0
$\begingroup$

I want to find the orthogonal matrix $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ which diagonalises the matrix $\begin{pmatrix} 0 & m\\ m & M \end{pmatrix}$.

The eigenvalues are easily found to be $\lambda = \frac{M}{2} \pm \frac{1}{2}\sqrt{M^{2}+4m^{2}}.$

However, I am having trouble finding the eigenvectors. I have the eigenvector equation

$$\begin{pmatrix} 0 & m\\ m & M \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \lambda \begin{pmatrix} a \\ b \end{pmatrix},$$

which gives me $mb = \left( \frac{M}{2} \pm \frac{1}{2}\sqrt{M^{2}+4m^{2}} \right)a$ and $ma+Mb = \left( \frac{M}{2} \pm \frac{1}{2}\sqrt{M^{2}+4m^{2}} \right)b.$

Could you help me out here?

The answer's supposed to be $\cos \theta = \frac{1}{2} \arctan \frac{2m}{M}$.

$\endgroup$
  • 1
    $\begingroup$ The LHS of the former equation should be $mb.$ $\endgroup$ – tommy xu3 Feb 6 '17 at 6:14
  • $\begingroup$ Ah! Yes! I see! Updating answer! $\endgroup$ – nightmarish Feb 6 '17 at 6:16
  • $\begingroup$ You’ve got a pair of linear equations in the unknowns $a$ and $b$. Solving this system should be at least as easy as finding the eigenvalues. $\endgroup$ – amd Feb 6 '17 at 7:45
  • $\begingroup$ $\cos(\theta) = \frac{1}{2}\arctan(2m/M)$ just looks wrong. You mean $\theta = \frac{1}{2}\arctan(2m/M).$ $\endgroup$ – spaceisdarkgreen Feb 6 '17 at 10:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.