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?


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 $$ -iY = X . $$

  • $\begingroup$ 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

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?


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