## A Hermitian Matrix Has Real Eigenvalues

When I studied math, I tended to find myself more interested in the “continuous” side of things rather than the discrete. This is calculus and analysis and such, in contrast to things like logic, abstract algebra, number theory, graphs and other things where everything is rather chunky. I suppose I was largely motivated by the use of calculus in physics, which was usually my main focus. It’s easy to see that calculus and continuous functions are essential when studying physics. It’s not as easy to argue for number theory’s direct applicability to an engineer.

Sometimes I feel a bit ashamed about this interest. Calculus and analysis are the next logical steps after real number algebra. One could argue I didn’t allow myself to expand my horizons~~! But, whatever, who cares. They’re still cool. OK? THEY’RE COOL.

It’s very easy to claim that you’re interested in something. It’s almost as easy to hear someone talk about how they’re a fan and try to call them out on it by asking about some trivia. This is often the crux of an argument in the whole “Fake Geek Girl” and Gamergate things.  Similarly, sometimes, it feels like everyone around me is nuts about Local Football Home Team, and I often find myself skeptical of the “purity” of their fanaticism. I need to remind myself that someone can enjoy something and not know everything about it. You can be interested in Watching Football Match and not know the names of everyone (or even anyone) on the team.

The same is true for something like math. It had better be true, since there’s always another thing we could define or discover, and all of the fields already developed aren’t completely understood by a single person. It’s fine to be more interested in one thing rather than another. If you take that too far, you’d end up criticizing people for not being interested in every single thing in existence equally.

