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I have this matrix $A=\left[ {\begin{array}{ccc} -3&3&-2\\ -7&6&-3\\ 1&-1&2\\ \end{array} } \right]$

I computed the characteristic polynomial $C_A(t)=-(t-2)^2(t-1)$

When I go to try and find the eigenvectors and generalized eigenvectors I compute $A-2I=\left[ {\begin{array}{ccc} -5&3&-2\\ -7&4&-3\\ 1&-1&0\\ \end{array} } \right]$

Using row reduction $\left[ {\begin{array}{ccc} 1&-1&0\\ 0&1&1\\ 0&0&0\\ \end{array} } \right]x=0$ gives me eigenvectors having the form $e=r\left[ {\begin{array}{c} 1\\ 1\\ -1\\ \end{array} } \right]$

$v=\left[ {\begin{array}{c} 1\\ 1\\ -1\\ \end{array} } \right]$ Is the only eigenvector up to a constant. So I know there is 1 generalized eigenvector and I simply chose this eigenvector to be my initial vector in the cycle.

At this point I know the Jordan form is $J=\left[ {\begin{array}{ccc} 2&1&0\\ 0&2&0\\ 0&0&1\\ \end{array} } \right]$

Now to find the generalized eigenvector I can take an augmented matrix $\left(\begin{array}{ccc|c} 5&3&-2&1\\ -7&4&1&1\\ 1&-1&0&-1 \end{array}\right)$ I row reduced this to the form $\left(\begin{array}{ccc|c} 1 & -1 & 0&-1\\ 0 & 1 & 1&2\\ 0&0&0&0 \end{array}\right)$

So I need a vector that satisfies $x_1-x_2=-1$ and $x_2+x_3=2$ I let $x_3=1$

Which gives me a vector $x=\left(\begin{array}{c} 0\\ 1\\ 1 \end{array}\right)$ which seems to work, so I have a basis for the generalized eigenspace for $\lambda=2$

then I need to find a solution for $(A-I)x=0$ for $\lambda=1$.

I row reduce $A-I=\left[\begin{array}{ccc} -4&3&-2\\ -7&5&1\\ 1&-1&1 \end{array}\right]$ to get $\left[\begin{array}{ccc} 1&0&-1\\ 0&1&-2\\ 0&0&0 \end{array}\right]$ and using $(A-I)x=0$ found that all eigenvectors for $\lambda=1$ have the form $v=s\left(\begin{array}{c} 1\\ 2\\ 1 \end{array}\right)$

So I can make a matrix $Q=\left[\begin{array}{ccc} 1&1&0\\ 2&1&1\\ 1&-1&1 \end{array}\right]$ s.t $A=QJQ^{-1}$ where the columns of $Q$ are a jordan basis for $J$

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    $\begingroup$ I stopped reading at the first unclear point: given the characteristic polynomial, how can you be sure that the Jordan normal form is what you have provided? Can't it be the diagonal matrix with $2,2,1$ in the diagonal? $\endgroup$ – A. Pongrácz Dec 8 '18 at 17:38
  • $\begingroup$ Yes I fixed this. I dont know that until I know the geometric multiplicity of that eigenvalue is 1. $\endgroup$ – AColoredReptile Dec 8 '18 at 17:42
  • $\begingroup$ Which you can figure out by solving the system of linear equations $(A-2I)x=0$. Try that! (Or compute the rank of $A-2I$ at least, that also helps.) $\endgroup$ – A. Pongrácz Dec 8 '18 at 17:48
  • $\begingroup$ I found an eigenvector. My problem is finding the generalized eigenvector. I tried to use this eigenvector I found , $v$ and then compute $(A-2I)(x)=v$. $\endgroup$ – AColoredReptile Dec 8 '18 at 17:51
  • $\begingroup$ You are not focusing on the main point of my comment: compute the rank, and if it is $2$, you need to find $2$ independent eigenvectors, not just one. In that case, the Jordan normal form is diagonal. (You still seem to work under the assumption that the Jordan form has a $2\times 2$ block, which it might not...) $\endgroup$ – A. Pongrácz Dec 8 '18 at 17:54
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Because you are looking for a Jordan canonical basis, $Ax = 2x+v$, where $x$ is the generalized eigenvector you're looking for. Equivalently, $(A-2I)x = v$. Applying $A-2I$ one more time to each side will give you the following statement: $(A-2I)^{2}x = 0$.

So, your generalized eigenvector is in the nullspace of $(A-2I)^2$, and it is not in the nullspace of $(A-2I)$. Can you finish from there?

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  • $\begingroup$ I believe I finished. $\endgroup$ – AColoredReptile Dec 8 '18 at 18:31

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