20 ms·

What is the inverse of a vector?

317 points by mkeeter 5y ago

billfruit 5y ago
Is there a book that give comprehensive treatment of euclidean geometry, but using vectors?
- pizza 5y ago
  David Hestenes’s Geometric Algebra for Physicists, maybe?
  - peeterjoot 5y ago
    That one is by Doran and Lasenby. Hestenes wrote: New Foundations for Classical Mechanics.
    - pizza 5y ago
      Ah, my bad, thanks
foobarbazbarfoo 5y ago
love your writing, can anyone recommend blogs like these
- shoto_io 5y ago
  Curious to understand: What do you exactly love about it?
- mferraro89 5y ago
  thanks! the 3blue1brown website has some great written content that you may enjoy: https://www.3blue1brown.com/#lessons https://www.3blue1brown.com/#lessons
sgt101 5y ago
I am curious, how could one apply these insights to inference over sets of vectors generated by creating embeddings over things like photos etc? I understand well the ideas of +/- for things like word2vec, but what would multiplication and inverse mean in this context?
Grustaf 5y ago
Ah, another geometric algebra evangelist? I can't figure out if GA actually adds anything substantial, or if it merely lets us write some equations in a more succinct fashion. But it certainly looks cool. As to vectors, obviously they have inverses, additive inverses. Since vectors don't have multiplication, there is no multiplicative inverse, but if you define new operations on them, well then that operation can have an inverse but that is not really "the inverse of a vector" anymore.
- thaumasiotes 5y ago
  > As vectors, obviously they have inverses, additive inverses. Since vectors don't have multiplication, there is no multiplicative inverse A vector is pretty much by definition also a matrix, and there is a standard way to multiply matrices. You can define several inverses of a vector that way, though you can't define a unique inverse. The standard inner product is of course also an exceptionally typical way to multiply vectors, but the concept of an inverse there doesn't make much sense.
  - monktastic1 5y ago
    > The standard inner product is of course also an exceptionally typical way to multiply vectors, but the concept of an inverse there doesn't make much sense. Not all vector spaces are equipped with an inner product. The point is that you can start with some simple axioms and build these more complicated things (inner product spaces, algebras over a field, geometric algebras, etc.).
    - thaumasiotes 5y ago
      > Not all vector spaces are equipped with an inner product. Any vector space over a field (usually part of the definition of a vector space) is equipped with the standard inner product, because multiplication and addition are part of the definition of a field.
      - Grustaf 5y ago
BlackFly 5y ago
For those who are interested, this sort of algebra would be known as the [Grassman algebra or the exterior algebra](https://en.wikipedia.org/wiki/Exterior_algebra https://en.wikipedia.org/wiki/Exterior_algebra). It becomes much more interesting if you use non-orthonormal bases (or non-euclidean geometry), since then you need to introduce a dual basis and distinguish between contravariant vectors and covariant vectors. When you add derivatives to the mix you end up in differential geometry.
- LotusFunctor 5y ago
  Grassmann algebra is a very important part of it, in fact you can reconstruct it in geometric algebra. More generally though, this algebra would be known as Clifford algebra.
- mathgenius 5y ago
  Yes, this is exterior algebra. It's also interesting to figure out how this works in ambient dimensions other than three. The author has a table of grades: 0 for scalars, 1 for vectors, 2 for "bivectors", 3 for "trivectors", and they count the number of bases for each of these grades as 1 3 3 1. These basis counts are the dimensions of the (vector space of) scalars, vectors, "bivectors", "trivectors". If you go to two ambient dimensions you get 1 2 1, and if you go to four ambient dimensions you get 1 4 6 4 1. It's Pascal's triangle.
  - LotusFunctor 5y ago
    Pascal's triangle, and also with a transparently power-set flavor to it :)
spekcular 5y ago
The article begins: >In this post we will re-invent a form of math that is far superior to the one you learned in school. The ideas herein are nothing short of revolutionary. and concludes: > I firmly believe that in 100 years, Geometric Algebra will be the dominant way of introducing students to mathematical physics. In the same way that Newton's notation for Calculus is no longer the dominant one, or that Maxwell's actual equations for Electromagnetism have been replaced by Heaviside's, textbooks will change because a better system has come along. These claims are wrong. There are three standard notation methods in physics: vectors, tensors, and differential forms. Geometric algebra is, as the article points out, a more powerful version of the usual vector notation. But it is deficient in various ways when compared to tensor notation (for calculations) and differential forms (e.g. if you want to work basis-free). [I'm oversimplifying a bit, but a full discussion is too long for a comment here.] Anyway, geometric algebra is not some esoteric secret. People know about it and have decided not to teach it, because the stuff that's already taught is better. [I picked up this specific phrasing from another user here, knzhou, which I think is a particularly good way of explaining it.]
- topaz0 5y ago
  This may be true in some places, but my undergraduate physics education spent a lot of time on standard Gibbs-style vector calculus. Taylor Classical Mechanics and Griffiths Electrodynamics especially depend on them. Maybe there is a case to be made that first years should start with differential forms, but until that happens I think geometric algebra could be a big improvement.
- csdvrx 5y ago
  > Geometric algebra is, as the article points out, a more powerful version of the usual vector notation. But it is deficient in various ways when compared to tensor notation (for calculations) and differential forms (e.g. if you want to work basis-free). [I'm oversimplifying a bit, but a full discussion is too long for a comment here.] I have no opinion about the claims, but I loved the article, as I quickly saw the gains from this algebra for my very basic needs. After checking out your 2 suggested alternatives, I'm not so convinced they are easier to understand.
- 5y ago
Garlef 5y ago
