Why are there no 3 dimensional "complex numbers"?
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I find the title of this video VERY misleading, as it transports a wrong interpretation of complex numbers and dimensionality. The "2-dimensional" part of |C is not the a+jb coordinates, but only the jb-vector. e^js gives a direction on the x-y-plane, the real part (scalar) is a 0-dimensional magnitude. It is better not to see |C as "two-dimensional", but as "1-planar".
Adding another axis thus adds two planes, the x-z plane (i) and the y-z plane (k). The vector of a Quaternion has three components, but those do not represent axis coordinates, but directions on planes made by combining 2 of the axes.
Stopping to think about complex number systems in axis dimensions and starting to think about them in planes from my experience is a major block in understanding how they can represent a point in the coordinate system they describe. That's why I think the video title is very bad. There are "three-dimensional complex numbers", they are the Quaternion with three-dimensional vectors. There are no three-component complex numbers as you cannot create a space with only two directions.
I once tried to extend C for the equation conj(z).z=-1 with canonical extension of complex operations. I only was able to prove that it can at least be an infinite vector space over the field C, since a solution "l" to the equation will have all non null polynomials over C have a non null value, because otherwise, the solution would be in C and thus not a solution of the extension equation
The reason being, the fundamental theorem of algebra, but then, maybe it's not valid in the extension to assume that a.b=0 means a=0 or b=0 in the extended set
First I thought (1,i,j) with i^2 = j , j^2=i , i*j=1 would do the job. But then I realised it is not an Algebra...
The problem with this is that i = i*j*j = j and thus i = j*j = i*i = 1, so it just collapses.
A more interesting idea is to try j^2=i. Then i and j are kind of like third roots of unity, and then you can try doing interesting things! Unfortunately, (i+j+1)*(i+j-2)=0, so it doesn’t fit into this framework, but it still gives a decent model for 120 degree rotations!
@@noahtaul sry, meant j^2=i - corrected it
@@Buridan84Thing I also needed some time to understand is that complex numbers don't work with axes, but with planes. e^js is a normalised direction on the x-y plane, with the real part (scalar) the magnitude.
2 dimensions means 1 plane, but another axis ("dimension") doesn't add only one plane (try to do a 3-dimensional system with only 2 planes - you always lose 1 direction), but 2: x-z (i) and y-z (k). e^is and e^ks give directions on those planes, which together with e^js of the complex numbers |C gives directions in all three possible planes.
The magnitudes of e^js as well as e^is and e^ks is 1 for all values of s. That's why you need a scalar (0-dimensional) in addition to both 1-planar complex numbers |C (e^(r+js)) as well as 3-planar Gaßmann-Hamilton Quarternions |H (e^(r+is_1+js_2+ks_3)).
I find the statement that there "are no 3-dimensional complex numbers" misleading. The scalar is always the 0-dimensional magnitude, so Quaternions of course ARE the "complex numbers with a 3-dimensional vector", albeit "3-planar" would be the better way to state it.
That's what Hamilton said
Why not?
Fun fact, you can define the Hamiltonian quarternions as complex numbers to the regular complex numbers. In the way that C is R[i] with i^2=-1 for i not in R, H is C[j] with j^2=-1 for j not in C. We then define k=i*j and we get the quarternions. We even get the distributivity and term-by-term imaginary-term anti-commutativity from this definition. This can be further extended, although it's not something I"ve personally explored.
Note: the notation P[x] for a set P and a "number" x denotes a new set where x is plugged into any polynomial with coefficients from P. In fact, the polynomial ring of a ring P is denoted as P[x] for x being a formal variable (i.e. we only care about the coefficients)
nice ! where can I find more about this ?
@@Regimeducamp en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction is a good place to start, but that's not the only way to complexify numbers. en.wikipedia.org/wiki/Classification_of_Clifford_algebras will produce, i think, every finite-dimensional associative algebra.
There is some overlap between the two ideas, but Cayley-Dickson construction will get you non-associative algebras, eventually. Otoh, lots of Clifford algebras have zero divisors iirc. It all depends on what kind of system you need, which type of hypercomplex number you should use.
Yeah but the degree of that field extension won't be equal to 3 its actually 4 but still everyone interested in algebra should know about quaternions
@@wojteksocha2002yes but quaternions are 4 dimensional
In usual notation A[x] denotes a commutative ring so the quaternions cannot be C[j] as they are not commutative.
