As I just posted before on Lost on rational and Jordan forms, I'm practising with jordan forms, rational canonical forms...
Well, now I'm stuck in a problem because I don't know where I'm wrong.
I have this matrix:
$$M=\begin{bmatrix} 4 & -1 & 1 & -1\\ 1 & 3 & 0 & 0\\ 0 & 1 & 2 & 1\\ 0 & 0 & 0 & 3\\ \end{bmatrix}$$
and I have to find a matrix $P$ such that $P^{-1}MP=C$ where $C$ represents the primary rational canonical form of M.
I've computed the characteristic polynomial ($(x-3)^4$) and the minimal polynomial ($(x-3)^3$). Therefore, the invariants factors and the elementary divisors are the same ($(x-3)^3$ and $(x-3)$) and the primary rational canonical form is the same as the rational canonical form, which is:
$$C=\begin{bmatrix} 3 & 0 & 0 & 0\\ 0 & 0 & 0 & 27\\ 0 & 1 & 0 & -27\\ 0 & 0 & 1 & 9\\ \end{bmatrix}$$
since $(x-3)^3=x^3-9x^2+27x-27$.
Now, in order to get $P$, I just need a basis which is going to be formed by $\{u,v,Mv,M^2v\}$ where $u$ is a vector such that $(3I-M)u=0$ and $v$ is a vector such that $(3I-M)u\neq0$, $(3I-M)^2u\neq0$ and linearly independent from $u$.
So, in this conditions, I chose $u=(0,1,1,0)$ and $v=(1,1,0,1)$. Then, my basis is $\{(0,1,1,0),(1,1,0,1),(2,4,2,3),(3,14,11,9)\}$. Finally:
$$P=\begin{bmatrix} 0 & 1 & 2 & 3\\ 1 & 1 & 4 & 14\\ 1 & 0 & 2 & 11\\ 0 & 1 & 3 & 9\\ \end{bmatrix}$$
However, $P^{-1}MP\neq C$, as you can check here.
I hope you can help me. Thanks in advance.