I have a $2\times 2$ real symmetric matrix: $$\begin{pmatrix} A & C \\ C & B \end{pmatrix} $$ and I know that the eigenvalues are: $$\lambda_{\pm} = \frac{1}{2}(A+B)\pm \frac{1}{2}\sqrt{(A-B)^{2}+C^{2}}$$ Define $x$ so that: $\lambda_{\pm} = e^{\pm x}$. I know its first eigenvector, associated to $\lambda_{+}$. It is given by: $$v = \cos \theta v_{1}+ \sin\theta v_{2}$$ where $v_{1}$ and $v_{2}$ form a basis for this two-dimensional space and the constant $\theta$ is defined by: $$\tan\theta = \frac{C}{e^{x}-A}.$$ Here is my problem. If I want to find an (orthogonal) eigenvector of the matrix associated to $\lambda_{-}$, I would simply say it is given by: $$w = \cos \Delta v_{1} + \sin\Delta v_{2}$$ where now: $$\tan\Delta = \frac{C}{e^{-x}-A}$$ However, the result is supposed to be: $$w = -\sin\theta v_{1} + \cos\theta v_{2}.$$
Why is that? I am trying to prove that the change $x \to -x$ implies $\cos\theta \to -\sin\theta$, but I really cannot prove it. Any help is useful!
Your $x$ doesn't exist unless $4AB-C^2=4$, which you obtain from the equality $\lambda_-=\frac1{\lambda_+}$. This just doesn't occur in general.
If instead of using your $e^{\pm x}$ you use $\lambda_{\pm}$, it is easy to check that $$ \frac C{\lambda_--A}=-\Big(\frac{C}{\lambda_+-A}\Big)^{-1}, $$ which is precisely the relation $\tan\theta=-\frac1{\tan\Delta}$ which guarantees orthogonality.
As for
the easier way to write a vector orthogonal to $$s\,v_1+t\,v_2,$$ if $v_1\perp v_2$, is $$-t\,v_1+s\,v_2.$$