"...we want associativity..."
*sad octonion noises*
sacrifice a property to start over with 2x dimensions
One detail that could use more explanation: at around 15:50, you bring up the fact that x has to satisfy a polynomial. You are using the fact that A is finite-dimensional here. That fact says that if you take the sequence of linear subspaces X_k = {expressions of degree k in x}, there must be some k at which X_k=X_{k+1}. This means that some degree k+1 polynomial in x is actually equal to a lower degree polynomial, which can be manipulated into saying that x satisfies some polynomial equation with real coefficients in A. [side note: typo where you say x is in A, but it should read x is in R]
You use this again in 25:25 or so, when you point out that another element must satisfy a polynomial equation.
Thank you, I got really confused at this stage.
Does that imply that if you define a space not composed of a polynomial, say, a*e^x or something, it is infinite dimensional (using Taylor series)?
@@pedroivog.s.6870 There do exist infinite dimensional fields extending the complex numbers. For example, the set of rational functions (ratios of two polynomials) forms an infinite dimensional field extending the complex numbers, where each complex number is identified with the corresponding constant function. We can understand the structure as an infinite dimensional vector space, where the basis vectors consist of the simple monomials x^n, along with the poles 1/(x-c)^n, where c can be any complex number and n is any natural number. Every rational function can be written uniquely as a finite linear combination of those functions, via partial fraction decomposition.
@@pedroivog.s.6870 forget about the Taylor series, it isn't useful in this context, even if it seems related.
To answer your question, yes. If x isn't the root of a polynomial over any field K, 1, x, x², x³, ... are lineary independent (almost by definition, this is an equivalence). Proof by contradiction, if they are dependent, there exists c_0, ..., c_n in K such that c_0 + c_1 x + .. + c_n x^n = 0, so x is the root of a polynomial.
Since we found an infinite free family, the dimension has to be infinite.
@@AlcyonEldara Wait, didn't you just say that x wasn't the root of a polynomial?
Note: an "algebra", here, is a vector space with some sort of multiplication defined on it.
Right, and multiplication has to be bilinear, so (ax)*(by+z) = (ab)(x*y) + a(x*z), where a and b are real numbers and x, y, and z are members of the algebra.
Technically it's a module, not a vector space. Vector spaces must by definition be over fields, but you can have modules and algebras over any ring.
@@MasterHigure How is it not a vector space?
@@EebstertheGreatIt is in this case. But you can have an algebra over the integers, and then it doesn't work to define it as a vector space with multiplication, as you can't have a vector space over the integers. So an algebra is a module with multiplication.
but we have a field, R
I feel like it's important to justify why any element of a finite dimensional algebra over R has to be a root of some nonzero polynomial. To do that, suppose that x is not a root of any nonzero polynomial. Then for all real numbers a0, ..., a_n (where at least one coefficient is nonzero) we have
a0 + a1*x + a2*x^2 + ... + an*x^n != 0
This means that {1, x, ..., x^n, ...} is linearly independant, and so A has to be infinite dimensional which is a contradiction
I think the other nontrivial fact used in this line was that not only is it the root of a real polynomial, but it's also the root of, at most, a quadratic real polynomial.
To get to this step, I would use the fact that the complex numbers are algebraically closed, and the fact that the reals are exactly the fix of complex conjugation, which is an order 2 automorphism of the complex numbers.
So firstly, we know that x is the root of some potentially higher degree polynomial; a0 + a1*x + a2*x^2 + ... + an*x^n = 0. Call the polynomial on the left hand side p. We know that p has real coefficients, so it (trivially) has complex coefficients. It is therefore a complex polynomial, and the complex numbers are algebraically closed, therefore p splits into linear factors in the complex numbers (i.e. all the factors are complex linear polynomials). Since p(x) = 0, and there are no zero divisors in A, one of these linear factors is 0; x-a = 0 for some complex a. Now we simply multiply this equation by x-conj(a):
(x-a)(x-conj(a))=0, x^2-(a+conj(a))x+a*conj(a)=0. When we conjugate a+conj(a) or a*conj(a), we get the same thing back (using conj having order 2, and addition + multiplication being commutative over the complex numbers). Since these coefficients are fixed under complex conjugation, they are real. Therefore x is a root of this real quadratic polynomial, so it's minimal polymomial in the reals is at most quadratic.
Thank you! I was about to write the same comment :)
@@stanleydodds9 ooh true. So the proof exposed in the video is not exactly from scratch since one has to use the fact that every polynomial has a root (in C), which is nontrivial. I wonder if there are other simple proofs that don't use this... Maybe a more "geometrical" one.
Yes, the part of the video starting at 15:50 is pretty messy. It looks like he was merely improvising while reading some notes.
It seems to me that using this fact and the fundamental theorem of algebra we can prove much more easily that a finite dimensional algebra over R with commutativity, associativity and no zero-divisors must be isomorphic to a subset of C.
1. Suppose we have this algebra, A. The following will work for any x ∈ A.
2. We must have a polynomial with real coefficients a0, ..., a_n, where a0 + a1*x + a2*x^2 + ... + a_n*x^n = 0. Otherwise, all of the powers of x must be linearly independent, which necessitates that the algebra is infinite-dimensional.
3. Since we have no zero-divisors, a factoring of the polynomial will produce its only roots. By the fundamental theorem of algebra, every non-zero, single-variable, degree n polynomial with complex coefficients factors into, counted with multiplicity, exactly n linear polynomials with complex coefficients over the same variable. Therefore, x must be a complex number.
I cant express enough my gratitud for answering this.
I always had this question in my mind but never dared to do any sort of research about it.
Professor Penn, i send you my best wishes for you all the way from Chile.
Greetings
Same here, I ended up with some questions about this after first watching 3B1B's video on quaternions and I'm glad my lingering queries have finally been addressed
En serio te llamás Elon SIMP jajajshjs poweon
Wena kuliao aguante el colo.
La chile o la puc?
how were you able to understand is.
@@26IME obvio que sí wn. Ni que fuera a usar un nombre de verdad xd
I would say that the quaternions _are_ the 3D equivalent of the 2D complex numbers. The 2D complex numbers are really the rotations and scaling of the 2D plane, not positions in the plane. And the quaternions are the rotations and scaling of 3D space in the same way. It just happens that there are 3 degrees of freedom in 3D rotation, as opposed to 1 number that measures 2D rotation. That 3 plus 1 for the scaling factor gives the 4-dimensional quaternions, just as the 1 dimensional 2D rotations plus 1 scaling factor gives the 2D complex numbers.
Not to be rude but your definition is not rigorous; there's no way to represent the space of quaternions with three dimensions just like there's no way to represent the space of complex numbers with one dimension. Just because the quaternions can represent rotations in 3D does not mean they're limited to operating in R^3.
I somewhat agree with both of you. Rigorously, the dim(H) is clearly 4. However in terms of some sort of operation space that the field provides rotation and displacement, the values would be 2 and 3 for C and H. I don't think that type of operation space has a formal definition, but I could easily picture someone creating one. It would likewise be interesting to no what you he field would look like for OS(new F)=4,5 and so on.
In physics all coordinate systems must be fictional. There is no motion in a particular direction but you can define perpendicularity as the square root of the opposite direction.
@@frankjohnson123 i mean C over C is a 1-dim vector space. but just being pedantic
@@zaheercoovadia4745 in the context of the present video the dimensionality is based on the reals
I studied maths at University years ago and never made it to the finish .. 30 years later rediscovering a passion for maths largely thanks to your channel and a handful of others. When I was taught Linear Algebra I thought "Ok, great" and never saw how it was useful. This video puts soooo much mathematics into linear algebra to prove something that seems totally unrelated. Starting to see how different areas of maths all tie together, despite them being taught individually at uni so many years ago.
i still cant see how "i" is useful at all....
@guitarszen help me now please sir, what can 'i' do that your usual numbers cant. theres nothing. I already know the answer, and u wont be able to tell me anything, because there is nothing.
@@magnuswootton6181 i^2=-1
@@Anonymous-df8it thats amazing, and I believe u, but my argument why i complain about it, is its probably useful for nothing, and it just seems like pointless trickery, why not just do normal maths, instead this nonsensical things that could be true, but arent actually that important to the function of things.