While the article is written very nicely, It seems that this is written out of a perspective of some missing knowledge. The basic object that the author seems to be interested in is that of an "algebra over a field" (https://en.wikipedia.org/wiki/Algebra_over_a_field https://en.wikipedia.org/wiki/Algebra_over_a_field). Specifically: Invertability of all elements with respect to the multiplication leads to the notion of division algebra and these have been studied for a long time. (https://en.wikipedia.org/wiki/Division_algebra https://en.wikipedia.org/wiki/Division_algebra) When studying math at german universities, alebras are something you'll encounter in your 2nd year (latest; but might already show up in 1st year analysis albeit with a different focus). Implicitly, division algebras show up a lot when students learn about field extensions, galois theory and the algebraic closure of a field (usually 3rd semester). A more general treatment of division algebras is not a common subject, though.
- vanderZwan 5y ago
  > It seems that this is written out of a perspective of some missing knowledge. Well, the author talks about what "we learned in school", not university, so that checks out but only because you two have different audiences in mind.
  - azalemeth 5y ago
    Indeed. I learnt about vector dot and cross products, basic linear algebra (including diagonalisation, simple Markov chains and similar), partial differentiation, grad div and curl, the volume of a parallel piped and "all that Jazz" in high school, as a 17-18 year old. I learnt about the divergence theorem, Stokes's theorem, multivariate integration, integrating factors and higher order ODEs and simple PDEs, Fourier transforms and other integral transformations and similar in the first year of university (studying Physics). This is not uncommon in the UK – but it is also not common either, and depends on exactly what A-level modules you did. My understanding is that it's quite rare to do exactly this in high school in the united states – but there, I think limits are taught much more heavily. I think having a clear, short statement of having "assumed knowledge" somewhere probably helps avoid these issues. (I thought the article was excellent, and beautifully illustrated!)
ur-whale 5y ago
Geometric algebra (Clifford Algebra) unfortunately came late historically. It's a shame, because the whole theory is a very useful (eg for engineering / applied math) superset of linear algebra. I really wish I had learned this first in my undergrad years, would have made a whole bunch of things way clearer from the get go: differential forms tensor calculus linear algebra etc From zero to geo is a very good video introduction to the topic: https://www.youtube.com/watch?v=2hBWCCAiCzQ&list=PLVuwZXwFua-0Ks3rRS4tIkswgUmDLqqRy&index=2 https://www.youtube.com/watch?v=2hBWCCAiCzQ&list=PLVuwZXwFua...
- an1sotropy 5y ago
  Yes, this. Though Zero to Geo is one of the links at the bottom of the article. It is really a shame that article does not clarify that, btw, what we've just derived is a re-derivation of a thing that has already been expressed and named, by Clifford, and well-characterized: https://en.wikipedia.org/wiki/Geometric_algebra https://en.wikipedia.org/wiki/Geometric_algebra Such a bummer to see very slick but very ahistorical articles.
  - mferraro89 5y ago
    Hi. I wrote whole section on the history of GA and what happened and why it isn't already the norm, but I chose to remove it because the article is already far too long, and I don't think that my intended audience (engineers, compsci people, university undergrads) would care about the history. Apologies that wasn't what you would have preferred.
    - an1sotropy 5y ago
      I think the article is great, and I think the interactive illustrations are sweet. Thanks for the taking the time to write it. As an educator, though, when I see presentations of existing ideas that present them as if they were new, I die a little. You're standing on the shoulders of giants whether or not you think so, and whether or not you say so. It's best to figure out who the giants are, and how you're standing on them. When you are up front about the connections to past scholars, you are giving the credit to those scholars that they deserve, and you are strengthening the storyline, and you are setting a good example for the people that look up to you. You can add a few sentences at the bottom saying, if you've gotten this far, congrats, you understand some basics of GA, and then link to other resources about the history and current applications of it.
jeffwass 5y ago
I stopped reading at this paragraph near the top : “In this post we will re-invent a form of math that is far superior to the one you learned in school. The ideas herein are nothing short of revolutionary.”
- mferraro89 5y ago
  why? too clickbaity?
OscarCunningham 5y ago
People might be interested in a similar post I wrote about dividing by a vector (https://oscarcunningham.com/4/dividing-by-a-vector/ https://oscarcunningham.com/4/dividing-by-a-vector/) although I came to a different answer as I was considering arbitrary vector spaces rather than just 3D space.
haunter 5y ago
Wow the interactive 3D illustrations are awesome. Works perfectly on touchscreen, feels very natural.
- mferraro89 5y ago
  thank you! It took a whole month to get the 3D illustrations working well.
deepsun 5y ago
English/American style of explanation fascinates me. First, they show some algebra formulas and mention dot product and cross product. But then they start introducing a definition of a vector! With images! Why, oh why do you need to waste yours and reader's time to introduce basic definitions, if any reader of the article definitely knows that? If they haven't, they wouldn't be able to read the first paragraph at all. PS: Russian style of explanation is more like: "Here's the essence of my idea, maybe with some leading pre-definitions, but definitely without basics. If you are here, you probably is as curious as I am to already know/heard of all the basics." In total, there's more material, because it's easier to write and read it, as author didn't need to explain 101s to PhDs.
- IshKebab 5y ago
  I definitely agree but there's no way this is "English/American style". It's because people have grand plans of making their article/book accessible to everyone, and they start off explaining e.g. what a vector is, but pretty soon realise the don't want to write an entire vector algebra textbook so they seamlessly give up and jump straight into Stoke's theorem or whatever. I read a Synopsys simulator manual that explained what double clicking was.
  - deepsun 5y ago
    Probably. But why would so many people want to make their article/book accessible to everyone. Let's accept the fact that some topics, like vector algebra, are just not that interesting to everyone.
    - fsloth 5y ago
      Vector algebra is at the heart of a fairly large industry - games. I don't think there can be enough of accessible and understandable content from that point of view.
