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I'm having trouble getting the geometric interpretation of complex eigenvectors and eigenvalues of a rotation matrix. I am given A= \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} The given eigenvalues are: D=\begin{bmatrix} 2i & 0 \\ 0 & -2i \end{bmatrix} and the eigenvectors are:\begin{bmatrix} i & -i \\ 1 & 1 \end{bmatrix} I need to know the geometric interpretation of these complex eigenvectors and values and also those of $A^{\frac{1}{N}}$.
I know that A= $2\cdot\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = 2\cdot\begin{bmatrix}\cos{\pi /2} & -\sin{\pi /2} \\ \sin{\pi /2} & \cos{\pi /2}\end{bmatrix}$, so $A^{\frac{1}{N}}=2^{\frac{1}{N}}\cdot\begin{bmatrix}\cos{\pi /2N} & -\sin{\pi /2N} \\ \sin{\pi /2N} & \cos{\pi /2N}\end{bmatrix}$, which is a rotation of $\frac{\pi}{2N}$ counterclockwise and scaling by $2^{\frac{1}{N}}$.
I did read the earlier post about geometric interpretation, and the fantastic animated explanation, but I still don't get it. I apologise, it sounds like I'm asking for answers (I kind of am yeah), but I need a worked example to get anything done at all, and David C. Lay's textbook was incomprehensible to me.
Thank you very much for your help.
I'd be very grateful if someone could draw out the basis vectors, show me how I tell how much they rotate, and in which direction, and then how much it scales.

Memiya
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  • So what is the question? – user619894 Mar 29 '21 at 14:59
  • I'd like a geometric interpretation of $A^{\frac{1}{N}}$, its eigenvalues and eigenvectors, in image form, please. – Memiya Mar 29 '21 at 15:00
  • Do you want a figure with complex eigenvectors? Hard to draw, and if you don't like https://math.stackexchange.com/questions/241097/geometric-interpretation-for-complex-eigenvectors-of-a-2%c3%972-rotation-matrix?rq=1, then I fear there is not much more that can be done. What exactly don't you follow there? – user619894 Mar 29 '21 at 15:05
  • Yes please.
    I didn't get why the eigenvector just kept spinning. I tried to use it to think of how the basis vectors would look like but I can't.
    The eigenvectors given to me also have i on top and 1 on the bottom, so I can't translate those into basis vectors.     – Memiya Mar 29 '21 at 15:06
  • We can't represent a complex number on the real axis, so we add an "imaginary" direction. So even a single complex number is represented by a 2d "vector" in the complex plane. Multiplying two complex numbers moves the position of this vector both by rotating and by stretching. Now if we have a complex vector, it is composed of of 2 of these "vectors" and each of them rotates in their own complex plane. – user619894 Mar 29 '21 at 15:17
  • I get the multiplying part, I don't get the second part. So we can't actually draw the basis vector on the complex plane? How would I interpret \begin{bmatrix} i \ 1 \end{bmatrix} on the complex plane? – Memiya Mar 29 '21 at 15:21
  • You don't. There are 2 complex planes, one for the "upper" component and one for the lower. – user619894 Mar 29 '21 at 15:33
  • So I can't actually draw it at all like real eigenvectors? So what would a geometric representation mean for complex eigenvectors? It seems that all that is changing are the coefficients in the complex plane; there's no geometric meaning with regards to the original matrix. – Memiya Mar 29 '21 at 15:53
  • The matrix has a nice geometric interpretation ( a rotation), the eigenvectors don't. – user619894 Mar 29 '21 at 18:35

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