@@magnuswootton6181 It shows up in electrical engineering iirc
Around 25:00, it gets a little unclear to me how the argument from factorizability of the satisfied polynomial follows from (xi)^m = x^m i^m. I guess P(x) is constructed from the x^m, but the connection is a little opaque.
ele ta roubando
tiny note at 18:52 : I think you assume commutativity when expanding (y-b)² which is a problem since quaternions aren't commutative. Not sure if that can be fixed by just adjusting the x²+2bx+c=0 equation.
Edit: Found my mistake, b is "a sub 1" and thus a real number and as such commutative with other elements from A.
26:29 I’m confused. The contradiction seemed to have proved the non-existence of such a polynomial satisfied by x. Since x is an object outside the complex numbers, it makes sense that it does not solve a polynomial with complex coefficients. But how does that disprove our assumption of xi=ix? Specifically, where did we use that assumption? There seems to be a big jump from (xi)^m = x^m i^m and the construction of p(z). Can someone help fill in the gap here in the comments?
Yeah that confused me too. I really don't see how being a root of a complex polynomial and commutativity is related
He didn’t really go into it, but here’s the idea: if you look at the Hamiltonians, it’s even true that j satisfies some polynomial equation, j^2+1=0. And that polynomial factors as (x+i)(x-i). But what we need commutativity for is to prove that this means that (j+i)(j-i)=0. This isn’t true if i and j don’t commute, because the cross terms don’t cancel! When you write x^2+1=(x-i)(x+i), you’re implicitly assuming that x commutes with all other coefficients involved, whereas the things you plug in for x might NOT.
And so if you have something that commutes with i and satisfies (z-a1)(z-a2)…(z-am), then plugging that thing in for z is acceptable, and hence we have a product equal to 0 and yada yada yada. Hope this helps!
If K is a commutative ring and x is an element of a K algebra B (non necessarily commutative) then you can define the evaluating map ev_x from the polynomial ring K[z] to B that maps p to p(x). This map is a ring homomorphism. Now if you suppose that B is a subring of some ring A, then you still can define the map ev_x from K[z] to A for every x in A, this is a linear map but it might not be a ring homomorphism, that is to say it might be possible that pq(x) and p(x)q(x) are distinct. In fact it is a ring homomophism if and only if x commutes with every element of K.
So in our case we have K=B=C the field of complex numbers, and what he omitted to say is that if xi=ix then x commutes with every elements of C which is due to the fact that C=R+R(i) and A is a R algebra. So the map ev_x is a ring homomorphism, hence if p=p1...pm then it is allowed to write p(x)=p1(x)...pm(x).
So, if I am understanding you correctly, if z commutes with the complex numbers, then (z-a1)(z-a2)... is equivalent to a polynomial p in C[z], Non-commutivity could potentially introduce non-complex coefficients when multiplying z and a_i in different orders. But also, we know that a product (z-a1)(z-a2)... has solutions z = a_i due to the no zero divisors assumption.
great video! hoping to see more fleshed out theoretical-oriented videos in the future! keep up the amazing work!
loved this kind of longer kind of video, especially in such an interesting topic. Keep up the great work
Cool! Would be nice to see a video where you do a similar process but allowing for non-associative algebras and thus deriving the octonions, then showing that there are no other division algebras.
Are sedenions, duals or double division algebras?
thank you!
@MichaelPennMath should do a video on the various Geometric Algebras, which are extensions of Clifford Algebras where all these operations are rigorously defined for mutli-vectors (sums of scalars, vectors, bivectors, etc) of arbitrary dimension and grade.
I am once again reminded of how I ran out of "math gas" once I reached differential equations
great video! you explain stuff very clearly. A playlist on galois theory would be a godsend
For geometry, you know, is the gate of science, and the gate is so low and small that we can only enter it as a little child. - William Kingdon Clifford
Waiting for a comment like this, now I can escape the comment section
One of my favorite videos thus far. Cool deep dive into an interesting topic
I started to watch this video on its release, but I hadn't had the knowledge to follow it. Now, after an abstract algebra course, I'm able to understand every step done. I'm happy to see my progress!
I think this could be one of the best videos of this channel. Thank you!
