Graphic representation of the complex eigenvector of a rotating matrix

Question

The eigenvector of the matrix

$$\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$

is

$$\begin{bmatrix}1\\i\end{bmatrix}$$

with its eigenvalue $-i$

$$-i\begin{bmatrix}1\\i\end{bmatrix}=\begin{bmatrix}-i\\1\end{bmatrix}$$

The question is how to graphically show this vector. The graphic representation has the real part in the x-axis and the imaginary part in the y-axis. Yet, we see $-i$ in the x component of the vector. I would like to "see" the eigenvector elongate, or contract, as a result of its multiplication by the eigenvalue.

Answer 1

Sorry, but it's unclear what exactly you mean by graphically show this vector.

A single complex number can be visualized by a vector in the complex plane. Note that the set $\mathbb{C}$ of complex numbers is a one-dimensional vector space $\mathbb{C}^1$ over itself (i.e. over $\mathbb{C}$, as a complex vector space). So in that sense it's more like a "complex number line" — akin to the real number line $\mathbb{R}$ as a one-dimensional vector space $\mathbb{R}^1$ over itself (i.e. over $\mathbb{R}$, as a real vector space). We visualize the complex plane as a plane, however, because $\mathbb{C}$ is two-dimensional as a real vector space over $\mathbb{R}$. And we as humans (at least, the vast majority of us) are much better at visualizing real dimensions, and only up to three of them.

And that's where the problem with your request lies. The space $\mathbb{C}^2$ is two-dimensional as a complex vector space. But to visualize it as such, we need two "complex number lines", i.e. two Argand planes is the two axes. In reality, we immediately think of each complex coordinate as $z=x+iy$ with two real components, so effectively we're thinking of $\mathbb{C}^2$ as a four-dimensional real vector space. Representing it geometrically requires drawing in four real dimensions — and there's nothing wrong with that mathematically, except for the fact that for most of us (myself included) it's really difficult to imagine that.

Answer 2

As $$ -i = e^{i3\pi/2} $$ multiplication by it means rotating the vector by $3\pi/2$ only and no length change.

Update:

Or try it yourself: link

Answer 3

I think probably the most straightforward way to plot complex eigenvectors is to draw up two Argand planes, and in the first plane, draw the two (or more, depending on the dimension in involved) complex coordinates of the first vector. Make certain to colour the vectors to distinguish between them (e.g. black for the first coordinate, red for the second coordinate). Then do the same for the image vector on the second Argand plane, keeping the colour scheme the same.

What are you looking for? Well, if we want a parallel vector, we would expect to see a common complex number $re^{i\theta}$ to multiply both complex numbers in the first plane to produce the complex numbers in the second plane. In other words, we expect to see that the second plane is a rotated and scaled version of the first (including preserving orientation between the different vectors).

This is about as visually clear as I think parallel complex vectors can be made.