How to calculate transition matrix to get Jordan form? Suppose $A$ is a linear transformation of a 3-dim vector space $V$, defined as
$$A(\epsilon_1,\epsilon_2,\epsilon_3)=(\epsilon_1,\epsilon_2,\epsilon_3)
\begin{pmatrix}
-10 & 12 & 7\\
-3 & 4 & 2\\
-13 & 15 & 9
\end{pmatrix},
$$here $\{\epsilon_i\}$ is a basis of $V$.
Is there a concise way to find the transition matrix to a new basis under which the linear operator $A$ has a matrix of Jordan form? And what's behind the solution?
 A: Here are a couple of observations: 


*

*The characteristic polynomial is $\lambda^3-3\lambda^2+3\lambda-1=0$ and therefore the only eigen value is $1$. 

*We observe that $A-I$ has rank $2$ and hence $\dim(\ker(A-I))=1$ from the Rank-Nullity theorem. 

*We know that the Jordan form for the matrix is $$J=\begin{bmatrix}1&1&0\\0&1&1\\0&0&1 \end{bmatrix}$$

*We are interested in finding a matrix $P$ such that $$P^{-1}AP=J=I+\begin{bmatrix} 0&1&0\\0&0&1\\0&0&0\end{bmatrix}$$
Setting $P=\begin{bmatrix} \biggl |& \biggl|&\biggl|\\x_1&x_2 &x_3\\\biggl|&\biggl|&\biggl|\end{bmatrix}$, and observing that, $AP=PJ$, can you solve the resulting system of vectors?
A couple of preliminaries: 


*

*Verify that $AP=\begin{bmatrix}\biggl| & \biggl| &\biggl|\\Ax_1& Ax_2 &Ax_3\\ \biggl| &\biggl| & \biggl|\end{bmatrix}$

*Verify that $PJ=\begin{bmatrix}\biggl| & \biggl| &\biggl|\\x_1& x_1+x_2 &x_2+x_3\\ \biggl| &\biggl| & \biggl|\end{bmatrix}$

The steps 1 and 2 were already known to the OP. I'll post more hints as I work out. 
