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QUESTION

Show that

$A=\left[\begin{array}{ccc}5&1&1\\-12&12&3\\-4&1&10\end{array}\right]$

has a triple eigenvalue 9 but that 9 has only a two dimensional
eigenspace $(4x=y+z)$ so $A$ cannot be diagonalised. Choose a
vector \textbf{b} not in the eigenspace and let
$\textbf{a}=(A-9I)\textbf{b}$

Show that \textbf{a} is an eigenvector. Choose a second
eigenvector \textbf{c} independent of \textbf{a} and form the
matrix $M$ which has as its columns the vectors \textbf{a,b,c}.
Calculate $M^{-1}$ and show that the matrix

$\Lambda=M^{-1}AM=\left[\begin{array}{ccc}9&1&0\\0&9&0\\0&0&9\end{array}\right]$

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ANSWER There are many choices for \textbf{b} but to make the
calculation as simple as possible, start with
$\textbf{b}=\left[\begin{array}{c}0\\0\\1\end{array}\right]$. Then

$\textbf{a}=(A-9I)\left[\begin{array}{c}0\\0\\1\end{array}\right]=
\left[\begin{array}{c}1\\3\\1\end{array}\right]$

Another simple choice is
$\textbf{c}=\left[\begin{array}{c}0\\1\\-1\end{array}\right]$ and
these choices give

\begin{eqnarray*}
& &
\left[\begin{array}{ccc}1&0&0\\-4&1&1\\-3&1&0\end{array}\right]
\left[\begin{array}{ccc}5&1&1\\-12&12&3\\-4&1&10\end{array}\right]
\left[\begin{array}{ccc}1&0&0\\3&0&1\\1&1&-1\end{array}\right]\\
&=&
\left[\begin{array}{ccc}5&1&1\\-36&9&9\\-27&9&0\end{array}\right]
\left[\begin{array}{ccc}1&0&0\\3&0&1\\1&1&-1\end{array}\right]\\
&=& \left[\begin{array}{ccc}9&1&0\\0&9&0\\0&0&9\end{array}\right]
\end{eqnarray*}

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