Eigenvectors of the derivative are functions of the form $e^{\lambda t}.$

We can verify this computationally:

But what does this mean? How is $e^{\lambda t}$ a vector?

Functions are vectors with an infinite number of components, and our traditional way of visualizing vectors no longer works with our traditional way of visualizing vectors once we have more than 3 components. So we'll adopt a different way of representing them visually:

It is probably strange to think of functions as vectors. What would something like $f(x)=\sin(x)$ have to do with a column vector like $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$? I suspect that the perceived difference between these is the form in which we visualize them.

The reason they seem different is because we can't visualize vectors with more than 3 components in the traditional way. Sure, vectors like these

can be plotted, where their entries are their $x$ and $y$ (and sometimes $z$) coordinates. But as soon as we need to plot a vector like

we run out of dimensions because there aren't enough axes to dedicate one axis to each component. Meanwhile, functions have an infinite number of components. So we will start with a new way to visualize column vectors that also for column vectors with more than 3 components.

With this new way of visualizing vectors, any number of components can be visualized. This includes vectors with infinitely many components. Click through the buttons below to see for yourself.

Scaling vectors with the old way of visualizing vectors makes them parallel but larger, smaller, or going in the opposite direction. On the other hand, scaling vectors with the new way of visualizing vectors makes all of the heights grow, shrink, or reflect on the $x$-axis together.

Pairing an eigenvector $e^{\lambda t}$ of the derivative with an eigenvector $\vec{v}$ of the matrix $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ yields a solution of the first order system

The pairing looks like this: