Thread by @3blue1brown

It's a very good question! What follows are many tweets attempting to answer in a possibly way-too-verbose manner. No pictures (sorry), but I'll trust in the readers' mind's eye.

https://twitter.com/saha_sidharth/status/1337519170011820032

Commentary from mathematicians is more than welcome at the bottom of the thread, which includes scattered thoughts on whether there's a purely geometric reason to expect any rotation to have complex eigenvalues, i.e. one that doesn't appeal to the fundamental theorem of algebra.

First, just answering the question, let's review the basics: What it means for an operator to have an eigenvector with eigenvalue λ is that the way this operator acts on that vector is to simply scale it by λ.

For real values of λ, scaling looks simple enough, just a stretching or squishing in the direction of that vector.

Phrased another way, eigenvectors remain in their own span as you apply the operator.

To understand complex eigenvalues, you want to first better understand the words "scale" and "span" once complex numbers enter the game.

This is important for understanding quantum mechanics, where states are described by complex vector spaces, and eigenvalues are deeply important to understanding the relevant operators on these spaces describing observables and dynamics.

Complex eigenvalues in particular play an important role there. For example, the all-important unitary operators have eigenvalues that sit on the unit circle in the complex plane, and skew-hermitian operators have purely imaginary eigenvalues.

Anyway, Multiplication by complex numbers looks not only like a stretching/squishing, but there's also a rotational component, right? So it might not come as a surprise that complex eigenvalues come up for operators with some kind of rotation to them.

For example, the 90-degree rotation matrix Rot = [[0, -1], [1, 0]] has eigenvalues i and -i, and it's no coincidence that the actions of i and -i in the complex plane look like 90-degree rotations, but let's unpack that.

One nice way to quickly see that the eigenvalues of Rot are i and -i is to know that the trace of a matrix (sum of the diagonal) gives the sum of eigenvalues, and the determinant gives their product.

So just looking at

Rot = [[0, -1],
[1, 0]],

you can read off that its eigenvalues sum to 0 and multiply to 1, which forces your hand to {i, -i}.

The question is what these mean geometrically, and if we want to think of an eigenvalue λ as being complex, we have to broaden the view of our relevant vector space to be over the complex numbers rather than the real numbers.

For example, 2d vectors might be represented with [a + bi, c + di], which involves four real parameters. This is rather frustrating for those of us who like clear mental pictures of what's going on, given that it seems we need four spatial dimensions to picture this.

All the more frustrating given that these complex vector spaces are fundamental to the universe.

The biggest intuitive difference between complex vector spaces and real vector spaces is what the words "scaling" and "span" refer to, because now it might not only be stretching/squishing of a vector, but rotating.

For example, if you "scale" the 2d complex vector [a+bi, c+di] by the imaginary unit i, you get [ai - b, ci - d], which is a vector with the same magnitude it had before but rotated by 90-degrees.

(Trouble visualizing this? You're not alone!)

So really "scalar" is a misnomer for complex vector spaces. Perhaps scalar-rotator or amplitwister would be better word choices if they weren't such a mouthful.

In general, the span of a complex vector, all the things you can get from "scaling" it by various complex numbers, looks like a 2d plane, a copy of the complex plane sitting inside the vector space.

This is entirely analogous to how the span of a real vector looks like a copy of the real number line sitting inside the full vector space. So when we say eigenvectors remain in their own space, for complex vector spaces, this means they remain inside some very special 2d plane.

But! There's one key difference in the analogy. For real vector spaces, any line running through the origin is the span of some vector. However, for complex vector spaces, most planes through the origin are _not_ the span of some vector.

These spans are quite special, and it's part of why a 2d vector space has a much richer structure than a 4d real vector space.

Now let's go back and think about that rotation operator Rot = [[0, -1], [1, 0]]. What does it do to the span of [1, 0], that special plane of points [a+bi, 0] that you get from "scaling" this vector by all possible complex numbers?

It completely moves this plane, rotating it to a plane that looks like [0, a+bi]. Analogous to how Rot acts on a 2d real vector space by rotating the horizontal axis onto the vertical axis, where these two axes intersect only at the origin...

...in the complex plane you might picture this (er, "picture") as a move in four dimensions that flips one plane 90-degrees onto another, but in such a way that the two planes only intersect at one point...okay maybe you can't picture this, but that's what happens.

This is all to say [1, 0] is not an eigenvector of this operator, and getting completely knocked off your own span is what it feels like not to be an eigenvector. It's very sad not to be an eigenvector, sort of the vector-equivalent of getting kicked out of your own home.

Similarly, [0, 1] is not an eigenvector, as it's span [0, c+di] gets moved to an entirely different plane [-c-di, 0].

However, take a look at [1, -i].