It’s all right to wonder if I should look into a certain topic, though. A few years ago, a colleague teaching introductory calculus to high school seniors mentioned that they were working on delta-epsilon proofs in the class. This blew me away! The concept of a limit is usually introduced in a pre-calculus class, or the beginning of a calculus class. I am under the impression that it’s usually a matter of, “yeah, this function gets really close to this point, but doesn’t actually hit it,” and that’s all a student really needs to know until they do some analysis. A delta-epsilon definition is a way to formally define a limit, so there is no uncertainty as to what is actually happening. “Gets really close” ends up being defined exactly — basically, it says, “For this function $f(x)$, gimme any number bigger than zero, even a SUPER tiny number, and I can give you a distance away from $x=c$ such that $latex$f(x)$is never further from a limiting value L than your number.” Okay, maybe that is not super enlightening. But on a side note, it’s fun to think about how much like a playground taunt that is. I was ready to argue that his students didn’t need to bother with delta-epsilon proofs, that they could learn to work with the fuzzy idea of a limit first and get along just fine in calculus, just as I had. But, I did start to doubt myself. Should I have learned the definition of a limit before trying to use it, in my hubris? In retrospect: no, that’s silly. Looking at epsilon-delta definition, I realize it would have taken me ages to get through it, taking away from valuable time spent thinking about the applications of calculus. But, that feeling is there. Should I have known about this? What does this have to do with Hermitian matrices, you demand, grabbing me by the shoulders and shaking, while my cravat soaks up your spittle. I had this same feeling this week, when thinking about a topic in linear algebra and physics. In quantum mechanics, matrices are used extensively to describe certain kinds of actions that could be done to a particle or a system. One of those actions could be to take a measurement, such as the amount of energy in the system. If you call your matrix H and your particle current state $\psi$, then you could represent a measurement of how much energy the system has by multiplying them together. $H\psi=\lambda\psi$ When you multiply them together, you can get a single number $\lambda$ times the particle state $\psi$ as a result. If measured the energy of the particle, then the value of $\lambda$ is that energy. You can’t just divide both sides by $\psi$ because a particle’s state is a vector of many possibilities, and division by vectors rather than numbers doesn’t mean a whole lot here. (You can multiply both sides by a vector transpose and get into something called Dirac notation, but don’t bother with that now.) The number $\lambda$ and the state$\psi$are called eigenvalues and eigenvectors of… Is this worth describing? If you don’t know this, it might be incomprehensible. It turns out that if H is Hermitian, meaning, it is self-adjoint: $H_{ij} = \bar{H_{ji}}$, then it always has real eigenvalues (as opposed to$\lambda$being a complex number). Physicists interpret this to mean the matrix definitely represents a physical measurement. Hermitian matrices aren’t the only ones with real eigenvalues, but they also have a property that lets you be sure you’ve measured your particle as being in a certain state. I’ve seen proofs that Hermitian matrices have real eigenvalues. Here are a couple. These start by assuming there is some eigenvalue/eigenvector pair, and using the fact that a vector magnitude is real at some point. Finding the eigenvalues of a matrix is a matter of solving an equation which involves a determinant: $\det(H-\lambda I) = 0$, where I is the matrix identity. I thought, could I use an expanded determinant to show that the eigenvalues have to be real? This isn’t so bad with a 2×2 matrix. With $H = \left[\begin{array}{cc} a & b+ci \\ b-ci & d\end{array}\right]$, the characteristic equation is $0 = \det(H-\lambda I)$ $= \left| \begin{array}{cc} a-\lambda & b+ci \\ b-ci & d-\lambda\end{array}\right|$ $= (a-\lambda)(d-\lambda) - (b^2 +c^2)$ $= \lambda^2 - (a+d)\lambda +ad - (b^2+c^2)$ You can show the two solutions for $\lambda$ have to be real by shoving all these numbers in the quadratic formula. The discriminant (remember that?) is positive because it ends up being a sum of squares. I’m bored. My thought after this point was to use mathematical induction. We’ve shown that an $n\times n$ Hermitian matrix has real eigenvalues. Let’s show that an $(n+1) \times (n+1)$ one does as a consequence. Maybe this is doable, but I haven’t done it. It occurred to me that all the cofactors of diagonal entries in a Hermitian matrix would be themselves Hermitian, and a proof by induction would rest on this. $H = \left[ \begin{array}{cccc}h_{11} & h_{12} & \dots & h_{1,n+1}\\ h_{21} & h_{22} & & h_{2,n+1} \\ \vdots & & \ddots & \vdots \\ h_{n+1,1} & \dots && h_{n+1,n+1} \end{array}\right]$ $= \left[ \begin{array}{cccc}h_{11} & h_{12} & \dots & h_{1,n+1}\\\\ \overline{h_{12}} & h_{22} & & h_{2,n+1} \\\\ \vdots & & \ddots & \vdots \\\\\overline{h_{1,n+1}} & \dots && h_{n+1,n+1} \end{array}\right]$ My thought was… can you construct a determinant using only cofactors of the diagonal entries? This turned out to be not helpful in an embarassing way. I asked myself, can you calculate a determinant by expanding over a diagonal, rather than a row or column? I was able to convince myself no, but the fact that I considered it at all seemed messed up. Shouldn’t that be something I should know by heart about determinants? Similar to a student using limits without knowing the delta-epsilon definition, I realized that I don’t have a solid grasp of what determinants even are, although I’ve used them extensively. It felt like I was missing a huge part of my mathematics education. I don’t think I had ever proven any of the properties of determinants I had used in a linear algebra course. I didn’t even know how to define a determinant. In high school, we learned how to calculate a 2×2 determinant. We then learned how to calculate determinants for larger matrices, using cofactors (although we didn’t use that word). But I didn’t (and still, don’t really) know what it was. This doesn’t look unusual. I’ve got three books on linear algebra next to my desk here. DeFranza and Gagliardi start by defining a 2×2 determinant as just ad-bc. It then tells how to calculate a 3×3 determinant, and then how to calculate larger determinants using cofactors. This seems in line with how I was taught (although this isn’t a coincidence. I took Jim DeFranza’s linear algebra class). The useful properties of determinants come later. Zelinsky starts off (on page 166 of 266!) with an entire chapter on all of the algebraic properties we’d like determinants to have. It waits 11 pages to actually give explicit formulas for 2×2, then 3×3 matrices. It isn’t until after that that it gives something that feels like a definition. Kolman starts with this definition right away: Let be an x n matrix. We define the determinant of A by $|\mathbf{A}| = \Sigma (\pm) a_{1j_1}a_{2j2}\dots a_{nj_n}$ where the summation ranges over all permutations $j_1j_2\dots j_n$ of the set $S = \{1,2,\dots, n\}$. The sign is taken as + or – according to whether the permutation is even or odd. Woah. This seems like something fundamental. I had only known determinants as, effectively, recurrence relations. This is a closed form statement of the matrix determinant. Why don’t I recognize this? Really, though, I can see why this might not be commonly taught. It’s even more cumbersome. But it feels like I missed the nuts and bolts of determinants while using them for so long. That definition seems ripe for a Levi-Cevita symbol. It’s probably not worth making most people trudge through millions of subscripts. That’s sort of the MO of mathematics, right? You make an easy-to-write definition or theorem, and then use that to save time and energy. Maybe I’ll try to show Hermitian matrices have real eigenvalues using that definition. Descartes’ rule of signs or Sturm’s theorem might help. But I’m sleepy. So maybe later. Advertisements ## The Martian Tripod Problem and Transcendental Functions I thought this week about a problem I originally considered about ten years ago. I imagined a source of a laser beam, mounted high up above the ground, shining straight down, and allowed to rotate upwards at a constant rate until it was shining horizontally. The point at which the laser beam touched the ground would travel from directly below the source to the horizon. The question is, if I know where the source of the laser is, and how quickly it is rotating, can I know exactly where the beam strikes the ground over time? The problem came about, I’m sure, as I was listening to Jeff Wayne’s Musical Version of the War of the Worlds, imagining beams of Martian heat rays mounted to towering tripods sweeping across the hull of the Thunder Child. Trigonometry: What’s the length of a side of a right triangle with a constant height? This is not so bad of a problem when only the geometry is considered. For now let’s call the angle that the laser makes with the “straight down” direction (the vertical) “omega-t”: $\omega t$. With $t$ a length of time since the laser started shining, we can see that $\omega$ is a sort of speed — when I multiply it by a length of time $t$, it gives a total angle, which is like a distance. The product $\omega t$ works the same way that 30 miles per hour times 2 hours is 60 miles. In physics we’d call this speed of rotation $\omega$ the angular speed, or angular velocity if you’re considering rotations in all three dimensions. For now, it doesn’t matter what the value of this speed of rotation is. Setting up the laser at an angle $\omega t$ from the vertical and a height $H$ from the ground, we find it shines at a point $H \sec(\omega t)$ away, and a horizontal distance $H \tan(\omega t)$ to the right. If you’re not so used to working with trig functions, you could get to the image above by first setting up the “classic” trig diagram, with the point a distance (hypotenuse) $H$ away: The above image has all the right ratios, but keeps the hypotenuse constant, not the height. Divide all the lengths by $\cos(\omega t)$, and remember that secant = 1/cosine (by definition). We’re already done. The point at which the laser beam hits the ground is $\bigg( H \tan (\omega t) , 0 \bigg)$ Tangent “blows up” to infinity at$\pi/2\$, which corresponds to the laser shining parallel to the ground. It intersects the flat ground an infinite distance away, at the horizon. Hopefully this agrees with your expectations; tangent is defined to act this way.