This is fascinating to me from the perspective of Clifford algebras / geometric algebra, since a clifford algebra describing an n-dimensional space is 2^(n-1)-dimensional. Hence, 2d complex numbers for 2d and 4d hamiltonians for 3d. But does that mean that the 8-dimensional geometric algebra for 4d space has... zero divisors? Can I rotate a 4d object in such a way that it... disappears?
Depends on your reference point. You can rotate a 4d object so that it disappears from our 3d view
@@ladyravendale1 Field of view notwithstanding, a 0 divisor seems to mean that i can find a nonzero transformation that results in actual 4-dimensional 0.
I really love Geometric Algebra, except I think the way it is taught is completely backwards; making it too hard to learn. Noting that multiplying two vectors creates a complex number, that is a sum of a dot and a wedge is an interesting fact.
But If you try to leverage this to make a coordinate-free algebra, it becomes far too hard to answer questions like: "What type is a 4D vector multiplied times a 3D-Bivector in 5D space". The main place where coordinate-free calculations make any sense is in dealing with round-off error; because the type of an object, depends on what components are zero. In 3D space (0.0000001 + 5 e0 + 4 e1 + 1 e3) is a vector by "common sense", but due to round-off errors, parts that don't belong in a vector might actually be non-zero. So code libraries get involved to design the library such that type errors can't arise from numerical precision.
@@jneen The octonions don't have zero divisors, however the sedenions (16-dimensional geometric algebra) do, for example (w+yz)∘(wx- xyz) = 0
@@TheJamesernator I see! Just read that the octonions aren't fully associative either, which would exclude them from consideration by this proof as well. Thanks for the tip!
درس رائع جدا كالعادة.
شكرا بروفيسور
13:45 To say that a' is the inverse of a you should also show that a'a=1. This is not hard because l_{a'} is surjective so there exists a'' such that a'a''=1 and then by associativity it follows that a = a(a'a'') = (a a') a'' = a''.
It can be shown even easier using the absence of zero-divisors. If aa' = 1 then aa'a = a then a(a'a - 1) = 0
A theorem we covered in an abstract algebra course I took in my undergrad math studies would answer this question, if we require that there be a norm, with the property that the product of norms of two of our 'new numbers' equal the norm of their product. (I think the theorem might have been due to Hurwitz; really not sure.)
I remember the content (if not the name or attribution) of the theorem, because its statement elicited an extreme "wtf" reaction!
It says that if the product of two sums of squares of N numbers, is again a sum of N squares, in such a way that the terms in the product are bilinear in the terms of the factors, then N is 1, 2, 4, or 8.
The structures corresponding to these four values, are the reals, the complex numbers, the quaternions (where multiplicative commutativity is dropped), and the octonions (where multiplicative commutativity and associativity are both dropped).
And so, adopting the norm constraint, rules out any other "dimension" of such structures; and 3 in particular.
I suspect that you, being into number theory, are at least passingly familiar with this theorem, and could patch the holes in my faulty memory.
Fred
Your delivery fits perfectly into my brain. Instant fan 👋🏻
10:05 I don't think you need to start over at a point this early. Instead, you can notice that all you really needed in order to prove that e is a left identity was that e acted like a right identity for just one nonzero element, but now that we have proved that e is a left identity for ALL elements (including a nonzero element), you can turn just this latter argument around to prove that e is a right identity:
We know that e is a left identity. Choose an a in A such that a =/= 0. We know that ea=a. Now choose any b in A. From the earlier equation, it follows that bea=ba (be-b)a=0 be=b. Because right-multiplying e yields the identity, e must be a right identity as well.
I really appreciate the work trying to present concepts somewhat more sophisticated than the typical math video on TH-cam. I think the only "trust gap" in the derivation concerns polynomials over reals having max order 2 if irreducible; if someone knows a proof of this without Galois theory, please let me know.
Let P be a irreducible real polynomial, and let suppose that the order of P is greater or equal than 3.
By the fundamental theorem of algebra, P has a complex root z.
If z is real, then X - z is a real polynomial of order 1 that divides P, which contradicts the fact that P is irreducible. If z is not real, then it is not equal to its conjugate, and we know that the conjugate of a root (of a real polynomial) is also a root. If we call ż such a conjugate, then (X-z)(X-ż) is a real polynomial of order 2 that divides P, which again rises a contradiction
@@hadrienduval8628 nice and simple, much appreciated
I try very hard to construct these set in 90s after i returned in 2002 but without good study; what i learned in good practice of matrix, geometry algebra, quaternions,
I m still working in complex analysis.
