# Matrix Spectra with Troublesome Trigonometric Term

• June 25th 2010, 01:07 AM
alexandicity
Matrix Spectra with Troublesome Trigonometric Term
Hello!

I’m trying to find the spectra (eigenvalues) of matrix

$A = $\left( \begin{array}{ccc} cos(x) & -sin(x) & 0 \\ sin(x) & cos(x) & 0 \\ 0 & 0 & 1 \end{array} \right)$$

I start by generating its characteristic polynomial which yields:

$(cos(x)-\lambda)(cos(x)-\lambda)(1-\lambda)+(sin(x))(sin(x))(1-\lambda)=0$
$(1-\lambda)[(cos(x)-\lambda)(cos(x)-\lambda)+sin(x)sin(x) ]$
$(1-\lambda)[cos(x)^2-2\lambdacos(x)+\lambda^2+sin(x)^2]$
Applying the Euler Formula:
$(1-\lambda)[-2\lambda cos(x) + \lambda^2 + 1]$
$(1-\lambda)(\lambda^2 - 2\lambda.cos(x) + 1)$

At this point, I want to extract roots, the solutions for eigenvalues $\lambda$. One, from the LH term, is clearly $\lambda = 1$. But that $-2cos(x)$ in the middle is stopping me from factorizing it via any way I can work out. I expect, since A was orthogonal, that the other two roots will be a complex conjugate pair but am not 100% on that – in any case, I can’t see how I can legally break that trig term into a complex one.

Any ideas?
Cheers!
Alexandicity
• June 25th 2010, 04:51 AM
tonio
Quote:

Originally Posted by alexandicity
Hello!

I’m trying to find the spectra (eigenvalues) of matrix

$A = $\left( \begin{array}{ccc} cos(x) & -sin(x) & 0 \\ sin(x) & cos(x) & 0 \\ 0 & 0 & 1 \end{array} \right)$$

I start by generating its characteristic polynomial which yields:

$(cos(x)-\lambda)(cos(x)-\lambda)(1-\lambda)+(sin(x))(sin(x))(1-\lambda)=0$
$(1-\lambda)[(cos(x)-\lambda)(cos(x)-\lambda)+sin(x)sin(x) ]$
$(1-\lambda)[cos(x)^2-2\lambdacos(x)+\lambda^2+sin(x)^2]$
Applying the Euler Formula:
$(1-\lambda)[-2\lambda cos(x) + \lambda^2 + 1]$
$(1-\lambda)(\lambda^2 - 2\lambda.cos(x) + 1)$

At this point, I want to extract roots, the solutions for eigenvalues $\lambda$. One, from the LH term, is clearly $\lambda = 1$. But that $-2cos(x)$ in the middle is stopping me from factorizing it via any way I can work out. I expect, since A was orthogonal, that the other two roots will be a complex conjugate pair but am not 100% on that – in any case, I can’t see how I can legally break that trig term into a complex one.

Any ideas?
Cheers!
Alexandicity

You did everything fine, as far as I can see...! Now, the discriminant of $t^2-2\cos x\,\,t+1$ is

$\Delta=-4\sin^2x=(2i\sin x)^2$ , so using the roots formula for a quadratic eq. we

get $t_{1,2}=\frac{2\cos x\pm 2i\sin x}{2}=\cos x\pm i\sin x=e^{\pm ix}$, and here you have your conjugate pair of roots. (Wink)

Tonio
• June 25th 2010, 05:18 AM
alexandicity
Hi Tonio

I get most of what you're saying and it looks good. However - one think I don't follow. You identify the discriminant as $-4sin^2x$. However, when I try to determine it:
$a=1$
$b=-2cos(x)$
$c=1$
$\triangle = b^2-4ac$

Now squaring $b$ gives $b^2 = (4cos^2(x))$, so $\triangle = 4cos^2(x)-4$ I'm not sure how it converted to that sine function?

My guess is I got something wrong with my trig manipulations!
• June 25th 2010, 05:32 AM
Ackbeet
Just a few random thoughts: the eigenvalue problem has to do with operators whose action on an eigenvector is equivalent to multiplying the eigenvector by a scalar. Now, the matrix you've exhibited is a rotation about the z axis through an angle x. Therefore, I would expect any ol' vector along the z axis to be an eigenvector with eigenvalue 1. Any other vector could be an eigenvector with eigenvalue 1 if you rotated through $x=2\pi k$ radians, with $k\in\mathbb{Z}$. In addition, any vector in the xy plane could be an eigenvector with eigenvalue -1, if you rotate through an angle $(2k+1)\pi$ with $k\in\mathbb{Z}$. I don't think there are any other eigenvector/eigenvalue pairs. At least, none come to mind, because the fact is, that aside from a vector along the z axis, this rotation is going to change the direction of any other vector, unless $x=2\pi k$ with $k\in\mathbb{Z}$.

I just say all this by way of providing a double-check mechanism on your calculations, which seem to be in good hands.
• June 25th 2010, 05:40 AM
tonio
Quote:

Originally Posted by alexandicity
Hi Tonio

I get most of what you're saying and it looks good. However - one think I don't follow. You identify the discriminant as $-4sin^2x$. However, when I try to determine it:
$a=1$
$b=-2cos(x)$
$c=1$
$\triangle = b^2-4ac$

Now squaring $b$ gives $b^2 = (4cos^2(x))$, so $\triangle = 4cos^2(x)-4$ I'm not sure how it converted to that sine function?

My guess is I got something wrong with my trig manipulations!

Ah, that trigonometry! $\cos^2x+\sin^2x=1\Longrightarrow \cos^2x-1=-\sin^2x$ ...

Tonio
• June 25th 2010, 05:44 AM
alexandicity
Ah - I should have seen that. Thanks a load!

Ackbeet - cheers for the background. I've not looked at rotations is much depth yet but will be moving on to it soon. I'll ponder your words!