Let $\alpha$ be a real parameter
We have the following matrix
$A_\alpha = \begin{bmatrix} 1& 0 & 0 \\ \alpha & -1 & -2\\ 0 & 0 & 1 \\ \end{bmatrix}$
1- Calculate the characteristic polynomial
2- Give a necessary and sufficient condition on the parameter $\alpha$ so that the matrix $A_\alpha$ accepts a diagonal form.
I have calculated the characteristic polynomial where
$$P(X) = det(A_\alpha - X \cdot Id_3)$$
$$P(X) = -(1-X)^2 \cdot(1+X)$$
The parameter $\alpha$ is not involved in the characteristic polynomial, there is no way to include it in the dimensions of the eigenspaces.
I know the fact that a matrix is diagonalizable if and only if it is symmetric.
There is no way in which the parameter $\alpha$ can make the matrix a symmetric matrix.
Providing that $\alpha$ took a value of 1 to find the diagonal matrix in the rest of the exercise, are there any other criteria on which we can base to define the set of values that $\alpha$ can take in order for the matrix to be diagonalizable.
The matrix $A_\alpha$ is always diagonalizable. If $\alpha\ne0$, then$$\{(0,1,0),(2,0,\alpha),(2,\alpha,0)\}\tag1$$is a basis of eigenvectors. This fails if $\alpha=0$, of course, since, in that case, the set $(1)$ is no longer linearly independent. But then you can take $\{(0,1,0),(0,-1,1),(1,0,0)\}$.