30:11 you can't use "the same steps" to achieve that, because any element of B shifted by a real number doesn't belong to B. B is not a subalgebra, it's only a subspace!
Actually what you want to show is that the middle coefficient of the polynomial x^2 + 2cx + d with the root x=b is equal to zero, so you can just scale b without shifting it. It's indeed the case, because otherwise b would be a complex number (it can be checked that b^2 is complex)
great video! this kind of stuff is why i fell in love with algebra :)
I feel like using "every irreducible polynomial over the reals has either degree one or two" here is kind of like putting the cart before the horse. Both the statement we are trying to prove and that statement are intimately related with the fundamental theorem of algebra (at least for the commutative case). It's also a huge thing to bring in from outside after we just went over some lengthy calculations proving some nice but basic algebra results.
I can understand your sentiment, and yes, it is a big thing to bring from the outside, but here's a perspective to consider: this video assumes the existence of C and H. The video is not proving that C and H exist. Rather, the video is essentially a uniqueness argument. C and H are finite-dimensional associative R-algebras without zero-divisors, and, up to isomorphism, they are the only ones. We're proving the "only ones" part here.
But yes, it is still a big fact to bring in the the fact that C is algebraically closed to prove facts about irreducible polynomials over R without mentioning that it follows from C being algebraically closed.
Thx much! 😊
I wondered for 3D perpendicular number algebra long time ago, but my engineering background I didn’t gave me the tools to follow the difficult (and yet logical) arguments of the proof.
I think this might be my favourite video on this channel
Change my mind regarding how we read "F[x]" aloud. (In the video this issue comes up around 25:12, admittedly very, very briefly, with C[z].)
We can all agree that, when x is an indeterminate and F is a field, F[x] with square brackets is the notation for the ring of polynomials with coefficients in F, whereas F(x) with parentheses is the field of rational functions of x with coefficients in F. To emphasize the difference: F[x] cannot be a field and it does not contain, to pick an easy example, 1/x, whereas F(x) must be a field and must contain 1/x.
What if x is not an indeterminate but an element of some field that has F as a subfield. I will switch from x to a to emphasize that we are no longer talking about an indeterminate. I see three cases: (1) a is algebraic, (2) a is transcendental, and (3) we don't know and/or it's not relevant. In case (1) it is an amazing and beautiful fact that we can define an isomorphism between F[a] and F(a). In cases (2) and (3), a might as well be an indeterminate; F[a] and F(a) can't be treated as isomorphic.
Here's where we may disagree. I believe the term "adjoin" should be reserved for F(a) and its meaning should be clearly limited to "the smallest field containing a." I believe that the notation F[a], referring to a ring of polynomials, should not be called--or read aloud as--adjunction. Part of the problem seems to be that there's no obvious way to read "F[x]" aloud. Dr. Borcherds, reads it "F of x" which I like because I like anything besides "F adjoin x".
Now here's where we may strongly disagree. Some people prefer to intentionally blur the distinction between F[a] and F(a), saying "if only you were smarter, you'd understand all the steps I'm skipping when I treat F[a] as equivalent to F(a) in certain circumstances." To me, the fact that a quotient ring works out to be a field when the ideal is maximal (the kind of ideal you get from a minimal polynomial) is an amazing fact, a beautiful fact, and a fact that is not at all obvious. I feel that people who intentionally blur the distinction between F[a] and F(a) are missing out on an opportunity to both see and also to teach the beauty of this fact.
Awesome video!
Good timing. I was recently been wondering about this, generalized to "n" dimensions.
I'm pretty proud that I got all of that without issue 😁
now I want to see where you used associativity to see why removing it allows for the octonions.
Yeah right??
Its so crazy that after watching many videos in this channel ylu start to get familiarized and understand a bit about math without getting lost like it was chinese.
Dont need tk be studying maths, yet you learn college level pure maths. Thats what educatuon should look like
Associativity was used implicitly whenever there were three or more things multiplied together, by not having to use lots of brackets to specify the order of multiplication - for example on the last board there is "-iji", "ijij", "-i^2 j^2", and "ijk"
I've always wondered about this...Thanks!