The Tripod Problem: Incorporating the speed of light

So that’s not super interesting. The real “tripod problem” was this: Where is the point of intersection if the speed of light isn’t infinite? If it takes some time for the laser beam to travel from the source to the ground, and the laser continues to rotate, then the location where the beam strikes will lag behind the orientation of the laser emitter.

This results in a “floppy” trajectory of the laser beam, drooping down to the ground

The behavior of the point where the laser strikes the ground is very different with the speed of light restriction. It never reaches the horizon in finite time — the behavior for long lengths of time, and far distances, is totally dominated by the speed of light. It should travel along the ground at a speed approaching c.

The way the pointer location moves depends little on the speed of light directly under the source, where the distance to the ground doesn’t depend strongly on the angle, and a wider angle of the laser covers a small length. As the laser approaches the horizontal, the length covered by each small change in angle increases. The light that will eventually strike very far distances looks more like a point source, since the small angles will be covered in a very short length of time. The trajectory of the beam itself, while it’s still in the air, will look more and more like an expanding circle with time.

When I first mentioned the tripod problem to a friend recently, he had the insight of saying that a laser’s point could definitely travel faster than the speed of light. He could shine a laser at one side of the moon, and then the other. A quick enough rotation on a far enough canvas could result in a pointer appearing to travel faster than the speed of light. Remember, this doesn’t violate anything in relativity. No object is traveling faster than light, rather, a series of events in which different photons strike the distant moon are occurring. This situation is very much like that when the tripod laser is pointed nearly downwards. The speed of the pointer is dominated not by the speed of light, but by the rate of the laser rotation because the distance the light has to travel doesn’t change much when the laser is pointed directly downwards (or from one side of the moon to the other).

This suggests that the speed of the pointer, at very far distances from the tripod, would approach from slower speeds if the laser were rotating slowly. But, if the laser were rotating quickly enough, could counterintuitively approach from faster speeds.

Anyway, let’s try to deal with the problem. Take a look at our trig diagram again.

When a certain portion of the laser beam is emitted at a time $t$ (and an angle $\omega t$), it has to travel a distance $H\sec(\omega t)$. Traveling this distance at speed c takes a length of time equal to

$\cfrac{H}{c}\ \sec(\omega t)$.

