For Which Value The Matrix is Diagonalizable? 
For which values of $a$ the matrix $\left(\begin{array}{ccc} 2 & 0 & 0 \\ 2 & 2 & a \\ 2 & 2 & 2 \end{array}\right)$ is diagonalizable:

*

*above $\mathbb{R}$

*above $\mathbb{C}$

We need to look at the characteristic polynomial which is $(x-2)^3-2a(x-2)=x^3-6x^2+2x(6-a)+4(a-2)$ , How do I find the eigenvalues one is $2$ and the other? and how so I continue?

$x^3-6x^2+2x(6-a)+4(a-2)$. Clearly it shows that  $x=2$ is a zero of the polynomial. Now , 
$x^3-6x^2+2x(6-a)+4(a-2)=0$
$\implies (x-2)(x^2-4x+4-2a)=0$
$\implies x=2 , x=\frac{4\pm\sqrt{16-4(4-2a)}}{2}=2\pm\sqrt{2a}.$
If $a=0$ then , all roots are $2$ and then the matrix is NOT diagonalizable(check!)
If $a\not=2$ then the matrix is diagonalizable. 
