Consider the following matrix:
$$
\left[ \begin{array}{ccc}
2 & 0 & 2 \\
0 & -2 & 0 \\
2 & 0 & -1 \end{array} \right]
$$
What would be the standard way to find the eigenvectors for this matrix? since it's Hermitian they must be orthogonal.
By inspection it can be found that the eigenvalue is $-2$ for $[0, 1, 0]^T$ and then $\left[ \begin{array}{ccc}
2 & 2 \\
2 & -1 \end{array} \right]$ can be analyzed further to find that of the other two eigenvalues, of which one is $-2$, one results in a degeneracy. Since we already know an eigenvector we can use the scalar product to find another orthogonal to it and thus create are basis.
But how could you do this if you don't see the eigenvector by inspection and just calculate the eigenvalues blindly. I just get $2x=-z$ as the plane of solutions, how do I proceed?
Best Answer
Solving directly for $\lambda = 3$, I get: $\left[ \begin{array}{ccc} 2&0&2\\ 0&-2&0\\ 2&0&-1\\ \end{array} \right] \left[ \begin{array}{ccc} x\\ y\\ z\\ \end{array} \right] = 3 \left[ \begin{array}{ccc} x\\ y\\ z\\ \end{array} \right]$.
By solving for $x, y, z$, we get $y=0$ and $2z=x$. These equations define a line, not plane, in $\mathbb{R}^3$. As such, the eigenvector for $3$ is given by any choice of $x$, such as $x=2$ which gives $\left[ \begin{array}{ccc} 2\\ 0\\ 1\\ \end{array} \right]$. You can check that it is indeed an eigenvector for $\lambda =3$.