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**Juneu436** Thanks, but why is $\displaystyle k=3,4, \text{ or } 5$?

Shouldn't it be $\displaystyle k=1,2, \text{ or } 3$?

So for question 1 I get:

$\displaystyle m(\lambda)=(\lambda+2)(\lambda-5)$

$\displaystyle \mathbf{J_1} = \begin{bmatrix}-2 & 0 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 0 & 0 \\0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$\displaystyle m(\lambda)=(\lambda+2)^2(\lambda-5)$

$\displaystyle \mathbf{J_2} = \begin{bmatrix}-2 & 1 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 0 & 0 \\0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$\displaystyle m(\lambda)=(\lambda+2)(\lambda-5)^2$

$\displaystyle \mathbf{J_3} = \begin{bmatrix}-2 & 0 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 0 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$\displaystyle m(\lambda)=(\lambda+2)^2(\lambda-5)^2$

$\displaystyle \mathbf{J_4} = \begin{bmatrix}-2 & 1 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 0 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$\displaystyle m(\lambda)=(\lambda+2)^2(\lambda-5)^3$

$\displaystyle \mathbf{J_5} = \begin{bmatrix}-2 & 1 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 1 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

$\displaystyle m(\lambda)=(\lambda+2)(\lambda-5)^3$

$\displaystyle \mathbf{J_6} = \begin{bmatrix}-2 & 0 & 0 & 0 & 0 \\0 & -2 & 0 & 0 & 0 \\0 & 0 & 5 & 1 & 0 \\0 & 0 & 0 & 5 & 1 \\ 0 & 0 & 0 & 0 & 5\end{bmatrix}$

Is this correct?

What about question 2?

Thanks