I have the matrix A = $ \begin{bmatrix} 3 & 4 & 3 \\ -1 & 0 & -1 \\ 1 & 2 & 3 \end{bmatrix} $ and want to find a matrix S such that $SAS^{-1}$ is an upper triangular matrix.
I already found that $ x = 2 $ is the eigenvalue with multiplicity 3. Then I tried finding the (generalized) eigenvectors. I got:
$ (A-2E_{3})v_{1}=0 \rightarrow v_{1}= [1,-1,1]^{T} $
$(A-2E_{3})v_{2}=v_{1} \rightarrow v_{2}= [1,0,0]^{T} $
$(A-2E_{3})v_{3}=v_{2} \rightarrow v_{3}= [-2,1,0]^{T} $
Now I thought I just set $ S= [v_{1},v_{2},v_{3}]$, use this to find $S^{-1}$ and solve the equation $SAS^{-1}$. But in this case I don't get an upper triangular matrix/ Jordan normal form as a result. Is my whole approach wrong?
I'm just starting off with the topic JCF and I'm having my problems with that, as you can see. So I would be very grateful for any help! Thanks in advance