  - mferraro89 5y ago
    Did this article at some point give up and jump straight into something too difficult?
SantiagoQ 5y ago
Semantically, the inverse of a vector is something that has no magnitude nor direction. I wonder what would that look like (mathematically), and what surfaces or fields would it create?
- Rayhem 5y ago
  "A vector is a thing with both magnitude and direction" isn't really a good definition. Cars have both -- an SUV is larger than a sedan, establishing a magnitude, and they obviously point in a direction -- but I don't think anyone would mistake them for pointy arrow vectors. If you use the more rigorous definition that a vector is an element of a space that obeys the vector space axioms it becomes easier to invert a vector "semantically" (a thing that doesn't obey the axioms) but quite a bit less useful. Cats don't obey the axioms, nor do punctuation marks.
jefftk 5y ago
> The similarities are so striking that we might think of them as "pseudovpseudovectors". But I won't write them this way because I think that obscures their true nature. Written this way it looks like a bivector only encapsulates three degrees of freedom! > Instead, I will use: ... Because it forces us to remember what those coefficients are attached to. Knowing that a bivector contains five degrees of freedom, can you figure out what the other two describe? I'm confused here and don't understand why they keep saying a bivector has five degrees of freedom. If you can uniquely identify one with three scalar coefficients, doesn't it only have three degrees of freedom?
- tgb 5y ago
  Yes, only three. As defined, two bivectors are equal if their areas are equal and if their oriented planes are equal. Therefore two more degrees of freedom are absorbed by taking rotations of the two vectors in the plane.Along with the rescaling the author noted, we're down to three from six.
  - jefftk 5y ago
    That makes complete sense to me. But then later on they say "The output is a Geometric with a scalar component s and a bivector component ⇒c, which has 1 + 5 = 6 degrees of freedom so this system is not lossy! It should permit an unambiguous inversion operation!" If a bivector only has 3 degrees of freedom then the total is 4, which seems like it would be lossy?
    - tgb 5y ago
      I was also wondering this. But note that x^y is always perpendicular to x, so really only has two degrees of freedom while you need three to recover y (knowing x). Add in the dot product part to make up for it.
  - kevinwang 5y ago
    Interesting, I was confused about the same thing. So the author is not correct when they say a bivector has 5 degrees of freedom?
codeflo 5y ago
The writing is cute and the animations are nice, but none of it makes any sense. I stopped reading at > It is important to remember that bivectors have a certain redundancy built into them in the sense that s a ⃗ ∧ b ⃗ = a ⃗ ∧ s b ⃗ s a ∧ b = a ∧s b . We can write them using 6 numbers or 3 numbers, but they actually convey 5 degrees of freedom. Three (real) numbers have three degrees of freedom, by definition. (And nothing about complex numbers was mentioned.) Is this a parody I don’t get? I feel like I have wasted ten minutes on nonsense.
- tgb 5y ago
  You're right about that being wrong, and the author makes the same mistake consistently, but otherwise it looks correct. Some steps have details elided where it maybe should have been noted that things were being skipped, but with correct results. I think it's wonderfully written and a great exposition.
  - codeflo 5y ago
    Thanks, that helps. When I notice errors in the stuff I already know about, it find it hard to trust the other information that’s new to me.
  - mferraro89 5y ago
    hey, if you have time to detail those mistakes I'd be happy to fix them in the text. Can you email me at mattferraro.dev@gmail.com
- Grustaf 5y ago
  He's not talking about the triplet or sextet, he means that a bivector has 5 degrees of freedom. That isn't correct either though, the basis consists of three unit bivectors, so they have at most 3 degrees of freedom.
- mferraro89 5y ago
  author here. I was mistaken about the 5 degrees of freedom bit. Bivectors have three. I'll fix the text tonight. I'm sorry you wasted ten minutes on my nonsense.
admin786 5y ago
THanks you ya
admin786 5y ago
Nice Article !
foxhop 5y ago
Here is something similar I wrote a long while back as notes for my future selfs: https://www.foxhop.net/vector-math-for-video-games https://www.foxhop.net/vector-math-for-video-games
- aktenlage 5y ago
  In what way would that be similar? The OP performs a theoretical derivation of geometrical algebra, you wrote a well documented python class with the most basic operations on 2d vectors.
anotheraccount9 5y ago
"It turns out that inverting a vector on its own isn't well defined."
fractal618 5y ago
The inverse of a any positive or negative vector with an amplitude greater than zero points directly into your soul relative to it's original amplitude and the how many regrets you have.
fouronnes3 5y ago
As a programmer it seems to me that the number one problem of math notation is that it's weakly typed. There's abuse and reuse of notation everywhere, which makes learning it needlessly difficult. I want a strongly typed fork of math notation. 90% of existing math notation would just be laughed at if it had to go through code review.
- Grustaf 5y ago
  That's a very common criticism, but I don't think mathematics would work if you insisted on being 100% explicit all the time. Clear and short notation, that is just unambiguous enough, is a very important factor, without it books wouldn't just be much longer, I'm not sure we'd even be able to understand it.
  - civilized 5y ago
    This was what Bourbaki's Elements and Whitehead/Russell's Principia Mathematica were about. These books are admired and influential but very few people actually read them. As you might expect, they're too long. They're for giving a different perspective to people who have already achieved the highest levels of sophistication in math.
- civilized 5y ago
  Ironically, this post is an abuse of the concept of "weak typing". There's nothing "weakly typed" about, say, the plus sign being used to add both numbers and sets, or a dot being used for both multiplication of numbers and the dot product of vectors. It just means those symbols dispatch based on the types of their arguments, which is perfectly consistent with strong typing (cf. the Julia language). The situation that leads to weak typing in computer algorithms -- when you get data from a file or another process and don't know in advance what type it's going to be -- is basically non-existent in blackboard mathematics. Rigorous mathematical papers always tell you what set a variable belongs to when it is introduced, as well as the domain and co-domain of any functions that are defined. This is the blackboard equivalent of strong typing.
