# Visualizing linear transformations on vector fields

I'm trying to figure out what it means to apply a linear transformation to a vector field geometrically. So I start with the easiest geometrically interesting transformation: a rotation.

Using StreamPlot in WolframAlpha I see that applying the transformation $T(x,y) = (-y,x) = \pmatrix{\cos(90^\circ) & -\sin(90^\circ) \\ \sin(90^\circ) & \cos(90^\circ)}\pmatrix{x \\ y}$ yields exactly what I'd expect: circular stream lines: However when I try a different angle, say $\pi / 6$, then I have $T(x,y)=\left(\frac{\sqrt{3}}{2}x-\frac12 y, \frac12 x + \frac{\sqrt{3}}{2}y\right)$. But the picture no longer looks like circles, but spirals: Clearly something's wrong with my interpretation here. Why do the stream lines look the way they do? How can I imagine linear transformations being applied to vector fields?

To visualize the transformation of a vector field you first have to visualize the vector field. In your case, you are rotating the identity field in $\mathbb R^2$.

Consider an arbitrary point in $\mathbb R^2$, say $(3,4)$. The identity vector field assigns the vector $3\hat i + 4\hat j$ to this point. You imagine an arrow of length $5$ being drawn with its tail at the point $(3,4)$. You perform a similar construction for each point in the plane. Since you cannot do that in a finite period of time you plot enough of these in your mind or on paper until you get a satisfactory picture of the vector field.

To visualize what happens under the transformation, you apply the transformation to the vectors you have drawn. You replace the vectors that were previously drawn with the new vectors you get from the transformation. That will give you an idea of what the new vector field will look like.

If you try this exercise by hand you will discover why the $60^{\circ}$ rotation looks different.

• +1 for mentioning that he was transforming the identity field Nov 20, 2015 at 20:10

(I'll keep it simple, so this explanation sacrifices a lot of generality.)

There are two ways to apply a linear transformation to a vector field: multiplication and conjugation. What you've observed is the difference between the two.

Multiplying a vector field (on the left) rotates all the arrows in-place. If you start with a vector field that looks like circles, and you rotate all the arrows a bit, they no longer go in circles; they either spiral into or out of the origin. This is the relationship between your two versions of $T$; you can write $T_2 = RT_1$ for some rotation $R$.

It sounds like you expected the vector field to look the same, just rotated in space. That would indeed be the case if you had conjugated $T$ by a rotation, getting $T_3=RT_1R^{-1}$. In fact, since 2D rotations commute with each other, you would have $T_3=T_1$: the two vector fields would look exactly the same.

I'll try complement the current answers with a bit of fluid mechanics intuition. (You can replace "particle" with "boat" in what follows.)

You are using a simple transformation but not the simplest vector field (the identity), so maybe this makes it a little harder to visualize.

If you applied no transformations to this vector field, it would look like rays coming out of the origin. A plot like the one you are using can be thought of as the motion of a particle driven by the ocean's current with velocity equal to the value of the vector field in that point. A particle feeling this velocity-field would be driven outwards with a speed proportional to the distance from its position to the origin (that is, with increasing speed).

More specifically, let $O$ be the origin. This velocity field at a point $P$ can be visualized as an arrow starting from $P$ and extending in the direction $OP$ up to the point $2P$, so that the distance from the origin to the tip of the arrow is twice the distance form $O$ to $P$.

In the first case (90 degrees), the perpendicular transformation is strong enough to make it tangential to the circle with radius $OP$ centered at the origin. This makes the particle in the velocity field move along the circle, which is what your plot shows.

In contrast, the second transformation (30 degrees) is not as strong: it only deviates the arrow a bit in a counterclockwise manner (still starting from $P$). A particle feeling this rotated vector field will be carried away in a $30$-degree angle from the line connecting its position to the origin, and with a velocity equal to the distance from the origin. This can be seen in the plot: the distance traveled by the particle stays for a longer way in a simliar direction, so the curvature of its trajectory decreases while its speed increases.

If you want a clearer visualization of the transformation you could start applying it to a constant vector field and see what happens, or maybe one that varies only with $x$.