Find the eigenvectors , generalized eigenvectors and the minimal polynomial of a certain matrix

Question

Given $A=$ $ \begin{pmatrix} -3 & 1 &-1 \\ -7 & 5 & -1 \\ -6 & 6 & -2 \\ \end{pmatrix} $ and also the characteristic polynomial $P_A(x)=(x+2)^2 \cdot (x-4)$

1. Find the eigenvectors , generalized eigenvectors and the minimal polynomial of $A$ is $A$ diagonalizable ?
2. let $T: \Bbb R^3 \to \Bbb R^3$ be $Tv=Av$ find $[T]_B$ $B$ is the given basis of $\Bbb R^3$ >from the generalized eigenvectors

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I started by finding the minimal polynomial and it is the same as $P_A$ this is because $A$ is not a scalar matrix and $P_A$ does not factor (sorry if the translation is incorrect hopefully it is understandable)

now I checked the eigenvectors to check if it is diagonalizable according to their geometric and algebraic multiplicity and got $(A-4I)=$ $ \begin{pmatrix} -7 & 1 &-1 \\ -7 & 1 & -1 \\ -6 & 6 & 6 \\ \end{pmatrix} $ from here the null space is $span\{v_1=$$ \begin{pmatrix} 0 \\ 1 \\ 1 \\ \end{pmatrix}\}$

$(A+2I)=$ $ \begin{pmatrix} -7 & 1 &-1 \\ -7 & 1 & -1 \\ -6 & 6 & 0 \\ \end{pmatrix} $ and the same from here the null space is $span\{v_2=$$ \begin{pmatrix} -1 \\ -1 \\ 0 \\ \end{pmatrix}\}$

from here we can see it is not diagonalizable

the generalized eigenvector: $(A+2I)^2=$$ \begin{pmatrix} 0& 0 &0 \\ -36 & 36 & 0 \\ 36 & -36 & 0 \\ \end{pmatrix} $

as before the null space is $span\{v_2=$$ \begin{pmatrix} -1 \\ -1 \\ 0 \\ \end{pmatrix}$ $v_2=$$ \begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix}\}$

for the second part our basis is $B=\{v_2,v_3,v_1\}$ but how do I continue from here ? how can I find $[T]_B$?

Thanks for any help and tips

EDIT-

My main question is part two, not on how to calculate eigenvectors and eigenvalues

I also tried $Tv_2=-2v_2=-2v_2+0v_3+0v_1$ where $-2$ is from the eigenvalue and $Tv_1=4v_1=0v_3+0v_2+4v_1$ for the same reasno but I got stuck at $Tv_3=Av_3$

Answer 1

Let $B=(A-\lambda_2 I)$, we know $Bv_2=0$ and the generalized eigenvector $v_3$ is obtained as $Bv_3=v_2$. Let $S=[v_1, v_2, v_3]$, we get $S^{-1}AS=J$ where $J$ is a matrix in the Jordan form.

In details, we start with $Bv_3=v_2$, we get $$ \begin{bmatrix} -1& 1&-1&-1 \\ -7& 7& -1&-1\\ -6& 6& 0&0\end{bmatrix} $$ The reduced row echelon form of the above augmented matrix is

$$ \begin{bmatrix} 1& -1&0&0 \\ 0& 0& 1&1\\ 0& 0& 0&0\end{bmatrix} $$

From the above, we have the following equations $$ \begin{align} x_1 - x_2 &= 0, \\ x_3 = 1. \end{align} $$

Let $x_2=t$, we get $x_1=t,x_3=1$, the generalized vector is $v_3=[1,1,1]^T$ (i.e. not unique). Let $S=[v_1,v_2,v_3]$, we get

$$ S^{-1}AS= \underbrace{\begin{bmatrix} 4& 0&0 \\ 0& -2& 1\\ 0& 0& -2\end{bmatrix}}_{J}. $$