Calculating the null space of a $n^{\text{th}}$ power of a $2\times 2$ matrix with an unkown variable. $$
    \left(\begin{matrix}
    0 & 1 \\
    x & 1 \\
   \end{matrix}\right)
$$ 
 Suppose I have a matrix above to the power of n, is it possible to use eigendecomposition in order to find all the eigenvectors in terms of x?
 A: Hint

$$A^n = P D^n P^{-1}$$ where $P$ is the matrix of the eigenvectors and $D$ the matrix of the eigenvalues

A: Hint: An eigenvector of $A$ for eigenvalue $\lambda$ is an eigenvector of $A^n$ for eigenvalue $\lambda^n$.
The process of finding eigenvalues and eigenvectors is pretty much the same with a symbolic parameter $x$.
The only tricky part is if $x = -1/4$.
A: A more general result is true. It is a well known result that if $A \in M_n(\mathbb{C})$, there exists $P \in GL_n(\mathbb{C})$ such that $PAP^{-1}$ is upper triangular. In particular, all eigenvalues of $A$ are in this matrices' diagonal, counted with multiplicity (if not inmediately clear, recall that eigenvalues are preserved by conjugation, and write the expression of a characteristic polynomial for a general upper triangular matrix). 
Now, if $A, B \in M_n(\mathbb{C})$ are upper triangular and $i \in [n]$, 
$$
(\lambda A)_{ii} = \lambda A_{ii}
$$
and 
$$
(AB)_{ii} = \sum_{m= 1}^ma_{im}b_{mi} = a_{ii}b_{ii}
$$
so if $q \in \mathbb{C}[X]$, then $q(A)_{ii} = q(A_{ii})$ for all $i$ (again, induction should do the trick). Moreover, if $q = \sum_{j = 1}^{d}c_jX^{j}$, then
$$
q(PAP^{-1}) = \sum_{j = 1}^{d}c_j(PAP^{-1})^{j} = \sum_{j = 1}^{d}c_jPA^jP^{-1} = P\left(\sum_{j = 1}^{d}c_jA^j\right)P^{-1} = Pq(A)P^{-1}  
$$
To sum up, since
$$
q(A) = q(P^{-1}PAP^{-1}P) = P^{-1}q(PAP^{-1})P
$$
the eigenvalues of $q(A)$ are the same of those of $q(PAP^{-1})$. Since the eigenavalues of $q(PAP^{-1})$ are the eigenvalues of $PAP^{-1}$ via $q$ and $PAP^{-1}$ has the same eigenvalues than $A$, 
$$
\operatorname{Spec}(q(A)) = q(\operatorname{Spec}(A)).
$$
In your particular case, $q = X^{n}$. Thus, it will suffice to find the eigenvalues of $A$ and take their $n$-th power. 
