How to find the complex eigenvectors of the rotation matrix
$\begin{bmatrix}
cos \theta &\sin \theta & 0\\
- \sin \theta & cos \theta & 0\\
0&0&1
\end{bmatrix}
I already have the eigenvalues
1, $cos \theta + i \sin \theta$, $cos \theta - i \sin \theta$
and I found in a book that the eigenvectors for each eigenvalue are
$v_1 = \begin{bmatrix} 0 \\0 \\1 \end{bmatrix}$, $v_2 = \begin{bmatrix} \frac{1}{\sqrt{2}} \\\frac{i}{\sqrt{2}} \\0 \end{bmatrix}$, $v_3 = \begin{bmatrix} \frac{1}{\sqrt{2}} \\\frac{-i}{\sqrt{2}} \\0 \end{bmatrix}$, respectively.
But I can't get to those results, because I don't understand very well trigonometric functions and their properties, identities and so.
Could someone please guide me on how to get to the corresponding eigenvectors? What's the procedure? Or could someone at least give me a clue on this, or recomend me some good book or web page?
Thanks in advance, and sorry if I failed using LaTeX code in this web.
Do it for. 2x2 matrix and then generalize.
>>7977525
You gonna repost this every night?
>>7977525
>sorry if I failed using LaTeX code in this web
>in this web
There's only one web. Unless...
Are you a time traveler with access to multiple "webs"/global information networks?
>>7977525
First off, latex on /sci/ doesn't get initiated by $, but by [.math] and [./math] (ommit the dots
Since you already have the eigenvalues, you can find the eigenvectors by calculating the kernel of [math] A-\lambda [/math]
with the first eigenvalue 1 it's trivial since [math] A-\lambda [/math] already has a full zero row, so we can immediately see, that (0,0,1) getting mapped to 0.
For the others it's a little more tricky, but only a little
[math]
A-\lambda_2 =
\begin{bmatrix}
-is & s & 0 \\
-s & -is & 0 \\
0 & 0 & 1-\lambda_2
\end{bmatrix}
[/math]
which leads to equations
[math] \mu_1 (-is) +\mu_2 s = 0 [/math]
[math] \mu_1 (-s) +\mu_2 (-is) = 0 [/math]
[math] \mu_3 (1-\lambda_2) = 0 [/math]
which have multiple solutions - one of them
[math] (\mu_1,\mu_2,\mu_3) = (1,i,0) [/math]
Since we want our eigenvectors to have norm 1 (so that we can have diagonalizing transformations with orthogonal matrizes), we have to divide the vector by it's norm, which is [math] \sqrt{2} [/math]
same principle goes for the last eigenvalue
>>7977547
Only until I could learn how to get those eigenvectors.
>>7977753
This is what I needed, thank you very much, anon
>>7977605
And yet again this tripfag shitposts instead of contributing to a potentially quality thread.
Why the fuck even de-anonymize yourself?
>>7977605
I'm the Lizard King, I can do anything...except for find those eigenvectors.
>>7978147
You don't know how to get those eigenvectors, right?
This post could become a completely discussion about rotation matrix, eigenvalues, eigenvectors, if only suckers like you contribute to an alternative reply instead of your shitposting reply...
>>7977525
Rotation matrices have the property R^T R = I
So Det(R^T)Det(R) = 1 => Det(R) = +/-1
The determinant of a matrix is the product of its eigenvalues
We also know that the trace is the sum of the eigenvalues
2cos theta + 1 = sum lambda
You could play a bit with this in addition to the usual techniques for finding eigenvalues
>>7978163
ignore this post, I read eigenvalues instead of eigenvectors
>>7978159
Nah I took Linear Algebra a couple of years ago. I'm fine. OP's question was answered, so who gives a fuck? If I'm going to call this tripfag out, it has to be somewhere. Here seemed fine.
>>7978164
That's useful information though. Thanks...
>>7978147
And yet another person doesn't know what a tripfag is.
>>7977525
Which complex numbers z, when multiplied by a given complex scalar q, yield z times a complex scalar?