  - 5y ago
da39a3ee 5y ago
> A scalar is a point on a number line. This is going to confuse readers. A point on a number line is a 1D vector; in other words it is a unit vector pointing along that number line, multiplied by something which scales its length. It’s the latter dimensionless and directionless quantity that’s the scalar.
- adrian_b 5y ago
  Nitpicking: a more correct view is that the set of points on a straight line is an affine space, so the points are neither scalars nor vectors, but elements of an affine space. The set of translations of the straight line is a vector space a.k.a. a linear space. So the vectors are the classes of equivalences of the differences between 2 points on the straight line (i.e. the differences between 2 pairs of points, where the distances are the same, are equivalent and they determine the same vector). While the vectors are classes of equivalence of the differences between 2 points, the scalars are classes of equivalence of the quotients of 2 (collinear) vectors, i.e. a scalar is the ratio between the signed magnitudes of 2 collinear vectors. If you choose a point on the straight line as the origin, you can choose as a representative of each class of equivalence that corresponds to a vector, the vector corresponding to the origin point together with another point. This gives a bijective mapping between vectors and those second points. If now you also choose a vector as being the unit vector, which will correspond with a second point besides the origin point, together with the origin point, then you can choose as a representative for each class of equivalence corresponding to a scalar the ratio between a vector and the unit vector, which will correspond to a third point, besides the origin and the point corresponding to the unit vector. So you obtain a bijective mapping between scalars and those third points. Because on a straight line there are bijective mappings between points, vectors and scalars (after choosing 1 origin point and a 2nd point as the extremity of a unit vector), they can be used interchangeably in most contexts, but it would be good to remember that all 3 are in fact different mathematical entities.
  - da39a3ee 5y ago
    Thanks for that! It was extremely clear.
felipeqq2 5y ago
A rotcev
- wisienkas 5y ago
  My thought exactly
unixhero 5y ago
Somehow I got that first math explanation.
gpderetta 5y ago
Great article. I had a similar kind of revelation when I learned about generalized linear models after failing to understand all the various statistical tests.
necovek 5y ago
I love the way this was presented (others have pointed out flaws with the article already). I'd appreciate a post from Matt Ferraro on how this is built. Bonus points for including nice syntax-highlighted code "widget" for a cross between maths/programming.
steve76 5y ago
> What is the Inverse of a Vector? Manifolds, right? A vector takes a value, adds dimensions, and expands the value by spatial definition. A manifold takes a value, adds dimensions, and constrains the value by surface projection. Vectors are values. Even though they are made up of a bunch of numbers, they are a value like any other input or output. If you have a vector function, continuity works with vector sums just like continuity between input and output. Maxwell uses this, from Gibbs, to describe electromagnetic fields so he could have an equation for empty space and get the speed of light. Sometimes it's more difficult to find a reason for discontinuity, and are forced to assume continuity. Lie algebra is when you want to describe something with symmetries. Whatever you are describing, such as the insides of an atom, can't be described with exact values, so you use "what things are symmetric" and from there can get a differential equation.
amatic 5y ago
I would say the title should be "reciprocal" of a vector. Right? The inverse of a function (f^-1) has an unfortunate notation equality with the reciprocal (x^-1), or multiplicatory inverse. Or am I wrong?
debbiedowner 5y ago
Doesn't seem self consistent. He defines the ab multiplication as dot product plus "extrusion"/bivector (which seems simpler to call convex combinations of 0,a,b,a+b). Then he says aa is a scalar, presumably because the "extrusion" is 0, but you can't have this identity be 0. Just because it's degenerate does not mean it's 0. And the "extrusion" while not a plane is a line in his definition.
fallingfrog 5y ago
It occurs to me that some of these axioms depend on how many dimensions the space has- in 4 dimensions, a vector would have 4 components, a bivector would have 6, a trivector would have 4, and a quadvector would have 1. And so on, in accordance with Pascal’s triangle.
pmarreck 5y ago
Would this potentially enable generalized matrix inversion?
cannabis_sam 5y ago
I would have loved to implement this in Haskell as an exercise in uni!
portpecos 5y ago
I'm taking a first semester physics course right now, and we're learning about Torque and Angular Momentum. I just finished a calculus course last semester. Can someone tell me how I would use τ=r∧F on a physics problem for Torque?
- ajkjk 5y ago
  You wouldn't really. It's the same concept as `τ = r × F`. the only difference is that it is useful to think of the 'type' of the output as being a bivector instead of a vector -- there's no sense in which it points 'out of the plane'; rather, it is a single vector in the vector space of (planes), with the same magnitude as r × F. The distinct gets a little more useful when you start dealing with covariance under coordinate transformations. There it becomes more meaningful, because the _vector_ given by r x F doesn't transform the same way as their cross product should. For an obvious example of why this is true: suppose r=x and F=y. Then r × F = z. If you change coordinates by mapping z -> 2z, then you would be doubling the torque that you computed .. which is wrong; the torque is unchanged. The bivector x^y is correctly unchanged by z -> 2z. Currently in physics courses (usually not until more advanced mechanics or relativity) the resolution to this is to wave ones' hands and declare that, no, torque is a 'pseudovector'. But it is really much easier to think about if you type it as a bivector in the first place.
bsedlm 5y ago
where has this been all my life!? on the plus side I found it now...
DreamScatter 5y ago
This article is really about geometric algebra, check out my geometric algebra software https://github.com/chakravala/Grassmann.jl https://github.com/chakravala/Grassmann.jl

