Solution: We will find the eigenvalues of the given matrix. For that, we need to find the characteristic polynomial $p(\lambda )$ of $A$, which is given by \begin{align*} p(\lambda ) & = \det \left( A - \lambda I\right) \\ & = \det \begin{pmatrix} 1 - \alpha - \lambda & \alpha \\ -\alpha & 1 + \alpha - \lambda \\ \end{pmatrix} \\ & = (1- \alpha -\lambda ) (1 + \alpha -\lambda ) + \alpha ^2 \\ & = (1-\lambda )^2 - \alpha ^2 + \alpha ^2 \\ & = (1-\lambda )^2. \end{align*} Therefore, $\lambda =1$ is an eigenvalue of $A$ with algebraic multiplicity $2$.

Now we will find the eigenvector corresponding to the eigenvalue $\lambda =1$. Note that \begin{align*} A - I = \begin{pmatrix} -\alpha & \alpha \\ -\alpha & \alpha \\ \end{pmatrix} \xrightarrow{R_2 = R_2 - R_1} \begin{pmatrix} -\alpha & \alpha \\ 0 & 0 \\ \end{pmatrix}. \end{align*} If $\alpha \neq 0$, then one of the eigenvector will be $\begin{pmatrix} 1 \\ 1 \\ \end{pmatrix}$, and hence the dimension of the eigenspace is $1$ which is less than the algebraic multiplicity. Hence, the matrix can not be diagonalizable.

If $\alpha =0$, then \[ A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}, \] which is a diagonal matrix and hence diagonalizable. Therefore, we proved that the given matrix $A$ is diagonalizable if and only if $\alpha = 0$.