Just to note, I haven't yet studied Jordan Normal Form or similar. I am working through a question that contains the following:
Say we have $$A=\begin{pmatrix} 0&1&0\\ 0&0&1\\ 1&0&0 \end{pmatrix}.$$ Its only eigenvalue is $1$ with eigenvector $v_1=(1,1,1)^T$, so it is not diagonalisable. Also the subspace $X=\{(x,y,z)^T:x+y+z=0\}$ is 2-dimensional, with basis $\{x_1,x_2\}$, and is $A$-invariant.
Why is it the case then that if we define $$P=\begin{pmatrix} |&|&|\\ v_1&x_1&x_2\\ |&|&| \end{pmatrix}$$ we have $$P^{-1}AP= \begin{pmatrix} 1&0&0\\0&a&b\\0&c&d \end{pmatrix},$$ without the need to complete any calculation? It is clear that the first column would be $(1,0,0)^T$ but I do not see how to rigorously determine that the top row must have the two zeroes.