I am given the matrix:
$\begin{pmatrix} a & b \\ b & -a \end{pmatrix}$
and I already calculated the eigenvalues $\lambda = \pm \sqrt{a^2+b^2}$. Now, I want to get the normalised eigenvectors, so that there norm is equal to 1. Unfortunately, I seem to be incapable of carrying out this cumbersome calculation. Does anybody know if there is an easy way to do this?
$$\lambda I-A=\begin{pmatrix}\lambda-a&-b\\-b&\lambda+a\end{pmatrix}$$
Substituting in the above and forming tha corresponding homogeneous system (and pay attention to the fact that only one equation is needed (why?)) when:
$$\lambda=\pm\sqrt{a^2+b^2}\;\;\implies\;\;\;(\pm\sqrt{a^2+b^2}-a)x-by=0\iff x=\frac b{\pm\sqrt{a^2+b^2}-a}y$$
Thus, we get two eigenvectors (one for each eigenvalues), for example
$$\binom{b}{\sqrt{a^2+b^2}-a}\;,\;\;\binom b{-\sqrt{a^2+b^2}-a}$$
Now, orthonormalize the above (why can you?) . Easier than this in the general case I can't do it.