Consider the matrix which for some $$\alpha,\beta \in \mathbb R$$ is given as $$A:=\begin{pmatrix} \alpha & \beta \\ \beta & -\alpha \end{pmatrix}.$$

This matrix is self-adjoint and correspondingly has real eigenvalues $$\pm \sqrt{\alpha^2+\beta^2}.$$

The eigenvectors are then given by

$$\left(\frac{\alpha \pm\sqrt{\alpha^2+\beta^2}}{\beta}, 1\right).$$

However, they are not orthonormal. Is there an easy way to get orthonormal eigenvectors?

• Gram-Schmidt would work. Commented Apr 12, 2021 at 17:10

The eigenvalues will be distinct if and only if $$\alpha,\beta$$ are not both equal to zero. When $$A$$ has distinct eigenvalues, the eigenvectors will automatically be orthogonal which means that it is sufficient to normalize the eigenvectors. We note that $$v = (\alpha \pm \sqrt{\alpha^2 + \beta^2}, \beta) \implies\\ \|v\| = (\alpha \pm \sqrt{\alpha^2 + \beta^2})^2 + \beta^2 = 2(\alpha^2 + \beta^2) \pm 2\alpha \sqrt{\alpha^2 + \beta^2}.$$ When $$\beta \neq 0$$, these we have $$\|v\| \neq 0$$. So, the normalized eigenvectors will be given by $$\frac v{\|v\|} = \left(\frac{\alpha \pm \sqrt{\alpha^2 + \beta^2}}{2(\alpha^2 + \beta^2) \pm 2\alpha \sqrt{\alpha^2 + \beta^2}}, \frac{\beta}{2(\alpha^2 + \beta^2) \pm 2\alpha \sqrt{\alpha^2 + \beta^2}} \right).$$ In the case that $$\beta = 0$$, the matrix is diagonal, which means that $$(1,0),(0,1)$$ is an orthonormal set of eigenvector.