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QUESTION

Show that

$A=\left[\begin{array}{ccc}3&amp;-2&amp;1\\3&amp;-1&amp;1\\4&amp;-5&amp;4\end{array}\right]$

has a triple eigenvalue but only a one dimensional eigenspace (the
line $x=\alpha,y=2\alpha,z=3\alpha$) so $A$ cannot be
diagonalised. Calculate $(A-2I)^2$ and $(A-2I)^3$.

Find a vector \textbf{u} satisfying

$(A-2I)^3\textbf{u}=\textbf{0}$ but $(A-2I)^2\textbf{u}\neq
\textbf{0}.$

Calculate

$\textbf{v}=(A-2I)\textbf{u}$

and

$\textbf{w}=(A-2I)\textbf{v}$.

Show that \textbf{w} is an eigenvector of $A$.

Now form the matrix $M$ the columns of which are the vectors
\textbf{w,v,u } respectively and calculate $M^{-1}AM$

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ANSWER

$A-2I=\left[\begin{array}{ccc}1&amp;-2&amp;1\\3&amp;-3&amp;1\\4&amp;-5&amp;2\end{array}\right]\
(A-2I)^2=\left[\begin{array}{ccc}-1&amp;-1&amp;1\\-2&amp;-2&amp;2\\-3&amp;-3&amp;3\end{array}\right]\
(A-2I)^3=\textbf{0}$

Choosing
\textbf{u}=$\left[\begin{array}{c}1\\0\\0\end{array}\right]$, then
\textbf{v}=$\left[\begin{array}{c}1\\3\\4\end{array}\right]$ and
\textbf{w}=$\left[\begin{array}{c}-1\\-2\\-3\end{array}\right]$.
With this choice of \textbf{u, v, w} then $$M^{-1}AM=
\left[\begin{array}{ccc}0&amp;4&amp;-3\\0&amp;3&amp;-2\\1&amp;1&amp;-1\end{array}\right]
\left[\begin{array}{ccc}3&amp;-2&amp;1\\3&amp;-1&amp;1\\4&amp;-5&amp;4\end{array}\right]
\left[\begin{array}{ccc}-1&amp;1&amp;1\\-2&amp;3&amp;0\\-3&amp;4&amp;0\end{array}\right]=
\left[\begin{array}{ccc}2&amp;1&amp;0\\0&amp;2&amp;1\\0&amp;0&amp;2\end{array}\right]$$



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