I have a matrix $B:= \begin{bmatrix}0 & 1\\-1 & -\lambda\end{bmatrix} $
I need to diagonalise it and work out the transition matrix.
I have worked out that the eigenvalues are $ \mu_± = \frac{-\lambda ± \sqrt{\lambda^2-4}}{2} $ hence the diagonal matrix is $\begin{bmatrix}\frac{-\lambda + \sqrt{\lambda^2-4}}{2} & 0\\0 & \frac{-\lambda - \sqrt{\lambda^2-4}}{2}\end{bmatrix} $
I cant seem to work out the eigenvectors for the corresponding eigenvalues, therefore cant construct the transition matrix.
We can write $$ B - \mu_+ I = \pmatrix{-\mu_+ & 1\\-1&1/\mu_+}\\ B - \mu_- I = \pmatrix{-\mu_- & 1\\-1 & 1/\mu_-} $$ from there it, should be easy to determine the eigenvectors in terms of $\mu_{\pm}$.
The clever algebra I mentioned: for example, $$ - \mu_+ - \lambda = \\ \frac{\lambda - \sqrt{\lambda^2 - 4}}{2} - \lambda = \\ \frac{-\lambda - \sqrt{\lambda^2 - 4}}{2} = \mu_- = \\ \frac{-\lambda - \sqrt{\lambda^2 - 4}}{2} \frac{-\lambda + \sqrt{\lambda^2 - 4}}{-\lambda + \sqrt{\lambda^2 - 4}} =\\ \frac 12 \frac{\lambda^2 - (\sqrt{\lambda^2 - 4})^2}{-\lambda + \sqrt{\lambda^2 - 4}} = \\ \frac{2}{-\lambda + \sqrt{\lambda^2 - 4}} = 1/\mu_+ $$