317 points by mkeeter 5y ago

billfruit 5y ago

Is there a book that give comprehensive treatment of euclidean geometry, but using vectors?

pizza 5y ago
David Hestenes’s Geometric Algebra for Physicists, maybe?
- peeterjoot 5y ago
  That one is by Doran and Lasenby. Hestenes wrote: New Foundations for Classical Mechanics.
  - pizza 5y ago
    Ah, my bad, thanks

foobarbazbarfoo 5y ago

love your writing, can anyone recommend blogs like these

shoto_io 5y ago
Curious to understand: What do you exactly love about it?
mferraro89 5y ago
thanks! the 3blue1brown website has some great written content that you may enjoy: https://www.3blue1brown.com/#lessons https://www.3blue1brown.com/#lessons

sgt101 5y ago

I am curious, how could one apply these insights to inference over sets of vectors generated by creating embeddings over things like photos etc? I understand well the ideas of +/- for things like word2vec, but what would multiplication and inverse mean in this context?

Grustaf 5y ago

Ah, another geometric algebra evangelist? I can't figure out if GA actually adds anything substantial, or if it merely lets us write some equations in a more succinct fashion. But it certainly looks cool. As to vectors, obviously they have inverses, additive inverses. Since vectors don't have multiplication, there is no multiplicative inverse, but if you define new operations on them, well then that operation can have an inverse but that is not really "the inverse of a vector" anymore.

thaumasiotes 5y ago
> As vectors, obviously they have inverses, additive inverses. Since vectors don't have multiplication, there is no multiplicative inverse A vector is pretty much by definition also a matrix, and there is a standard way to multiply matrices. You can define several inverses of a vector that way, though you can't define a unique inverse. The standard inner product is of course also an exceptionally typical way to multiply vectors, but the concept of an inverse there doesn't make much sense.
- monktastic1 5y ago
  > The standard inner product is of course also an exceptionally typical way to multiply vectors, but the concept of an inverse there doesn't make much sense. Not all vector spaces are equipped with an inner product. The point is that you can start with some simple axioms and build these more complicated things (inner product spaces, algebras over a field, geometric algebras, etc.).
  - thaumasiotes 5y ago
    > Not all vector spaces are equipped with an inner product. Any vector space over a field (usually part of the definition of a vector space) is equipped with the standard inner product, because multiplication and addition are part of the definition of a field.
    - Grustaf 5y ago

BlackFly 5y ago

For those who are interested, this sort of algebra would be known as the [Grassman algebra or the exterior algebra](https://en.wikipedia.org/wiki/Exterior_algebra https://en.wikipedia.org/wiki/Exterior_algebra). It becomes much more interesting if you use non-orthonormal bases (or non-euclidean geometry), since then you need to introduce a dual basis and distinguish between contravariant vectors and covariant vectors. When you add derivatives to the mix you end up in differential geometry.

LotusFunctor 5y ago
Grassmann algebra is a very important part of it, in fact you can reconstruct it in geometric algebra. More generally though, this algebra would be known as Clifford algebra.
mathgenius 5y ago
Yes, this is exterior algebra. It's also interesting to figure out how this works in ambient dimensions other than three. The author has a table of grades: 0 for scalars, 1 for vectors, 2 for "bivectors", 3 for "trivectors", and they count the number of bases for each of these grades as 1 3 3 1. These basis counts are the dimensions of the (vector space of) scalars, vectors, "bivectors", "trivectors". If you go to two ambient dimensions you get 1 2 1, and if you go to four ambient dimensions you get 1 4 6 4 1. It's Pascal's triangle.
- LotusFunctor 5y ago
  Pascal's triangle, and also with a transparently power-set flavor to it :)

spekcular 5y ago

The article begins: >In this post we will re-invent a form of math that is far superior to the one you learned in school. The ideas herein are nothing short of revolutionary. and concludes: > I firmly believe that in 100 years, Geometric Algebra will be the dominant way of introducing students to mathematical physics. In the same way that Newton's notation for Calculus is no longer the dominant one, or that Maxwell's actual equations for Electromagnetism have been replaced by Heaviside's, textbooks will change because a better system has come along. These claims are wrong. There are three standard notation methods in physics: vectors, tensors, and differential forms. Geometric algebra is, as the article points out, a more powerful version of the usual vector notation. But it is deficient in various ways when compared to tensor notation (for calculations) and differential forms (e.g. if you want to work basis-free). [I'm oversimplifying a bit, but a full discussion is too long for a comment here.] Anyway, geometric algebra is not some esoteric secret. People know about it and have decided not to teach it, because the stuff that's already taught is better. [I picked up this specific phrasing from another user here, knzhou, which I think is a particularly good way of explaining it.]

topaz0 5y ago
This may be true in some places, but my undergraduate physics education spent a lot of time on standard Gibbs-style vector calculus. Taylor Classical Mechanics and Griffiths Electrodynamics especially depend on them. Maybe there is a case to be made that first years should start with differential forms, but until that happens I think geometric algebra could be a big improvement.
csdvrx 5y ago
> Geometric algebra is, as the article points out, a more powerful version of the usual vector notation. But it is deficient in various ways when compared to tensor notation (for calculations) and differential forms (e.g. if you want to work basis-free). [I'm oversimplifying a bit, but a full discussion is too long for a comment here.] I have no opinion about the claims, but I loved the article, as I quickly saw the gains from this algebra for my very basic needs. After checking out your 2 suggested alternatives, I'm not so convinced they are easier to understand.
5y ago

