For which $\beta \in \mathbb{C}$ is the matrix $A=\bigl(\begin{smallmatrix} 0&1\\1&\beta \end{smallmatrix}\bigr)$ diagonalisable?

Question

I have got a question refering to the following problem.

Let $K=\mathbb{C}$. For which $\beta \in \mathbb{C}$ is this matrix diagonalisable? $$A=\pmatrix{0&1\\1&\beta}$$

I think that it is not diagonalisable in any case, because the determinant of $A-xI_n$ is $-1$. Am I right?

Answer 1

The characteristic polynomial is $P(x)=x^2-\beta x-1$ whose discriminant is $\Delta=\beta^2+4$

If $\beta\neq\pm 2i$ The eigenvalues are distinct and the matrix is diagonalisable.

If $\beta=\pm 2i$ the root $2$ is double and the matrix is not diagonalisable.