\documentclass[a4paper,12pt]{article}
\begin{document}


{\bf Question}

The following matrices occur in the $\pi$-molecular orbital theory
of conjugated hydrocarbons. The eigenvalues of the matrices
determine the energies of the  molecular orbitals (this theory
will be met in the second year physical chemistry course).
\begin{description}
\item[(a)]
Find the eigenvalues and eigenvectors of the following matrices:\
\  (i)\ Allyl radical $\left( \begin{array}{ccc}
     0 & 1 & 0\\
     1 & 0 & 1\\
     0 & 1 & 0
 \end{array}  \right)$;\
(ii)\ Cyclopropenyl $\left( \begin{array}{ccc}
     0 & 1 & 1\\
     1 & 0 & 1\\
     1 & 1 & 0
 \end{array}  \right)$;
\item[(b)]
Find the eigenvalues of the following matrices:

(i)\ Cyclobutadiene $\left( \begin{array}{cccc}
     0 & 1 & 0 & 1\\
     1 & 0 & 1 & 0\\
     0 & 1 & 0 & 1\\
     1 & 0 & 1 & 0
 \end{array}  \right)$;\
(ii)\ Butadiene $\left( \begin{array}{cccc}
     0 & 1 & 0 & 0\\
     1 & 0 & 1 & 0\\
     0 & 1 & 0 & 1\\
     0 & 0 & 1 & 0
 \end{array}  \right)$.

\item[(c)]
Show that the vectors

$$ \left( \begin{array}{rrrr}
     2\\
     {-1}\\
     {-1}\\
     0
 \end{array}  \right),\ \
\left( \begin{array}{rrrr}
     0\\
     1\\
     {-1}\\
     0
 \end{array}  \right),\ \
\left( \begin{array}{rrrr}
     1\\
     1\\
     1\\
     {\sqrt 3}
 \end{array}  \right),\ \
\left( \begin{array}{rrrr}
     1\\
     1\\
     1\\
     {-\sqrt 3}
 \end{array}  \right),
 $$
are all eigenvectors of the matrix

$$ Trimethylene methane \left( \begin{array}{cccc}
     0 & 0 & 0 & 1\\
     0 & 0 & 0 & 1\\
     0 & 0 & 0 & 1\\
     1 & 1 & 1 & 0
 \end{array}  \right)$$

Find the eigenvalues corresponding to these eigenvectors. Does the
matrix have any other eigenvectors?
\end{description}



{\bf Answer}

\begin{description}
\item[(a)]
\begin{description}
\item[(i)]
$\left|\begin{array}{ccc} -\lambda & 1 & 0\\ 1 & -\lambda & 1\\ 0
& 1 & -\lambda
\end{array} \right|=-\lambda(\lambda^2-1)-(-\lambda)+0=-\lambda(\lambda^2-2)=0$

so $\lambda=0, \pm \sqrt{2}$.

\begin{description}
\item[\underline{$\lambda = 0$}]
Solve $\left(\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0
\end{array}\right) \left(\begin{array}{c} x\\ y\\ z
\end{array}\right)=\left(\begin{array}{c} 0\\ 0\\ 0
\end{array}\right)$

or $\left.\begin{array} {rcl} y & = & 0\\ x+z & = & 0\\ y & = & 0
\end{array}\right\}$\ $y=0$,\ let\ $x=\alpha$ so $z=-\alpha$.

Suitable eigenvector: $\left(\begin{array}{c} \alpha\\ 0\\ -\alpha
\end{array} \right).$

\item[\underline{$\lambda = \sqrt{2}$}]
Solve $\left(\begin{array}{ccc} -\sqrt{2} & 1 & 0\\ 1 & -\sqrt{2}
& 1\\ 0 & 1 & -\sqrt{2}
\end{array}\right) \left(\begin{array}{c} x\\ y\\ z
\end{array}\right)=\left(\begin{array}{c} 0\\ 0\\ 0
\end{array}\right)$

or $\left.\begin{array} {rcl} -\sqrt{2}x+y & = & 0\\ x-\sqrt{2}y+z
& = & 0\\ y-\sqrt{2}z & = & 0
\end{array}\right\}$\ let\ $x=\beta,\ y=\sqrt{2}\beta, z=\beta$.