Garlef 5y ago

While the article is written very nicely, It seems that this is written out of a perspective of some missing knowledge. The basic object that the author seems to be interested in is that of an "algebra over a field" (https://en.wikipedia.org/wiki/Algebra_over_a_field https://en.wikipedia.org/wiki/Algebra_over_a_field). Specifically: Invertability of all elements with respect to the multiplication leads to the notion of division algebra and these have been studied for a long time. (https://en.wikipedia.org/wiki/Division_algebra https://en.wikipedia.org/wiki/Division_algebra) When studying math at german universities, alebras are something you'll encounter in your 2nd year (latest; but might already show up in 1st year analysis albeit with a different focus). Implicitly, division algebras show up a lot when students learn about field extensions, galois theory and the algebraic closure of a field (usually 3rd semester). A more general treatment of division algebras is not a common subject, though.

vanderZwan 5y ago
> It seems that this is written out of a perspective of some missing knowledge. Well, the author talks about what "we learned in school", not university, so that checks out but only because you two have different audiences in mind.
- azalemeth 5y ago
  Indeed. I learnt about vector dot and cross products, basic linear algebra (including diagonalisation, simple Markov chains and similar), partial differentiation, grad div and curl, the volume of a parallel piped and "all that Jazz" in high school, as a 17-18 year old. I learnt about the divergence theorem, Stokes's theorem, multivariate integration, integrating factors and higher order ODEs and simple PDEs, Fourier transforms and other integral transformations and similar in the first year of university (studying Physics). This is not uncommon in the UK – but it is also not common either, and depends on exactly what A-level modules you did. My understanding is that it's quite rare to do exactly this in high school in the united states – but there, I think limits are taught much more heavily. I think having a clear, short statement of having "assumed knowledge" somewhere probably helps avoid these issues. (I thought the article was excellent, and beautifully illustrated!)

ur-whale 5y ago

Geometric algebra (Clifford Algebra) unfortunately came late historically. It's a shame, because the whole theory is a very useful (eg for engineering / applied math) superset of linear algebra. I really wish I had learned this first in my undergrad years, would have made a whole bunch of things way clearer from the get go: differential forms tensor calculus linear algebra etc From zero to geo is a very good video introduction to the topic: https://www.youtube.com/watch?v=2hBWCCAiCzQ&list=PLVuwZXwFua-0Ks3rRS4tIkswgUmDLqqRy&index=2 https://www.youtube.com/watch?v=2hBWCCAiCzQ&list=PLVuwZXwFua...

an1sotropy 5y ago
Yes, this. Though Zero to Geo is one of the links at the bottom of the article. It is really a shame that article does not clarify that, btw, what we've just derived is a re-derivation of a thing that has already been expressed and named, by Clifford, and well-characterized: https://en.wikipedia.org/wiki/Geometric_algebra https://en.wikipedia.org/wiki/Geometric_algebra Such a bummer to see very slick but very ahistorical articles.
- mferraro89 5y ago
  Hi. I wrote whole section on the history of GA and what happened and why it isn't already the norm, but I chose to remove it because the article is already far too long, and I don't think that my intended audience (engineers, compsci people, university undergrads) would care about the history. Apologies that wasn't what you would have preferred.
  - an1sotropy 5y ago
    I think the article is great, and I think the interactive illustrations are sweet. Thanks for the taking the time to write it. As an educator, though, when I see presentations of existing ideas that present them as if they were new, I die a little. You're standing on the shoulders of giants whether or not you think so, and whether or not you say so. It's best to figure out who the giants are, and how you're standing on them. When you are up front about the connections to past scholars, you are giving the credit to those scholars that they deserve, and you are strengthening the storyline, and you are setting a good example for the people that look up to you. You can add a few sentences at the bottom saying, if you've gotten this far, congrats, you understand some basics of GA, and then link to other resources about the history and current applications of it.

jeffwass 5y ago

I stopped reading at this paragraph near the top : “In this post we will re-invent a form of math that is far superior to the one you learned in school. The ideas herein are nothing short of revolutionary.”

mferraro89 5y ago
why? too clickbaity?

OscarCunningham 5y ago

People might be interested in a similar post I wrote about dividing by a vector (https://oscarcunningham.com/4/dividing-by-a-vector/ https://oscarcunningham.com/4/dividing-by-a-vector/) although I came to a different answer as I was considering arbitrary vector spaces rather than just 3D space.

haunter 5y ago

Wow the interactive 3D illustrations are awesome. Works perfectly on touchscreen, feels very natural.

mferraro89 5y ago
thank you! It took a whole month to get the 3D illustrations working well.

deepsun 5y ago

English/American style of explanation fascinates me. First, they show some algebra formulas and mention dot product and cross product. But then they start introducing a definition of a vector! With images! Why, oh why do you need to waste yours and reader's time to introduce basic definitions, if any reader of the article definitely knows that? If they haven't, they wouldn't be able to read the first paragraph at all. PS: Russian style of explanation is more like: "Here's the essence of my idea, maybe with some leading pre-definitions, but definitely without basics. If you are here, you probably is as curious as I am to already know/heard of all the basics." In total, there's more material, because it's easier to write and read it, as author didn't need to explain 101s to PhDs.

IshKebab 5y ago
I definitely agree but there's no way this is "English/American style". It's because people have grand plans of making their article/book accessible to everyone, and they start off explaining e.g. what a vector is, but pretty soon realise the don't want to write an entire vector algebra textbook so they seamlessly give up and jump straight into Stoke's theorem or whatever. I read a Synopsys simulator manual that explained what double clicking was.
- deepsun 5y ago
  Probably. But why would so many people want to make their article/book accessible to everyone. Let's accept the fact that some topics, like vector algebra, are just not that interesting to everyone.
  - fsloth 5y ago
    Vector algebra is at the heart of a fairly large industry - games. I don't think there can be enough of accessible and understandable content from that point of view.
- mferraro89 5y ago
  Did this article at some point give up and jump straight into something too difficult?