Any portion of the heat ray travels in a straight line. Although the beam as a whole is curved, we’re still assuming it’s always traveling directly away from the source (no diffraction, etc.). The laser travels in the same direction and therefore strikes the ground at the point

$\big( H\tan(\omega t),0 \big)$

at a later time

$t^\prime = t + \cfrac{H}{c}\ \sec(\omega t)$

This seems great. We have the basis for a complete understanding of the position of the laser pointer (or toasted Edwardian human) given some time. A portion of the laser, emitted at time $t$, will strike the ground a horizontal distance $H\tan(\omega t)$ away not at $t$ but at $t^\prime$ above. This allows us to find the location corresponding to any time of emission in the interval

$0 \leq t < \cfrac{\pi}{2\omega}$.

If we were satisfied with this, the game plan would be to pick a time of emission $t$, determine how long that portion of the beam traveled, and then pair up the resulting $t^\prime$ with the distance $H\tan(\omega t)$.

I’m not satisfied, though. I’d like to get a trajectory of the laser pointer: a location as a function of the actual time $t^\prime$ rather than the time that portion of the laser was emitted, $t$. In order to do this, we’d need to replace the $t$ in the tangent function with an equivalent function of $t^\prime$. In order to do that, we’d need to solve

$t^\prime = t + \cfrac{H}{c}\ \sec(\omega t)$

for $t$. Good luck.

What we’ve got above is a transcendental equation. This means it is not composed of a finite number of additions, subtractions, multiplications and divisions of our variable $t$ and the constants, as well as rational powers of these. In most cases, and I’m pretty sure in this one, we can’t solve a transcendental equation exactly for the input variable. We cannot write $t$ in terms of $t^\prime$.

It seems like the best we could do, if we wanted to create an animation with a step by step progression of the position of the pointer, is to prepare ahead of time. Pick an emission time $t$, find the value of the tangent function to find the distance, find the value of the strike time $t^\prime$, and record that pair. Then do this many, many times to create a table with more values than we expect someone to ask us for. We could find the position as a function of time with as much precision as we wanted, supposing we were willing to put the effort in.

I wanted a closed form solution to the problem, a trajectory $x(t^\prime)$, and it seems more than out of reach. This annoyed me, until a friend (hey, there, buddo!) reminded me that “closed-form” is just a matter of what I’m allowing as a definition. In fact, like I mentioned in the last post, all of the trig functions are themselves transcendental: They can be written as Taylor polynomials, but these are polynomials of infinite length. The secant in the equation above can be estimated using

$1 + \cfrac{1}{2}\ x^2 + \cfrac{5}{24}\ x^4 + \cfrac{61}{720}\ x^6 + \cfrac{277}{8064}\ x^8 +\dots$

The problem with this, though, is that this isn’t much better. It would still take an infinite amount of time to achieve the exact value of secant given most x’s. The only reason I’m more comfortable using this and the other trig functions is because I’ve been trained to use this name for them, and rely on calculators or tables to give me the values whenever I need them. Anyone using a trigonometric function table is benefiting from someone else’s hard work to overprepare. When we use a calculator, we are relying on an estimation that is as precise as the manufacturer (or sometimes the user) dictates. One could make this estimation with a Taylor series, or with a more efficient method, but the calculator still wont give an exact decimal value.

Any single irrational number, whether it is the solution to an algebraic or a transcendental equation, is another instance of this. I’ve gotten used writing things like $\sqrt{2}$ or $\pi$ as representations of numbers with clear definitions. These numbers have exact values, but it’s hopeless for me to try writing them down. In a very definite sense, these numbers elude us. I could write or use them to any finite precision I wanted, with millions and millions of digits, so long as I were willing to come up with and use an efficient algorithm to find them, or if I were were willing to wait or work a very long time, or both. But, I still wouldn’t have the “exact” value, just one that was plenty good enough for whatever application I had in mind.

These examples remind me: it’s convenient to have named functions like “Cosine” to cover a mathematical idea, but we can’t let this name cover up the meaning of that idea. There are an infinite number of angles whose cosine is a transcendental value. We’re able to use cosine because we can always (right?) reach a higher precision than is necessary in a physical application. I’ve gotten used to working with cosine, and mentally separated it from the solution to the tripod problem, because someone gave it a name that I’ve adopted.

So, I guess I should name the solution. Let’s call the composed function

$x(t^\prime) = H\tan\big(\omega t(t^\prime)\big)$

where $t$ and $t^\prime$ are related by

$t^\prime = t + \cfrac{H}{c}\ \sec(\omega t)$

the heat ray function. We could create a huge table for x(t), someone could come up with an efficient algorithm for calculating values of x, and in the future we could use these to invade infinite planes with laser pointers more effectively.