Suitable eigenvector: $\left(\begin{array}{c} \beta\\
\sqrt{2}\beta\\ \beta
\end{array} \right).$

\item[\underline{$\lambda = -\sqrt{2}$}]
Solve $\left(\begin{array}{ccc} \sqrt{2} & 1 & 0\\ 1 & \sqrt{2} &
1\\ 0 & 1 & \sqrt{2}
\end{array}\right) \left(\begin{array}{c} x\\ y\\ z
\end{array}\right)=\left(\begin{array}{c} 0\\ 0\\ 0
\end{array}\right)$

or $\left.\begin{array} {rcl} \sqrt{2}x+y & = & 0\\ x+\sqrt{2}y+z
& = & 0\\ y+\sqrt{2}z & = & 0
\end{array}\right\}$\ let\ $x=\gamma,\ y=-\sqrt{2}\gamma, z=\gamma$.

Suitable eigenvector: $\left(\begin{array}{c} \gamma\\
-\sqrt{2}\gamma\\ \gamma
\end{array} \right).$
\end{description}
\item[(ii)]
\begin{eqnarray*} \left|\begin{array}{ccc} -\lambda & 1 & 1\\ 1 & -\lambda &
1\\ 1 & 1 & -\lambda
\end{array} \right| & = & \left|\begin{array}{ccc} -(\lambda+1) & 0 & (\lambda+1)\\
1 & -\lambda & 1\\ 1 & 1 & -\lambda
\end{array} \right| (R_1'=R_1-R_3)\\
& = & (1+\lambda) \left|\begin{array}{ccc} -1 & 0 & 1\\ 1 &
-\lambda & 1\\ 1 & 1 & -\lambda
\end{array} \right|\\ & = & (1+\lambda)(-\lambda^2+\lambda+2)\\ & = &
(1+\lambda)(2-\lambda)(1+\lambda)=0 \end{eqnarray*} So
$\lambda=2,\ -1,\ -1$.

\begin{description}
\item[\underline{$\lambda=2$}]
Solve $\left(\begin{array}{ccc} -2 & 1 & 1\\ 1 & -2 & 1\\ 1 & 1 &
-2
\end{array}\right) \left(\begin{array}{c} x\\ y\\ z
\end{array}\right)=\left(\begin{array}{c} 0\\ 0\\ 0
\end{array}\right)$

or $\left.\begin{array} {crcl} (1) & -2x+y+z & = & 0\\ (2) &
x-2y+z & = & 0\\ (3) & x+y-2z & = & 0
\end{array}\right\}$
$(2)+2(3) \Rightarrow 3x-3z=0$.  Let $x=\alpha$\ so that
$z=\alpha$. Substitute $x=\alpha=z$ into $(1)$ to get
$y=2x-z=2\alpha - \alpha = \alpha$.

Suitable eigenvector: $\left(\begin{array}{c} \alpha\\ \alpha\\
\alpha \end{array} \right)$

\item[\underline{$\lambda=-1$}] (Repeated eigenvalue)

Solve $\left(\begin{array}{ccc} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1
\end{array}\right) \left(\begin{array}{c} x\\ y\\ z
\end{array}\right)=\left(\begin{array}{c} 0\\ 0\\ 0
\end{array}\right)$

So $x+y+z=0$. If $x=\beta$ then $y+z=-\beta$ (thus we have a 2
parameter family of eigenvectors). Let $z=\gamma$ so that
$y=-(\beta+\gamma)$.

Suitable eigenvector: $\left(\begin{array}{c} \beta\\
-(\beta+\gamma)\\ \gamma
\end{array} \right).$

\end{description}
\end{description}

\item[(b)]
\begin{description}
\item[(i)]
$$\left|\begin{array}{rrrr} -\lambda & 1 & 0 & 1\\ 1 & -\lambda &
1 & 0\\ 0 & 1 & -\lambda & 1\\ 1 & 0 & 1 & -\lambda \end{array}
\right| = \left|\begin{array}{rrrr} -\lambda & 0 & \lambda & 0\\ 1
& -\lambda & 1 & 0\\ 0 & 1 & -\lambda & 1\\ 1 & 0 & 1 & -\lambda
\end{array} \right| (R_1'=R_1-R_3)$$
\ \ \ \ \ (Expand determinant along first row)
\begin{eqnarray*}
& = & (-\lambda)(-1)^{1+1} \left|\begin{array}{ccc} -\lambda & 1 &
0\\ 1 & -\lambda & 1\\ 0 & 1 & -\lambda
\end{array} \right|+0+(\lambda)(-1)^{1+3} \left|\begin{array}{ccc} 1 & -\lambda &
0\\ 0 & 1 & 1\\ 1 & 0 & -\lambda
\end{array} \right|+0\\
& = &
-\lambda[-\lambda(\lambda^2-1)-(-\lambda)]+\lambda[-\lambda+\lambda(-1)]\\
& = & \lambda^2(\lambda^2-4)
\end{eqnarray*}
So $\lambda=0,\ 0,\ 2,\ -2$.