When you apply Rot to it, you get [i, 1]. This is also what you'd get by multiplying it by i. In fact, the span of [1, -i], that special copy of the complex plane sitting in our full space defined by all points (a + bi)*[1, -i], remains completely invariant when you apply Rot.

Moreover, the action of R on this plane is to rotate it 90-degrees in place.

It feels very nice to be an eigenvector. If your heels are at the origin and your head is at v, and you stick your arms out in the direction of i*v, to be an eigenvector is to know that you'll remain comfortably in the plane of your body and arms as the operator acts on you.

The vector [1, i] also feels quite at home. But to this vector, the rotation Rot feels like being multiplied by -i instead of by i.

Still equally comfortable, though, remaining happily in its own span. In the image above, this would mean it feels like getting pushed away from the direction your arms are pointed.

But all this is just showing you _that_ the rotation matrix Rot has some eigenvectors, but it doesn't really tell you why we should expect it.

We have this matrix with nothing but real numbers in it, and it acts on the unspecial not-a-span-of-anything plane {[a, b] : a, b ∈ ℜ} by 90-degrees. Is it obvious that once we extend to complex numbers, there's some special span-of-a-vector plane that gets rotated in place?

One answer, for those comfortable with eigenvalues, is to just solve the equation det(λ * I - Rot) = 0. The fundamental theorem of algebra guarantees that this will have a complex number solution, which means _any_ operator in a complex vector space has at least one eigenvector.

This is very different from real vector spaces, where plenty of operators (like rotations) might not have eigenvectors.

But the original question asked for a "geometric" picture, so is there any more geometric reason that an action which rotates one of the unspecial-not-a-span-of-anything planes should necessarily leave one of the special-span-of-some-vector planes invariant?

Honestly, I'm not quite sure how to justify this without appealing to the fundamental theorem of algebra (which is fine to appeal to, but it's a fun puzzle to strive for a different perspective).

However, here's what I suspect might be the start of a more purely geometric justification.

It's possible that what follows is entirely confusing, way too complicated, and adds more fog than clarity, but, well, you made the choice to read this far. Consider what follows more a set of working notes than a coherent explanation.

Again, my apologies for being too lazy to draw pictures for all this right now, it's hard to overstate how much time it would take to do it effectively. Instead, give your mind's eye a little massage, because we're about to ask a lot from it (poor thing.)

The idea is to consider the family of vectors v_θ = [1, e^{i*θ}] for various angles θ, along with how Rot acts on these vectors.

If you want to picture this, we might imagine the 3d slice of the full vector space that looks like [x, y + zi] for values x, y and z, each one pictured as the point (x, y, z) in 3d space.

In this view, the family v_θ = [1, e^{i*θ}] forms a circle around the point (1, 0, 0) which is perpendicular to the x-axis.

At the beginning, Rot(v_0) = Rot([1, 1]) = [1, -1], so (v_0, Rot(v_0)) "spans" the xy plane in the real number sense of the word span. Similarly, v_pi = [1, -1] gets taken to Rot(v_pi) = [1, 1], so (v_pi, Rot(v_pi)) also span-in-the-real-sense the xy plane.

However, they form a basis of the xy-plane (in a real-vector sense) with a different orientation.

So in some (admittedly vague) way, the pair (v_θ, Rot(v_θ)) swaps orientation as θ ranges from 0 to pi.

It seems hardly coincidental that at the halfway point, v_{pi/2} = [1, i], we hit a pint where {v_θ, Rot(v_θ)} sits on a spanning plane, which is to say v_{pi/2} is an eigenvector of Rot.

Is there a way to make this less fuzzy?

Take the 4-tuples of vectors (v_θ, Rot(v_θ), i*v_θ, Rot(i*v_θ)), and consider their orientation in the following sense: Treat them as 4d real vectors and consider the sign of the determinant of a matrix with these vectors as columns.

This determinant is positive at θ=0, and negative at θ=pi, so must be 0 somewhere along the way, so at that point all four must be linearly dependent (in the same 3d subspace).

But in fact this means they must all be in a 2d subspace. Why? Well, v is orthogonal to Rot(v), and also to i*v. If Rot(i*v) is confined to be in the same 3d subspace as {v, Rot(v), i*v}, then because it's orthogonal to i*v it must be in the same plane as {v, Rot(v)}.

But that means i*v must be in that same plane as well.

(Surely there's a quicker way to see that...)

So somewhere along the way, the plane of (v, Rot(v)) is the same as the plane of (v, i*v). At that point, v is an eigenvector.

I think this works. Do you find it satisfying? Needlessly confusing? Is there a way to clean it into a more succinct view?

*er, read that as why a 2d _complex_ vector space has a richer structure than a 4d real vector space.