Zero as an Eigenvalue of stress tensor Given, $\sigma=\left[ \begin{array}{ccc}
0 & 0 & \alpha\beta^2 \\
0 & 0 & -\alpha\beta \\
\alpha\beta^2 & -\alpha\beta & 0 \end{array} \right]$, I am interested in the eigenvalues as well as eigenvectors of the stress tensor. $\lambda_1=0, \lambda_2=\alpha\beta\sqrt{1+\beta^2}, \lambda_3=-\alpha\beta\sqrt{1+\beta^2}$. But i need to construct a system of mutually orthogonal vectors using the eigenvectors. I know that if I have $\vec{x_1}, \vec{x_1}, \vec{x_2}$, then $\vec{x_1}, \vec{x_2}, \vec{x_1}\times \vec{x_2}$ form a set of mutually orthogonal vectors. If $\vec{x_1}, \vec{x_1}, \vec{x_1}$, then $\vec{x_1}, \vec{x_2}, \vec{x_3}$ form a set of mutually orthogonal vectors, where $\vec{x_2}, \vec{x_3}$ are any orthogonal vectors lying in any plane perpendicular to $\vec{x_1}$. But how to find the eigenvectors and a set of mutually orthogonal vectors in the present case?
 A: $\sigma$ is symmetric, hence eigenvectors corresponding to distinct eigenvalues are automatically orthogonal, hence you have little choice in the matter other
than scaling.
Assuming that $\alpha, \beta \neq 0$, then just by inspection we can find a solution to $\sigma x = 0$, for example $x=(\alpha \beta, \alpha \beta^2,0)^T \in \ker \sigma$ (that is corresponding to the zero eigenvalue).
A: Assuming $\alpha\beta\ne 0$ we have:
$$\left( \begin{array}{ccc}
0 & 0 & \alpha\beta^2 \\
0 & 0 & -\alpha\beta \\
\alpha\beta^2 & -\alpha\beta & 0 \end{array} \right)\left( \begin{array}{c}
 \frac{1}{\beta} \\
 1 \\
0 \end{array} \right) =0 \left( \begin{array}{c}
 \frac{1}{\beta} \\
 1 \\
0 \end{array} \right)$$
$$\left( \begin{array}{ccc}
0 & 0 & \alpha\beta^2 \\
0 & 0 & -\alpha\beta \\
\alpha\beta^2 & -\alpha\beta & 1 \end{array} \right)\left( \begin{array}{c}
 \frac{-\beta}{\sqrt{1+\beta^2}} \\
 \frac{1}{\sqrt{1+\beta^2}} \\
1 \end{array} \right) =-\alpha \beta \sqrt{1+\beta^2} \left( \begin{array}{c}
 \frac{-\beta}{\sqrt{1+\beta^2}} \\
 \frac{1}{\sqrt{1+\beta^2}} \\
1 \end{array} \right)$$
$$\left( \begin{array}{ccc}
0 & 0 & \alpha\beta^2 \\
0 & 0 & -\alpha\beta \\
\alpha\beta^2 & -\alpha\beta & 0 \end{array} \right)\left( \begin{array}{c}
 \frac{\beta}{\sqrt{1+\beta^2}} \\
 \frac{-1}{\sqrt{1+\beta^2}} \\
1 \end{array} \right) =\alpha \beta \sqrt{1+\beta^2} \left( \begin{array}{c}
 \frac{\beta}{\sqrt{1+\beta^2}} \\
 \frac{-1}{\sqrt{1+\beta^2}} \\
1 \end{array} \right)$$
If $\alpha\beta=0$ the matrix is the zero matrix. In such a case any vector is an eigenvector with corresponding eigenvalue zero.
