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{\bf Question}

Given that $$A = \left( \begin{array}{rrr} 0 &amp; 1 &amp; 0 \\ 1 &amp; 0 &amp; 1
\\ 0 &amp; 1 &amp; 0 \end{array} \right)$$

and $B$ is such that $AB = BA$, show that $B$ must have the form
$$ \left( \begin{array}{ccc} a &amp; b &amp; c \\ b &amp; {a+c} &amp; b
\\ c &amp; b &amp; a \end{array} \right)$$ where $a$, $b$ and $c$ are
arbitrary. (HINT: Let $$B = \left( \begin{array}{ccc} a &amp; b &amp; c \\
d &amp; e &amp; f \\ g &amp; g &amp; i \end{array} \right)$$ and calculate both
$AB$ and $BA$.  Compaire the entries in these two matricies to
show that $d = f = h = b$ etc...)

{\bf Answer}

$AB=\left(\begin{array} {ccc} 0 &amp; 1 &amp; 0\\ 1 &amp; 0 &amp; 1\\ 0 &amp; 1 &amp; 0
\end{array} \right) \left(\begin{array} {ccc} a &amp; b &amp; c\\ d &amp; e &amp; f\\ g &amp; h &amp;
i \end{array} \right)= \left(\begin{array} {ccc} d &amp; e &amp; f\\ {a+g}
&amp; {b+h} &amp; {c+i}\\ d &amp; e &amp; f
\end{array} \right)$

$BA=\left(\begin{array} {ccc} a &amp; b &amp; c\\ d &amp; e &amp; f\\ g &amp; h &amp; i
\end{array} \right)\left(\begin{array} {ccc} 0 &amp; 1 &amp; 0\\ 1 &amp; 0 &amp;
1\\ 0 &amp; 1 &amp; 0
\end{array} \right) = \left(\begin{array} {ccc} b &amp; {a+c} &amp; b\\ e
&amp; {d+f} &amp; e\\ h &amp; {g+i} &amp; h
\end{array} \right)$

and so $d=h=f=b$.

Also: $e=a+c=g+i=a+g=c+i$ when
\begin{eqnarray*} a+c = c+i &amp; \Longrightarrow &amp; a=i \\
                  a+c = a+g &amp; \Longrightarrow &amp; c=g
\end{eqnarray*}
Therefore, $d,\ e,\ f,\ g,\ h$ and $i$ are uniquely determined by
a choice of $a,\ b$ and $c$.  Hence $B$ must have the form:

$$B=\left(\begin{array} {ccc} a &amp; b &amp; c\\ b &amp; {a+c} &amp; b\\ c &amp; b &amp;
a
\end{array} \right).$$



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