For wich values of $\alpha \in \mathbb{R}$ does the matrix $A$ is diagonalizable: $$A= \left( \begin{array}{ccc} 1+\alpha & -\alpha & \alpha \\ 2+\alpha & -\alpha & \alpha-1 \\ 2 & -1 & 0 \end{array} \right)$$
I obtained this caracteristic polynomtal $$ -\lambda ^3 + \lambda ^2+ \lambda -1$$ and the eigenvalues are $$\lambda=1$$ and $$\lambda=-1$$ The eigenvectors are $N(A+I)=(0,0,1)^T$ and $N(A-I)=(2,3,1)^T$. Sorry for the misunderstands, I'm new here
$$A= \left( \begin{array}{ccc} 1+\alpha & -\alpha & \alpha \\ 2+\alpha & -\alpha & \alpha-1 \\ 2 & -1 & 0 \end{array} \right)$$
$$A - \lambda I= \left( \begin{array}{ccc} 1+\alpha -\lambda& -\alpha & \alpha \\ 2+\alpha & -\alpha - \lambda & \alpha-1 \\ 2 & -1 & -\lambda \end{array} \right)$$
The characteristic polynomial is $$ -\lambda^3+\lambda^2+\lambda -1 = -(\lambda + 1)(\lambda-1)^2.$$
And the eigenvalues are $\lambda_1 = 1$ and $ \lambda_2 = -1$
$$A - I= \left( \begin{array}{ccc} \alpha & -\alpha & \alpha \\ 2+\alpha & -\alpha - 1 & \alpha-1 \\ 2 & -1 & -1 \end{array} \right).$$
The corresponding eigenspace $ V_1 = \ker(A- I) = \{v = (v_1,v_2,v_3)^T \in \mathbb{R}^3: (A-I)v = 0 \}$ is
if $\alpha = 0$ : $$ \ker(A- I) = \{ (v_1,2v_1 - v_3,v_3) \vert v_1,v_3 \in \mathbb{R}\} = < (1,2,0) \; (0,-1,1)> $$
if $\alpha\neq 0:$
$$ \ker(A- I) = \{ (2v_3,3v_3,v_3) \vert v_3 \in \mathbb{R}\} = <(2,3,1)>$$
$$A + I= \left( \begin{array}{ccc} 2 + \alpha & -\alpha & \alpha \\ 2+\alpha & -\alpha +1 & \alpha-1 \\ 2 & -1 & 1 \end{array} \right).$$
The corresponding eigenspace $ V_2 = \ker(A + I) = \{v = (v_1,v_2,v_3)^T \in \mathbb{R}^3: (A+I)v = 0 \}$ is
$$ \ker(A+ I) = \{ (0,v_2,v_2) \vert v_2 \in \mathbb{R}\} = <(0,1,1)> .$$
$$ \ker(A+ I) = \{ (0,v_2,v_2) \vert v_2 \in \mathbb{R}\}.$$
If $ \alpha = 0$ then we have a basis of eigenvectors and A is diagonalisable if $\alpha \neq 0$ then $A$ is not diagonalisable.