Finding vectors which satisfy the cyclic decomposition theorem . Let $f: \Bbb{R}^3 \to \Bbb{R}^3$ be a linear operator defined by $$f(x,y,z)=(3x-4y-4z,-x+3y+2z,2x-4y-3z)$$
Find non-zero vectors $v_1, \ldots,v_n$ satisfying the cyclic decomposition theorem.(i.e $\Bbb{R}^3=W_0 \bigoplus z(v_1,f) \bigoplus \ldots z(v_n,f)$ ) 
I found that $f$ is non diagonalizable so I cannot use the fact that $\pi_f(x)=P_{v_1}(x)$ where $\pi_f(x)$ is the minimal polynomial of $f$ and $P_{v_1}(x)$ is the $f$-annihilator of $v_1$ . The characteristic polynomial of $f$ is $-(x-1)^3$. 
I don't have any idea on how to continue, how do I determine the vectors?
 A: Let $I$ denotes the identity operator. The only eigenvalue of $f$ is $\lambda=1$. The corresponding eigenspace has dimension $2$. Eigenvectors of $f$ are of the form $u=(2y+2z,y,z)^T$. For these eigenvectors, you will find that the equation $(f-\lambda I)(w)=u$ is solvable only when $u$ is a multiple of some $u_2\neq0$. Let $(f-\lambda I)(w_2)=u_2$ and $u_1$ be an eigenvector that is independent of $u_2$. Now $\mathbb{R}^3=\{0\}\oplus Z(u_1;f) \oplus Z(w_2;f)$ is a cyclic decomposition.
A: Let's rename the linear operator as $T$ and the non-zero vectors corresponding to the cyclic decomposition as $x^1,x^2,\ldots,x^r$(force of habit :)).
We'll take $W_0 = \{0\}$(with this, we will always have $p_1$(the $T$-annihilator of $x^1$)$=p$(the minimal polynomial), which greatly helps our calculation).
You've determined that the characteristic polynomial, $f=(t-1)^3$(I prefer to keep the leading coefficient as 1). Verify that the minimal polynomial is $p=(t-1)^2=p_1$.
We have $\deg(p_1)=\dim(Z(x^1;T))=2$, and hence $\dim(Z(x^2;T))=1$. We have no more cyclic factors of $x^i$ in our cyclic decomposition.
Since $\dim(Z(x^2;T))=1$, $x^2$ must be an eigenvector of $T$.
Verify that the only two linearly independent eigenvectors of $T$(which correspond to the eigenvalue $1$) are $v^1=\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$ and $v^2=\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$. Both of these are permitted vectors for $x^2$. We'll take $\boxed{x^2=\begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}}$
Let's look at how we can determine $x^1$. The following must be true regarding $x^1$:
(i). It must be linearly independent of $x^2$
(ii). It must generate a cyclic subspace of dimension precisely $2$, i.e. $Z(x^1;T)=span(\{x^1,T(x^1)\})$, in particular, $T^2(x^1)$ must be a linear combination of $x^1$ and $T(x^1)$(and consequently higher composition orders of $T$ as well).
(iii). We have $p_1=p=(t-1)^2$, i.e. $(T-2I)^2$ must be the $T$-annihilator of $x^1$.
Verify that $\boxed{x^1=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}}$ is allowed.
Note: $x^1,x^2$ are certainly not unique, there can be many(even infinitely many) possible candidates for them. Only the number of cyclic factors and $T$-annihilators are unique(and hence the later is also referred to as invariant factors).
