I need to show that $A$ is a diagonizable matrix: $$A \in M_{n\times n}(\mathbb{R}) = \begin{bmatrix}0 & a & b\\ a & 0 & b\\ b & a & 0\end{bmatrix}; \space a\neq b; \space a,b \neq 0$$
Since I can't find the eigenvectors I think the best way to approach this is to see if, with the help of the characteristic polynomial, I can prove that $A$ has three different eigenvalues.
$$|A - \lambda I_{n}| = 0 \Leftrightarrow$$ $$\Leftrightarrow -\lambda (\lambda^{2} - ab) -a (-a \lambda -b^{2}) + b(a^{2} + b\lambda) = 0 \Leftrightarrow$$ $$ \Leftrightarrow -\lambda^{3} + (ab + a^{2} + b^{2})\lambda + ab^{2} + a^{2}b = 0$$
But I didn't come to any conclusion... Maybe I'm missing on how to factorize this polynomial, or maybe there is a better approach to the problem... Can some one guide me?
The statement is wrong with $a = 1, b = -2$, whose Jordan form is \begin{align*} \begin{pmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{pmatrix}, \end{align*} hence it's not diagonalizable.