\item[(ii)]
$$\left|\begin{array}{rrrr} -\lambda & 1 & 0 & 0\\ 1 & -\lambda &
1 & 0\\ 0 & 1 & -\lambda & 1\\ 0 & 0 & 1 & -\lambda \end{array}
\right| = (-\lambda)(-1)^{1+1}\left|\begin{array}{rrr} -\lambda &
1 & 0\\ 1 & -\lambda & 1\\ 0 & 1 & -\lambda
\end{array} \right| + (-1)^{1+2} \left|\begin{array}{rrr} 1 &
1 & 0\\ 0 & -\lambda & 1\\ 0 & 1 & -\lambda
\end{array} \right|+0+0$$
\ \ \ \ \ (Expand determinant along first row)
\begin{eqnarray*}
& = & -\lambda[-\lambda(\lambda^2-1)-(-\lambda)]-[\lambda^2 -1]\\
& = & \lambda^4 -3\lambda^2+1 = 0
\end{eqnarray*}

Think of this as a quadratic in $\lambda^2$, so that

$$\lambda^2=\frac{3 \pm \sqrt{9-4}}{2}=\frac{3 \pm \sqrt{5}}{2}.$$

Hence $\displaystyle \lambda=\pm \sqrt{\frac{3 \pm
\sqrt{5}}{2}}=\pm \left(\frac{1 \pm \sqrt{5}}{2} \right)$

$\displaystyle \left( \rm{You\ should\ check\ that}
\left(\frac{1+\sqrt{5}}{2} \right)^2=\frac{3+\sqrt{5}}{2} \right)$
\end{description}

\item[(c)]
$$\left(\begin{array}{cccc} 0 & 0 & 0 & 1\\0 & 0 & 0 & 1\\0 & 0 &
0 & 1\\1 & 1 & 1 & 0 \end{array} \right) \left(\begin{array} {r}
2\\-1\\-1\\0 \end{array} \right)=\left(\begin{array} {c}
0\\0\\0\\0 \end{array} \right)=0\left(\begin{array} {r}
2\\-1\\-1\\0 \end{array} \right)\ \ \rm{eigenvalue}=0$$

$$\left(\begin{array}{cccc} 0 & 0 & 0 & 1\\0 & 0 & 0 & 1\\0 & 0 &
0 & 1\\1 & 1 & 1 & 0 \end{array} \right) \left(\begin{array} {r}
0\\1\\-1\\0 \end{array} \right)=\left(\begin{array} {c} 0\\0\\0\\0
\end{array} \right)=0\left(\begin{array} {r} 0\\1\\-1\\0
\end{array} \right)\ \ \rm{eigenvalue}=0$$

$$\left(\begin{array}{cccc} 0 & 0 & 0 & 1\\0 & 0 & 0 & 1\\0 & 0 &
0 & 1\\1 & 1 & 1 & 0 \end{array} \right) \left(\begin{array} {r}
1\\1\\1\\ \sqrt{3} \end{array} \right)=\left(\begin{array} {c}
\sqrt3\\ \sqrt3\\ \sqrt3\\3
\end{array} \right)=\sqrt3\left(\begin{array} {r} 1\\1\\1\\ \sqrt3
\end{array} \right)\ \ \rm{eigenvalue}=\sqrt3$$

$$\left(\begin{array}{cccc} 0 & 0 & 0 & 1\\0 & 0 & 0 & 1\\0 & 0 &
0 & 1\\1 & 1 & 1 & 0 \end{array} \right) \left(\begin{array} {r}
1\\1\\1\\-\sqrt3 \end{array} \right)=\left(\begin{array} {c}
-\sqrt3\\ -\sqrt3\\ -\sqrt3\\3
\end{array} \right)=-\sqrt3 \left(\begin{array} {r}
1\\1\\1\\-\sqrt3
\end{array} \right)\ \ \rm{eigenvalue}=-\sqrt3$$

An n x n matrix has at most n orthogonal eigenvectors.

The eigenvectors of this 4x4 matrix are mutually orthogonal (you
should check this!) and there are four of them - the maximum
number possible.  Hence any other eigenvector of the matrix will
be a linear combination of these four orthogonal vectors:

$$r\left(\begin{array}{r}2\\-1\\-1\\0\end{array}\right)
+s\left(\begin{array}{r}0\\1\\-1\\0\end{array}\right)
+t\left(\begin{array}{r}1\\1\\1\\ \sqrt3\end{array}\right)
+u\left(\begin{array}{r}1\\1\\1\\ -\sqrt3\end{array}\right)$$

for any scalars $r,\ s,\ t,\ u$.
\end{description}



\end{document}