Find $\lambda$ and $\theta$ such that it validates this matricial equation

Question

Find $\lambda$ and $\theta$ such that it validates the matricial equation

$$ \left( \begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array} \right) % \left( \begin{array}{cc} \cos \theta \\ \sin \theta \end{array} \right) = \lambda\left( \begin{array}{cc} \cos \theta \\ \sin \theta \end{array} \right) $$

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What I have tried:

$$\left\{ \begin{array}{ll} \cos \theta + 2\sin \theta &= \lambda \cos \theta\\ 2\cos \theta + 3\sin \theta &= \lambda \sin \theta.\end{array} \right. $$

$$\left\{ \begin{array}{ll} \ -2\cos \theta -4 \sin \theta &= -2\lambda \cos \theta\\ 2\cos \theta + 3\sin \theta &= \lambda \sin \theta.\end{array} \right. $$

$$-\sin \theta = -2\lambda \cos \theta + \lambda \sin \theta$$

$$ (\lambda + 1)\sin \theta = 2 \lambda \cos \theta$$

I can´t find $\lambda$ and $\theta$

Answer 1

Here are the steps:

1- Find the eigenvalues of the coefficient matrix which are $2\pm \sqrt{5}$.

2- Find the associated eigenvectors with the values.

3- equalling the vectors with your sine and cosine vector, find the $\theta$.

Answer 2

A clean way to do this is to multiply both sides of the expression on the left with $\left(\begin{array}{cc} -\sin \theta & \cos \theta \end{array} \right)$

$$\left(\begin{array}{cc} -\sin \theta & \cos \theta \end{array} \right) \left( \begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array} \right) % \left( \begin{array}{cc} \cos \theta \\ \sin \theta \end{array} \right) = \lambda\left(\begin{array}{cc} -\sin \theta & \cos \theta \end{array} \right)\left( \begin{array}{cc} \cos \theta \\ \sin \theta \end{array} \right) $$

$$2\sin{\theta}\cos{\theta} + 2\cos^2{\theta} - 2\sin^2{\theta}= 0$$

$$\sin{2\theta}+2\cos{2\theta}=0$$

Solve for theta, plug in and back out $\lambda$, delete extraneous roots if present.

Answer 3

For given $\theta$ this is an eigenvalue problem.

The characteristic equation of your matrix is $p(\lambda) = (1-\lambda)(3 - \lambda) -4 = \lambda^2 - 4\lambda -1$

And so the eigenvalues are

$ \lambda = \dfrac{4 \pm \sqrt{16+4}}{2} = 2 \pm \sqrt{5}$

It can be shown (though I used computer algebra) that the eigenvectors are

$$v_1 = \begin{bmatrix} 1\\1 \end{bmatrix} v_2 = \begin{bmatrix} \dfrac{2}{1+ \sqrt{5}} \\ \dfrac{2}{1-\sqrt{5}}\end{bmatrix} $$

The polar coordinate representation of a vector allows you to find the angle. Since there are two vectors, there should be two $\theta$, and so

$$\theta_i = \tan ^{-1}(\dfrac{y_i}{x_i})$$