SantiagoQ 5y ago

Semantically, the inverse of a vector is something that has no magnitude nor direction. I wonder what would that look like (mathematically), and what surfaces or fields would it create?

Rayhem 5y ago
"A vector is a thing with both magnitude and direction" isn't really a good definition. Cars have both -- an SUV is larger than a sedan, establishing a magnitude, and they obviously point in a direction -- but I don't think anyone would mistake them for pointy arrow vectors. If you use the more rigorous definition that a vector is an element of a space that obeys the vector space axioms it becomes easier to invert a vector "semantically" (a thing that doesn't obey the axioms) but quite a bit less useful. Cats don't obey the axioms, nor do punctuation marks.

jefftk 5y ago

> The similarities are so striking that we might think of them as "pseudovpseudovectors". But I won't write them this way because I think that obscures their true nature. Written this way it looks like a bivector only encapsulates three degrees of freedom! > Instead, I will use: ... Because it forces us to remember what those coefficients are attached to. Knowing that a bivector contains five degrees of freedom, can you figure out what the other two describe? I'm confused here and don't understand why they keep saying a bivector has five degrees of freedom. If you can uniquely identify one with three scalar coefficients, doesn't it only have three degrees of freedom?

tgb 5y ago
Yes, only three. As defined, two bivectors are equal if their areas are equal and if their oriented planes are equal. Therefore two more degrees of freedom are absorbed by taking rotations of the two vectors in the plane.Along with the rescaling the author noted, we're down to three from six.
- jefftk 5y ago
  That makes complete sense to me. But then later on they say "The output is a Geometric with a scalar component s and a bivector component ⇒c, which has 1 + 5 = 6 degrees of freedom so this system is not lossy! It should permit an unambiguous inversion operation!" If a bivector only has 3 degrees of freedom then the total is 4, which seems like it would be lossy?
  - tgb 5y ago
    I was also wondering this. But note that x^y is always perpendicular to x, so really only has two degrees of freedom while you need three to recover y (knowing x). Add in the dot product part to make up for it.
- kevinwang 5y ago
  Interesting, I was confused about the same thing. So the author is not correct when they say a bivector has 5 degrees of freedom?

codeflo 5y ago

The writing is cute and the animations are nice, but none of it makes any sense. I stopped reading at > It is important to remember that bivectors have a certain redundancy built into them in the sense that s a ⃗ ∧ b ⃗ = a ⃗ ∧ s b ⃗ s a ∧ b = a ∧s b . We can write them using 6 numbers or 3 numbers, but they actually convey 5 degrees of freedom. Three (real) numbers have three degrees of freedom, by definition. (And nothing about complex numbers was mentioned.) Is this a parody I don’t get? I feel like I have wasted ten minutes on nonsense.

tgb 5y ago
You're right about that being wrong, and the author makes the same mistake consistently, but otherwise it looks correct. Some steps have details elided where it maybe should have been noted that things were being skipped, but with correct results. I think it's wonderfully written and a great exposition.
- codeflo 5y ago
  Thanks, that helps. When I notice errors in the stuff I already know about, it find it hard to trust the other information that’s new to me.
- mferraro89 5y ago
  hey, if you have time to detail those mistakes I'd be happy to fix them in the text. Can you email me at mattferraro.dev@gmail.com
Grustaf 5y ago
He's not talking about the triplet or sextet, he means that a bivector has 5 degrees of freedom. That isn't correct either though, the basis consists of three unit bivectors, so they have at most 3 degrees of freedom.
mferraro89 5y ago
author here. I was mistaken about the 5 degrees of freedom bit. Bivectors have three. I'll fix the text tonight. I'm sorry you wasted ten minutes on my nonsense.

admin786 5y ago

THanks you ya

admin786 5y ago

Nice Article !

foxhop 5y ago

Here is something similar I wrote a long while back as notes for my future selfs: https://www.foxhop.net/vector-math-for-video-games https://www.foxhop.net/vector-math-for-video-games

aktenlage 5y ago
In what way would that be similar? The OP performs a theoretical derivation of geometrical algebra, you wrote a well documented python class with the most basic operations on 2d vectors.

anotheraccount9 5y ago

"It turns out that inverting a vector on its own isn't well defined."

fractal618 5y ago

The inverse of a any positive or negative vector with an amplitude greater than zero points directly into your soul relative to it's original amplitude and the how many regrets you have.

fouronnes3 5y ago

As a programmer it seems to me that the number one problem of math notation is that it's weakly typed. There's abuse and reuse of notation everywhere, which makes learning it needlessly difficult. I want a strongly typed fork of math notation. 90% of existing math notation would just be laughed at if it had to go through code review.

Grustaf 5y ago
That's a very common criticism, but I don't think mathematics would work if you insisted on being 100% explicit all the time. Clear and short notation, that is just unambiguous enough, is a very important factor, without it books wouldn't just be much longer, I'm not sure we'd even be able to understand it.
- civilized 5y ago
  This was what Bourbaki's Elements and Whitehead/Russell's Principia Mathematica were about. These books are admired and influential but very few people actually read them. As you might expect, they're too long. They're for giving a different perspective to people who have already achieved the highest levels of sophistication in math.
civilized 5y ago
Ironically, this post is an abuse of the concept of "weak typing". There's nothing "weakly typed" about, say, the plus sign being used to add both numbers and sets, or a dot being used for both multiplication of numbers and the dot product of vectors. It just means those symbols dispatch based on the types of their arguments, which is perfectly consistent with strong typing (cf. the Julia language). The situation that leads to weak typing in computer algorithms -- when you get data from a file or another process and don't know in advance what type it's going to be -- is basically non-existent in blackboard mathematics. Rigorous mathematical papers always tell you what set a variable belongs to when it is introduced, as well as the domain and co-domain of any functions that are defined. This is the blackboard equivalent of strong typing.
- 5y ago

da39a3ee 5y ago

> A scalar is a point on a number line. This is going to confuse readers. A point on a number line is a 1D vector; in other words it is a unit vector pointing along that number line, multiplied by something which scales its length. It’s the latter dimensionless and directionless quantity that’s the scalar.

