let $A=\begin{pmatrix} 1 &a &a \\ a&2 &0 \\ 1&0 &2 \end{pmatrix}$ $a \in\mathbb{R}$ find for what values of $a$ is diagonalizable
Solution. The characteristic polynomial is $ (2-t)(t^2-3t-a^2-a+2)$
it is enough to show that $t^2-3t-a^2-a+2$ has roots in $\mathbb{R}$
it has to be true that
$9-4(-a^2-a+2)\geq 0\Leftrightarrow a^2+a+1/4 \geq 0$ wich is true because $a^2+a+1/4=(a+1/2)^2\geq 0$
That means for every $a\in\mathbb{R}$ the matrix $A$ is diagonalizable.
I feel like I have miss something here, is this correct?