How to diagonalize a matrix with a non-numeric element?

Question

For what $\alpha$ is the matrix A diagonalizable, where
$$ A =\begin{pmatrix} 1 & 0 & 0 & \\ \alpha-3 & 6-2\alpha & 6-3\alpha \\ 2-\alpha & 2\alpha -4 & 3\alpha-4 \\ \end{pmatrix} $$ How does one approach to these kind of matrices, as I get stuck at some useless polynomials?

Answer 1

You've asked in the comments how to find the eigenvalues, so I'll show that method here. \begin{align*} A &= \begin{pmatrix} 1 & 0 & 0 \\ \alpha - 3 & 6 - 2 \alpha & 6 - 3 \alpha \\ 2 - \alpha & 2 \alpha - 4 & 3 \alpha - 4 \end{pmatrix} \\ \implies \det(A - \lambda I) &= \left|\begin{matrix} 1 - \lambda & 0 & 0 \\ \alpha - 3 & 6 - 2\alpha - \lambda & 6 - 3\alpha \\ 2- \alpha & 2\alpha-4 & 3 \alpha - 4 - \lambda\end{matrix}\right| \\ &= (1-\lambda)\Big((6-2\alpha-\lambda)(3\alpha-4-\lambda)-(2\alpha-4)(6-3\alpha)\Big) \\ &= (1-\lambda)\Big(\lambda^2 -(6-2\alpha + 3\alpha-4)\lambda + (6-2\alpha)(3\alpha-4)-(2\alpha-4)(6-3\alpha)\Big) \\ &= (1-\lambda)\Big(\lambda^2 - (2+\alpha)\lambda + 2\alpha\Big) \\ &= (1-\lambda)(2-\lambda)(\alpha-\lambda). \end{align*} Thus the eigenvalues are $1$, $2$ and $\alpha$.

Diagonalisation is certainly possible if $\alpha \ne 1,2$ so that we have distinct eigenvalues, and you should check the cases of $\alpha = 1,2$ to see if you obtain linearly independent eigenvectors.