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I have the real-valued matrix

$$ \begin{bmatrix} 1 & -5& -7 \\ 1 & 4 & 2\\ 0 &1& 4 \end{bmatrix}$$

I want to find P such that $$ P^{-1}AP=\begin{bmatrix} 3 & 1& 0 \\ 0 & 3 & 1\\ 0 &0& 3 \end{bmatrix}=J(3,3)$$

I've found the minimal polynomial $(t-3)^3$ and the characteristic polynomial $(3-t)^3$. I've found the eigenvalue $\lambda=3$, but I'm not sure how to find the eigenvectors. I found one eigenvector of $(1,1,-1)^T$ but when I try and apply the techniques I've learnt to solve situations with an eigenvalue of $A.M=3$ I keep finding the same eigenvector.

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  • $\begingroup$ If the minimal polynomial is $(t-3)^3$, the matrix is not diagonalizable, which you might also notice because the target is a Jordan matrix that consists of a single $3\times3$ block. You’re not going to find three linearly independent eigenvectors and will have to look for generalized eigenvectors instead. $\endgroup$ – amd Apr 20 '17 at 18:17
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You have found $u=(1,1,-1)^T$ such that $Au=3u$.

Now try to find :

  • $v$ such that $Av=u+3v$
  • $w$ such that $Aw=v+3w$

Because in the base $(u,v,w)$ this is what your matrix looks like. Remember that the columns of $P^{-1}AP$ are respectively $(Au,Av,Aw)$ in the base $(u,v,w)$.

You find $v=(2-z,-1-z,z)^T$, for instance $z=0$ leads to $v=(2,-1,0)^T$.

And then $w=(-1-z,-z,z)^T$, for instance $z=0$ leads to $w=(-1,0,0)^T$.

And we have $P=(u,v,w)=\begin{pmatrix} 1 & 2 & -1 \\ 1 & -1 & 0 \\ -1 & 0 & 0\end{pmatrix}$

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