# Matrices, Transition matrix

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.

• Do you really need the eigenvectors to get the transition matrix? – Jef L Nov 26 '14 at 22:44
• @JefLaga Isn't the transition matrix constructed from the eigenvectors? – mike Nov 26 '14 at 22:45
• Note that $\lambda = \pm 2$, the matrix is not diagonalizable. – Omnomnomnom Nov 27 '14 at 0:39

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, $$- \mu_+ - \lambda = \\ \frac{\lambda - \sqrt{\lambda^2 - 4}}{2} - \lambda = \\ \frac{-\lambda - \sqrt{\lambda^2 - 4}}{2} = \mu_- = \\ \frac{-\lambda - \sqrt{\lambda^2 - 4}}{2} \frac{-\lambda + \sqrt{\lambda^2 - 4}}{-\lambda + \sqrt{\lambda^2 - 4}} =\\ \frac 12 \frac{\lambda^2 - (\sqrt{\lambda^2 - 4})^2}{-\lambda + \sqrt{\lambda^2 - 4}} = \\ \frac{2}{-\lambda + \sqrt{\lambda^2 - 4}} = 1/\mu_+$$
• on the RHS, in the bottom right of each matrix, how did you get that value, as i got $-\lambda ± \mu_±$ – mike Nov 26 '14 at 23:31
• Thank you that makes full sense, however, how do i use this to solve for eigenvectors, i keep spinning around a circle and getting $x=x$ where $x$ corresponds to the first value in the eigenvector – mike Nov 27 '14 at 1:23
• Not sure where you're having trouble. However, try $$\pmatrix{1\\ \mu_+}$$ – Omnomnomnom Nov 27 '14 at 1:28