0
$\begingroup$

I am given

$A = \left[\begin{array}[c]{rr} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{array}\right]$

from which I calculated

$λ = \cos\theta \pm i\sin\theta$

the eigenvalues are thus imaginary but I want to calculate the eigenvectors

$A-\lambda I = \left[\begin{array}[c]{rr} \pm i\sin\theta & -\sin\theta\\ \sin\theta & \pm i\sin\theta\end{array}\right]$ $\left[\begin{array}[c]{r} y \\ z \end{array}\right]$ $=$ $\left[\begin{array}[c]{r} 0\\ 0\end{array}\right]$

When I try to find the eigenvector(s) I keep getting things like $0 = 0$... which is pretty useless. Does this mean there are no eigenvectors or that the eigenvector is $\left[\begin{array}[c]{r} 0\\ 0\end{array}\right]$ or that I'm doing something wrong?

$\endgroup$
1
$\begingroup$

As you found eigenvalues, there are eigenvectors.

Let us find one eigenvector:

$$ \begin{cases}\cos\theta X - \sin\theta Y = (\cos\theta + i\sin\theta)X\\ \sin\theta X + \cos\theta Y = (\cos\theta + i\sin\theta)Y\end{cases} \iff - \sin\theta Y = i\sin\theta X\\ \Leftarrow iY = X $$ Here, check that both equations are equivalent: hence your eigenvalue is good!

Assume that $ Y=1 $ (you only need one eigenvector) gives $X= i$. Now do the same with the other eigenvalue, you find the relation $$ -iY = X . $$

$\endgroup$
  • $\begingroup$ So then I get $\left[\begin{array}[c]{r} 1\\ ±i\end{array}\right]$ and $\left[\begin{array}[c]{r} ±i\\ 1\end{array}\right]$ , right? $\endgroup$ – mangopancake Mar 30 '14 at 21:56
  • $\begingroup$ only $(1, \pm i)$! $\endgroup$ – mookid Mar 30 '14 at 21:57
  • $\begingroup$ because $i(1,i) = -(-i, 1)$ $\endgroup$ – mookid Mar 30 '14 at 21:57
  • $\begingroup$ Ah yes! They are essentially the same. Then I can pick either one, can't I? $\endgroup$ – mangopancake Mar 30 '14 at 22:00
  • $\begingroup$ yes you can. You only one for each eigenvalue here (because there are 2 different eigenvalues and the matrix is 2x2) $\endgroup$ – mookid Mar 30 '14 at 22:01
1
$\begingroup$

Case 1: $\theta\neq 0$

What you should have from the first equation is that

$$(\pm i\sin\theta) y -(\sin\theta) z = 0.$$

This gives you an expression for $z$ in terms of $y$ ($z=\pm iy$), which then gives you your eigenvectors ($(y,\pm iy)$). This occurrence of $y$ in the eigenvector is not surprising. If you scale an eigenvector by a number, it stays an eigenvector so typically we just drop the $y$ and have $(1,\pm i)$.

Case 2: $\theta = 0$

If $\theta = 0$, your original matrix is just the identity matrix. What eigenvalue(s) and eigenvectors does this have?

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.