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QUESTION

Show that

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

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

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

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 $$M^{-1}AM=
\left[\begin{array}{ccc}0&4&-3\\0&3&-2\\1&1&-1\end{array}\right]
\left[\begin{array}{ccc}3&-2&1\\3&-1&1\\4&-5&4\end{array}\right]
\left[\begin{array}{ccc}-1&1&1\\-2&3&0\\-3&4&0\end{array}\right]=
\left[\begin{array}{ccc}2&1&0\\0&2&1\\0&0&2\end{array}\right]$$



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