Homework Help: Relation between complex eigenvalues and rotations

1. Jan 28, 2016

TheMathNoob

1. The problem statement, all variables and given/known data
I have the following matrix:
0 0 0 1
1 0 0 0 = A
0 1 0 0
0 0 1 0
and the vector
v = (1,0,0,0)

If I perform Av, this gives:
Av=(0,1,0,0)
And If I keep multiplying the result by A like A*A*(Av), the outcome will be something like
j= (0,0,1,0)
k=(0,0,0,1)
l=(1,0,0,0)

The professor made a relation between these outcomes and the unit circle. In the unit circle the positive x axes is labelled with 1, the -x axes with-1, the positive y axes with i and negative y axes with -i. Then, He said that the eigen values are 1,-1,i,-1 because something rotates 90 degrees. He also related e^(pi*i) with rotations. I am just wondering if you can explain me all that or the exact place where I can find information.

2. Relevant equations

3. The attempt at a solution

2. Jan 28, 2016

Staff: Mentor

The connection, possibly, is that for the matrix A above, $Iv, Av, A^2v, A^3v$ is a cycle that starts repeating again with $A^4v$, similar to the way that a rotation of a vector by $\pi/2$ repeats.

The matrix $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ rotates a vector by $\pi/2$ counterclockwise. The two eigenvalues are $\lambda = \pm i$, and the corresponding eigenvectors are $\vec{v} = \begin{bmatrix} \pm i \\ 1 \end{bmatrix}$.

3. Jan 28, 2016

TheMathNoob

Right, I have been reading and watching videos about this issue. The hard thing to understand are the vectors that multiply A. In this case
you can represent your an eigen vector by an imaginary and real part. In this case, your first eigen vector, let's call it v, can be represented as following:
v=(0,1)+i(1,0)
so the multiplicand vectors that get this behavior when multiplied by A are the basis (0,1) and (1,0), my question now would be how can I know what the eigen values are if I know that a pattern like that emerges.

Example
v=(1,0,0)
r=(0,1,0)
s=(0,0,1)

4. Jan 29, 2016

geoffrey159

I don't know if this help but I have a geometric interpretation of your problem.
You can notice that $A^2$ is a symmetry since $A$ is orthogonal and $A^4 = I$.
You can find the plane of symmetry by solving $A^2 X = X$ and it's orthogonal by solving $A^2 X = -X$.
The plane of symmetry is the span of $\{(1,0,1,0),(0,1,0,1)\}$ and it's orthogonal the span of $\{(1,0,-1,0),(0,1,0,-1)\}$. By the way these vectors form a nice orthogonal family 2 by 2 in the planes, and also as a whole. Therefore, as a whole, they form an orthogonal basis of $\mathbb{R}^4$, and 2 by 2, an orthogonal basis of the each planes.
You will notice that in these planes, $A$ performs 90 degrees rotations of the basis vectors since it transforms one basis vector into plus or minus the other depending on the plane.
I hope this will help.

5. Jan 29, 2016

geoffrey159

Sorry I realize this is very unclear. What's going on is that if $P$ is the plane of symmetry of $A^2$, then $A$ is a $\pi/2$ rotation in $P^\perp$ but a reflection about the axis directed by $(1,1,1,1)$ in $P$.

This is because orthogonal endomorphisms in the plane are either rotations or reflections. If $(e_1,e_2)$ is the orthogonal basis of $P$ mentioned above, $(e_3,e_4)$ the orthogonal basis of $P^\perp$, then you get that $Ae_1 = e_2$, $Ae_2 = e_1$. So clearly A is a reflection in $P$ about the axis at half angle between $e_1$ and $e_2$. This axis is directed by $e_1 + e_2 = (1,1,1,1)$.
Also $Ae_3 = e_4$, $A^2e_3 = -e_3$, $A^3 e_3 = -e_4$, $A^4 e_3 = e_3$. So $A$ is a $\pi / 2$ rotation in $P^\perp$.

I think this is correct this time