Parameters such that the matrix is diagonizable Given the matrix 
$$A= \begin{pmatrix}
1 & 0 & 0\\
k & 2 &0 \\
 m& n & 1
\end{pmatrix}$$
where $k, m, n \in \mathbb{R}$ find the values of $k, m, n $ such that the matrix is diagonizable. 
Solution
The eigenvalues of $A$ are the solutions of the equation:
$$\begin{vmatrix}
1-\ell & 0 &0 \\
 k& 2-\ell &0 \\
 m& n &1-\ell
\end{vmatrix}=0 \Leftrightarrow  \left ( 1-\ell \right )^2 \left ( 2-\ell \right )=0 $$
meaning that the eigenvalues of $A$ are $\ell_1=1$ (double) and $\ell_2=2$ (simple).
In order the matrix to be diagonizable we must have that $\dim \mathcal{V}(1)=2$. Also from the equation of the dimension and rank we have:
$$\dim \mathcal{V}(1)+{\rm r}(A-\mathbb{I}_3)=3$$
implying that the rank of the $A-\mathbb{I}_3$ is $1$. Now on to find $\mathcal{V}(1)$. In order to do so we have to solve the system $(A-\mathbb{I}_3)X=0$ which is easily reduced down to the system:
$$\left\{\begin{matrix}
kx+y &=0 \\ 
 mx+ ny&=0 
\end{matrix}\right.$$
And I got stuck at this point. I don't get enough equations so that I can actually determine the values of the parameters. Any help?

The system of equations that define the kernel of $A-I$ must have rank $1$ for $\ker A-I$ to have dimension 2, because of the rank-nullity theorem. This means the condition is  the vectors $(k,1)$ and $(m,n)$ must be colinear. Explicitly:
$$kn=m.$$
You should be able to check that, given that relation, the minimal polynomial of $A$ is $(x-1)(x-2)$.
