# matrix with sin and cos? help wanted.

• Feb 8th 2009, 12:39 PM
fiksi
matrix with sin and cos? help wanted.
Hi guys, this question bothers me...

I have a matrix A=

cos(theta) sin(theta)
-sin(theta) cos(theta)

So it is a 2x2 matrix. I am asked to show that e to the power of (i theta)
and e to the power of (-i theta) are eigenvalues. Then to find eigenvectors and simple formula for A to pwr of n.

Now, I guess I should proceed finding determinant of A(with - lambda from left top to bottom right)=0. However, i can't find my way out of this. I get smth like lambda squared + sin squared theta+ cos squared theta - 2lambda cos theta=0.

Any help is appreciated!
• Feb 8th 2009, 02:08 PM
Moo
Hello,
Quote:

Originally Posted by fiksi
Hi guys, this question bothers me...

I have a matrix A=

cos(theta) sin(theta)
-sin(theta) cos(theta)

So it is a 2x2 matrix. I am asked to show that e to the power of (i theta)
and e to the power of (-i theta) are eigenvalues. Then to find eigenvectors and simple formula for A to pwr of n.

Now, I guess I should proceed finding determinant of A(with - lambda from left top to bottom right)=0. However, i can't find my way out of this. I get smth like lambda squared + sin squared theta+ cos squared theta - 2lambda cos theta=0.

Any help is appreciated!

$A-\lambda I_2=\begin{pmatrix} \cos \theta-\lambda & \sin \theta \\ -\sin \theta & \cos \theta-\lambda \end{pmatrix}$

The determinant is :
$p(\lambda)=(\cos \theta-\lambda)^2+\sin^2 \theta$

Find $\lambda$ such that $(\cos \theta-\lambda)^2+\sin^2 \theta=0$
Now, you can see that this is true if and only if $\sin^2 \theta=-(\cos \theta-\lambda)^2=(i [\cos \theta-\lambda])^2$

Hence $\sin \theta=\pm i (\cos \theta-\lambda)$
$\pm i \sin \theta=\cos \theta-\lambda$
$\lambda=\cos \theta \pm i \sin \theta$

which gives the values you want, thanks to the identity $\cos \theta+i \sin \theta=e^{i \theta}$ ( $\cos \theta-i \sin \theta=\cos(-\theta)+i \sin (-\theta)=e^{-i\theta}$, because cos is an even function and sin is an odd function)

:)
• Feb 8th 2009, 02:34 PM
HallsofIvy
Quote:

Originally Posted by Moo
Hello,

$A-\lambda I_2=\begin{pmatrix} \cos \theta-\lambda & \sin \theta \\ -\sin \theta & \cos \theta-\lambda \end{pmatrix}$

The determinant is :
$p(\lambda)=(\cos \theta-\lambda)^2+\sin^2 \theta$

Find $\lambda$ such that $(\cos \theta-\lambda)^2+\sin^2 \theta$

You mean $\lambda$ such that $(\cos \theta-\lambda)^2+\sin^2 \theta= 0$, of course.

Quote:

Now, you can see that this is true if and only if $\sin^2 \theta=-(\cos \theta-\lambda)^2=(i [\cos \theta-\lambda])^2$

Hence $\sin \theta=\pm i (\cos \theta-\lambda)$
$\pm i \sin \theta=\cos \theta-\lambda$
$\lambda=\cos \theta \pm i \sin \theta$

which gives the values you want, thanks to the identity $\cos \theta+i \sin \theta=e^{i \theta}$ ( $\cos \theta-i \sin \theta=\cos(-\theta)+i \sin (-\theta)=e^{-i\theta}$, because cos is an even function and sin is an odd function)

:)
• Feb 9th 2009, 02:28 AM
fiksi
Quote:

Originally Posted by Moo
Hello,

$A-\lambda I_2=\begin{pmatrix} \cos \theta-\lambda & \sin \theta \\ -\sin \theta & \cos \theta-\lambda \end{pmatrix}$

The determinant is :
$p(\lambda)=(\cos \theta-\lambda)^2+\sin^2 \theta$

Find $\lambda$ such that $(\cos \theta-\lambda)^2+\sin^2 \theta=0$
Now, you can see that this is true if and only if $\sin^2 \theta=-(\cos \theta-\lambda)^2=(i [\cos \theta-\lambda])^2$

Hence $\sin \theta=\pm i (\cos \theta-\lambda)$
$\pm i \sin \theta=\cos \theta-\lambda$
$\lambda=\cos \theta \pm i \sin \theta$

which gives the values you want, thanks to the identity $\cos \theta+i \sin \theta=e^{i \theta}$ ( $\cos \theta-i \sin \theta=\cos(-\theta)+i \sin (-\theta)=e^{-i\theta}$, because cos is an even function and sin is an odd function)

:)

Yes, it ain't hard actually... just seems to involve a bit of recognizing. The thing is that we haven't really covered this stuff well, or any matrices with cos and sin problems. And I'm not too sure bout all the identities either, even so prolly a long time ago.

So, how would the eigenvectors and "simple" formula turn out? Hope not anything ugly... I don't want to get stuck in anything ugly. Thx for help, though! I wish this was just a normal example I am able to do outright... but it is not.

thx again