I have the eigenvector matrix like this
$\begin{bmatrix} 1 & 1 & 1 \\ 0 & \frac{-b + \sqrt{(b^{2} + 8a^2)}}{2a} & -\frac{b + \sqrt{(b^2 + 8a^2)}}{2a} \\ -1 & 1 & 1 \end{bmatrix}$
Are they orthogonal?
If not, should I use the Gram-Schmidt orthogonalization?
The matrix $\begin{bmatrix} 1 & 1 & 1 \\ 0 & \frac{-b + \sqrt{(b^{2} + 8a^2)}}{2a} & \frac{-b - \sqrt{(b^2 + 8a^2)}}{2a} \\ -1 & 1 & 1 \end{bmatrix}$ is orthogonal if, and only if, its column vectors $$ v_1=\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \quad v_2= \begin{bmatrix} 1 \\ \frac{-b + \sqrt{(b^{2} + 8a^2)}}{2a} \\ 1 \end{bmatrix} \quad v_3= \begin{bmatrix} 1 \\ \frac{-b -\sqrt{(b^2 + 8a^2)}}{2a} \\ 1 \end{bmatrix} $$ are orthogonal vectors. It is easy to verify that $$ \langle v_1, v_2\rangle =0 \quad \mbox{ and } \quad \langle v_1, v_3\rangle =0 . $$ A bit more laborious, but not difficult, is to calculate $v_2\cdot v_3$. In fact, $$ v_2\cdot v_3 = 1\cdot 1+ \left( \frac{-b + \sqrt{(b^{2} + 8a^2)}}{2a} \right)\cdot \left( \frac{-b - \sqrt{(b^{2} + 8a^2)}}{2a} \right) + 1\cdot 1 \\ = 1+ \frac{b^2 - (b^{2} + 8a^2)}{4a^2} + 1 \\ = 2+ \frac{- 8a^2}{4a^2} \\ =0 $$