There's such a problem about finding the transition matrix.
Let $\mathbf A=\begin{bmatrix}2&6&-15\\1&1&-5\\1&2&-6\end{bmatrix}$, find non-singular matrix $\bf P$ such that $\mathbf P^{-1}\mathbf{AP}$ is the Jordan form of $\bf A$.
I can understand the fist few steps.
Jordan form $\mathbf J=\begin{bmatrix}-1&0&0\\0&-1&1\\0&0&-1\end{bmatrix}$. Let $\mathbf P=(\mathbf\alpha_1,\mathbf\alpha_2,\mathbf\alpha_3)$, then $\mathbf J=\mathbf P^{-1}\bf AP$ gives $\mathbf{AP}=\bf PJ$.
The following part is where I can't understand. Somehow we obtained the equations $$(\mathbf A+\mathbf I_3)\mathbf\alpha_1=0,~(\mathbf A+\mathbf I_3)\mathbf\alpha_2=0,~(\mathbf A+\mathbf I_3)\mathbf\alpha_3=\mathbf\alpha_2.$$ Where do these come from?
There is a method that is helpful when the matrix entries and the eigenvalues are all integers. The relations you put at the end are built in. If there are to be fractions, these come from the determinant of $P$ at the end; almost everything is thus integers, reducing errors.
Anyway, you have characteristic polynomial $(x+1)^3 $ with minimal polynomial $(x+1)^2.$ Therefore the largest Jordan block is size two. We begin with this. We need to choose some $(A+I)^2 \alpha_3 = 0, $ which is always true, but then we need $(A+I) \alpha_3 \neq 0. $ Looking at $A+I$ I felt the numbers would stay small if I chose $\alpha_3 =(1,0,0)^T,$ meaning the column vector that is the transpose of the row I typed. We define $\alpha_2 = (A+I) \alpha_3$ which came out $(3,1,1)^T.$ This is automatically an eigenvector because $(A+I) \alpha_2 = (A+I)^2 \alpha_3.$
We are not quite done. We need the smaller Jordan block, meaning we need $\alpha_1$ to be an eigenvector that is independent of $\alpha_2.$ When I just looked for eigenvectors, I got (basis) $(5,0,1)^T $ and $(-2,1,0) .$ I decided to take $\alpha_1 = (-2,1,0) .$
That's it. The way Jordan form is done in the U.S. has the extra diagonal just above the main diagonal. So, you make the matrix $P$ in order $(\alpha_1, \alpha_2, \alpha_3)$ as the columns. You then need the determinant of $P$ to find $P^{-1}$ from the adjoint matrix of $P.$ Then, confirm that $P^{-1} P = $ and calculate $J = P^{-1} A P$
This is a valuable skill, ought to be practiced for its own sake