adrian_b 5y ago
Nitpicking: a more correct view is that the set of points on a straight line is an affine space, so the points are neither scalars nor vectors, but elements of an affine space. The set of translations of the straight line is a vector space a.k.a. a linear space. So the vectors are the classes of equivalences of the differences between 2 points on the straight line (i.e. the differences between 2 pairs of points, where the distances are the same, are equivalent and they determine the same vector). While the vectors are classes of equivalence of the differences between 2 points, the scalars are classes of equivalence of the quotients of 2 (collinear) vectors, i.e. a scalar is the ratio between the signed magnitudes of 2 collinear vectors. If you choose a point on the straight line as the origin, you can choose as a representative of each class of equivalence that corresponds to a vector, the vector corresponding to the origin point together with another point. This gives a bijective mapping between vectors and those second points. If now you also choose a vector as being the unit vector, which will correspond with a second point besides the origin point, together with the origin point, then you can choose as a representative for each class of equivalence corresponding to a scalar the ratio between a vector and the unit vector, which will correspond to a third point, besides the origin and the point corresponding to the unit vector. So you obtain a bijective mapping between scalars and those third points. Because on a straight line there are bijective mappings between points, vectors and scalars (after choosing 1 origin point and a 2nd point as the extremity of a unit vector), they can be used interchangeably in most contexts, but it would be good to remember that all 3 are in fact different mathematical entities.
- da39a3ee 5y ago
  Thanks for that! It was extremely clear.

felipeqq2 5y ago

A rotcev

wisienkas 5y ago
My thought exactly

unixhero 5y ago

Somehow I got that first math explanation.

gpderetta 5y ago

Great article. I had a similar kind of revelation when I learned about generalized linear models after failing to understand all the various statistical tests.

necovek 5y ago

I love the way this was presented (others have pointed out flaws with the article already). I'd appreciate a post from Matt Ferraro on how this is built. Bonus points for including nice syntax-highlighted code "widget" for a cross between maths/programming.

steve76 5y ago

> What is the Inverse of a Vector? Manifolds, right? A vector takes a value, adds dimensions, and expands the value by spatial definition. A manifold takes a value, adds dimensions, and constrains the value by surface projection. Vectors are values. Even though they are made up of a bunch of numbers, they are a value like any other input or output. If you have a vector function, continuity works with vector sums just like continuity between input and output. Maxwell uses this, from Gibbs, to describe electromagnetic fields so he could have an equation for empty space and get the speed of light. Sometimes it's more difficult to find a reason for discontinuity, and are forced to assume continuity. Lie algebra is when you want to describe something with symmetries. Whatever you are describing, such as the insides of an atom, can't be described with exact values, so you use "what things are symmetric" and from there can get a differential equation.

amatic 5y ago

I would say the title should be "reciprocal" of a vector. Right? The inverse of a function (f^-1) has an unfortunate notation equality with the reciprocal (x^-1), or multiplicatory inverse. Or am I wrong?

debbiedowner 5y ago

Doesn't seem self consistent. He defines the ab multiplication as dot product plus "extrusion"/bivector (which seems simpler to call convex combinations of 0,a,b,a+b). Then he says aa is a scalar, presumably because the "extrusion" is 0, but you can't have this identity be 0. Just because it's degenerate does not mean it's 0. And the "extrusion" while not a plane is a line in his definition.

fallingfrog 5y ago

It occurs to me that some of these axioms depend on how many dimensions the space has- in 4 dimensions, a vector would have 4 components, a bivector would have 6, a trivector would have 4, and a quadvector would have 1. And so on, in accordance with Pascal’s triangle.

pmarreck 5y ago

Would this potentially enable generalized matrix inversion?

cannabis_sam 5y ago

I would have loved to implement this in Haskell as an exercise in uni!

portpecos 5y ago

I'm taking a first semester physics course right now, and we're learning about Torque and Angular Momentum. I just finished a calculus course last semester. Can someone tell me how I would use τ=r∧F on a physics problem for Torque?

ajkjk 5y ago
You wouldn't really. It's the same concept as `τ = r × F`. the only difference is that it is useful to think of the 'type' of the output as being a bivector instead of a vector -- there's no sense in which it points 'out of the plane'; rather, it is a single vector in the vector space of (planes), with the same magnitude as r × F. The distinct gets a little more useful when you start dealing with covariance under coordinate transformations. There it becomes more meaningful, because the _vector_ given by r x F doesn't transform the same way as their cross product should. For an obvious example of why this is true: suppose r=x and F=y. Then r × F = z. If you change coordinates by mapping z -> 2z, then you would be doubling the torque that you computed .. which is wrong; the torque is unchanged. The bivector x^y is correctly unchanged by z -> 2z. Currently in physics courses (usually not until more advanced mechanics or relativity) the resolution to this is to wave ones' hands and declare that, no, torque is a 'pseudovector'. But it is really much easier to think about if you type it as a bivector in the first place.

bsedlm 5y ago

where has this been all my life!? on the plus side I found it now...

DreamScatter 5y ago

This article is really about geometric algebra, check out my geometric algebra software https://github.com/chakravala/Grassmann.jl https://github.com/chakravala/Grassmann.jl