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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.

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  • $\begingroup$ Do you really need the eigenvectors to get the transition matrix? $\endgroup$ – Jef L Nov 26 '14 at 22:44
  • $\begingroup$ @JefLaga Isn't the transition matrix constructed from the eigenvectors? $\endgroup$ – mike Nov 26 '14 at 22:45
  • $\begingroup$ Note that $\lambda = \pm 2$, the matrix is not diagonalizable. $\endgroup$ – Omnomnomnom Nov 27 '14 at 0:39
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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, $$ - \mu_+ - \lambda = \\ \frac{\lambda - \sqrt{\lambda^2 - 4}}{2} - \lambda = \\ \frac{-\lambda - \sqrt{\lambda^2 - 4}}{2} = \mu_- = \\ \frac{-\lambda - \sqrt{\lambda^2 - 4}}{2} \frac{-\lambda + \sqrt{\lambda^2 - 4}}{-\lambda + \sqrt{\lambda^2 - 4}} =\\ \frac 12 \frac{\lambda^2 - (\sqrt{\lambda^2 - 4})^2}{-\lambda + \sqrt{\lambda^2 - 4}} = \\ \frac{2}{-\lambda + \sqrt{\lambda^2 - 4}} = 1/\mu_+ $$

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  • $\begingroup$ on the RHS, in the bottom right of each matrix, how did you get that value, as i got $-\lambda ± \mu_± $ $\endgroup$ – mike Nov 26 '14 at 23:31
  • $\begingroup$ You can simplify the expression with some clever algebra. The easier method, however, is to remember that the matrix has to be singular, which in this case implies that one row is a multiple of the other. $\endgroup$ – Omnomnomnom Nov 26 '14 at 23:44
  • $\begingroup$ could you possibly show the clever algebra, as i attempted it several times and couldn't work it out $\endgroup$ – mike Nov 26 '14 at 23:57
  • $\begingroup$ 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 $\endgroup$ – mike Nov 27 '14 at 1:23
  • $\begingroup$ Not sure where you're having trouble. However, try $$\pmatrix{1\\ \mu_+}$$ $\endgroup$ – Omnomnomnom Nov 27 '14 at 1:28

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