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?

What is important is that when looking for eigenvectors for the $l=1$ eigenvalue, the equation system is not degenerate. If it is degenerate means that at least one vector is not a true eigenvector but is a generalized eigenvector residing in the same generalized eigenspace as the true one. And then diagonalization is not possible.
The necessary and sufficient condition is thus that your last equation system be not degenerate. row1: $[k,1]$ can not be allowed to be a scalar multiple of $[m,n]$
row operation removing $n$ times first row from second row gives $[m-nk,0]$.
So the requirement is $m-nk \neq 0$
