Let $$A=\left(\begin{matrix}4&-5&2 \\ 5&-7&3\\ 6&-9&4 \end{matrix}\right)$$ And I found B, A's Jordan form to be: $$B=\left(\begin{matrix}0&1&0 \\ 0&0&0\\ 0&0&1 \end{matrix}\right)$$, so that there is invertible matrix S, so that $S^{-1}AS=B$.
How do I find S?
Thanks in advance!!!
There are a number of methods...perhaps the simplest is to take the generalized eigenspace $\text{Null}((A-\lambda I)^d)$ and pick a random vector $v$ in the space. It will almost certainly be in $\text{Null}((A-\lambda I)^d) \setminus \text{Null}((A-\lambda I)^{d-1})$. Then $\{v, (A-\lambda I)v, \dotsc, (A-\lambda I)^{d-1}v\}$ are linearly independent (this requires a short proof). There's your basis for that particular Jordan block. Repeat for each block.