normalised eigenvectors: $\left( \begin{array}{c} \frac{1}{\sqrt3}
\\ \frac{1}{\sqrt3} \\ \frac{1}{\sqrt3} \end{array} \right)$ , $\left(
\begin{array}{c} \frac{1}{\sqrt6} \\ \frac{-2}{\sqrt6} \\ \frac{1}{\sqrt6}
\end{array} \right)$ and $\left( \begin{array}{c} \frac{1}{\sqrt2} \\
0 \\ \frac{-1}{\sqrt2} \end{array} \right) \rm {respectively}.$

Take the orthogonal matrix $R = \left(
\begin{array}{ccc} \frac{1}{\sqrt3} &amp; \frac{1}{\sqrt6} &amp; \frac{1}{\sqrt2} \\
 \frac{1}{\sqrt3} &amp; \frac{-2}{\sqrt6}  &amp; 0 \\ \frac{1}{\sqrt3} &amp; \frac{1}{\sqrt6}
  &amp; \frac{-1}{\sqrt2} \\ \end{array} \right)$

The original matrix diagonalises to $R^TAR =
\left(\begin{array}{ccc} 2 &amp; 0 &amp; 0 \\
 0 &amp; -1  &amp; 0 \\ 0 &amp; 0  &amp; {-1} \\ \end{array} \right)$


\item[(iii)]
Trimethylene methane $\left( \begin{array}{cccc} 0 &amp; 0 &amp; 0 &amp; 1 \\
0 &amp; 0 &amp; 0 &amp; 1 \\ 0 &amp; 0 &amp; 0 &amp; 1 \\ 1 &amp; 1 &amp; 1 &amp; 0 \\ \end{array}
\right)$ Eigenvalues:$0$, $0$, $\sqrt3$, $-\sqrt3$.

Four mutually orthogonal eigenvectors $\left(
\begin{array}{c} 2 \\ -1 \\ -1 \\ 0 \end{array} \right)$, $\left( \begin{array}{c}
0 \\ 1 \\ -1 \\ 0 \end{array} \right)$, $\left(
\begin{array}{c} 1 \\ 1 \\ 1 \\ \sqrt3 \end{array} \right)$, $\left( \begin{array}{c}
1 \\ 1 \\ 1 \\ -\sqrt3 \end{array} \right)$ which normalise to
$\left( \begin{array}{c} \frac{2}{\sqrt6}
\\ \frac{-1}{\sqrt6} \\ \frac{-1}{\sqrt6} \\ 0  \end{array} \right)$ , $\left(
\begin{array}{c} 0 \\ \frac{1}{\sqrt2} \\ \frac{-1}{\sqrt2} \\ 0
\end{array} \right)$, $\left( \begin{array}{c} \frac{1}{\sqrt6} \\
\frac{1}{\sqrt6} \\ \frac{1}{\sqrt6} \\ \frac{1}{\sqrt2}
\end{array} \right)$ and $\left( \begin{array}{c} \frac{1}{\sqrt6} \\
\frac{1}{\sqrt6} \\ \frac{1}{\sqrt6} \\ \frac{-1}{\sqrt2}
\end{array} \right)$ respectively .

Take the orthogonal matrix $R = \left(
\begin{array}{cccc} \frac{2}{\sqrt6} &amp; 0 &amp; \frac{1}{\sqrt6} &amp; \frac{1}{\sqrt6} \\
 \frac{-1}{\sqrt6}&amp; \frac{1}{\sqrt2} &amp; \frac{1}{\sqrt6}  &amp; \frac{1}{\sqrt6} \\ \frac{-1}{\sqrt6}
 &amp; \frac{-1}{\sqrt2} &amp; \frac{1}{\sqrt6} &amp; \frac{1}{\sqrt6} \\ 0 &amp; 0 &amp; \frac{1}{\sqrt2} &amp;
\frac{-1}{\sqrt2} \\ \end{array} \right)$

The original matrix diagonalises to $R^TAR =
\left(\begin{array}{cccc} 0 &amp; 0 &amp; 0 &amp; 0 \\
 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0  &amp; {\sqrt3} &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; -\sqrt3 \\ \end{array} \right)$
\end{description}

\end{document}
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