I watched this video two days ago when it came out... but I just now thought of something this suggests about the relationship between mathematics and reality. The non-existence of a three-dimensional complex mathematical space lines up nicely with the post-Einstein understanding of reality as a four-dimensional space-time continuum which also can be easily translated into two dimensions via light cone diagrams... and how it's easier both visually and mathematically for physicists to flatten everything down to two dimensions from four. The extent to which mathematics is fundamental to reality rather than simply being representative of it is something I've been fascinated by for a long time.
Well... it's not clear that the four-dimensionality of space-time has anything to do with the quaternions. They're quite different. Quaternions (unit ones at least) are more associated with rotations in *3D* space (a three-dimensional group, we knocked out a dimension by saying the quaternions have unit norm), and the corresponding transformations on 4D are a six-dimensional group.
But... Even so... I've still wondered for a long time if there is any connection here.
Great video. Thank you
Nice demonstration, thanks. However, at 25:00, I didn't get how (xi)^m=x^m i^m translates into x being satisfied by some polynomial in C ?
Then, at 34:38, you claim without prooving it that ij is independent from 1, i and j (although it is not too difficult, e.g. assuming ij=a +b i +c j and multiplying by i )
I didn't get it either so I'm commenting in the hope I get a notification if anyone answers
He uses the analogous fact at 15:56 without any explanation. But it follows from A being finite dimensional over R: if you consider the powers of x (1, x, x^2, etc...) you will find a linear combination of them equal to 0 at some point.
@@user-ne1nw6hw2q I don't get this, and even less the relationship with (xi)^m=x^m * i^m.
I'm not entirely sure, but I think it's possible to factor a polynomial in this way with alphas only if z commutes with complex numbers. For some reason x commutes with reals, so if it commutes with i, then it commutes with all complex numbers in A, then the factoring is possible, and the idea that x=alpha_j works. I'm still not sure where the fact that all elements in A commute with reals comes from, probably they must commute with elements in R, and they commute with e(multiplicative identity), so it commutes with R'=span{e}. I would like to know if these thoughts are correct.
@@Eye-vp5de you're correct, that is exactly the reason why everything in A commutes with R. Almost by the definition of an algebra
Perhaps have a part 2 video about relaxing the associativity and ending up to something isomorphic to the octonions when you go higher dimension than 4. Or do we end up with something else?
You get the octonions if you no longer have associativity but retain "alternativity", which means you have something like associativity if two of the numbers are the same: (xx)y = x(xy) and y(xx) = (yx)x. en.wikipedia.org/wiki/Alternative_algebra
There's a sort of ladder called the Cayley-Dickson construction that produces algebras of dimension 2^n, but they lose "nice" properties every step of the way. Beyond the octonions you have zero divisors.
so nice ! at 25:40 it should be pointed out that x is the root of a complex polynomial bc A is finite dimensional (like others said) AND the hypothesis xi=ix is necessary to factorize P (the coefficients must commute with the unknown to have all these factorization properties !)
26:00 I was pretty stumped about where you used the x*i=i*x premise until I realized that it's required for the polynomial to be factored into linear factors.
Thanks for this. I got really lost at that part
Very well explained! the other question I always have is the expression of Sin(10) which would be, using cardano formula as well as the fact Sin(30) = 0.5, you would have 2*Sin(10) = (-0.5+i*sqrt(3)/2)^(1/3)+(-0.5-i*sqrt(3)/2)^(1/3), which almost seems you are getting some simple algebra expression for Sin(10). It is actually not and I am confused as to why this cannot be further simplified, such as finding its magnitude because it always circles back to Sin(10)! Seems the only way to find Sin(10) is via infinite series😅
16:27 I think it's worth showing why x couldn't be transcendental, because it seems like we've assumed it will be algebraic. The reason is because if dim A=n, then considering 1, x, x^2, x^3, ..., x^n, we have n+1 vectors in A, so we must have a linear dependence in A. Thus, there exists c_k such that c_0+c_1x+...+c_nx^n=0; in other words, x is a root of some polynomial of degree at most n, so x is algebraic.
Thank you very much! I was puzzled about this exactly.
My starting point for this is to notice that a cut down of the H~Quaternions without k would not be closed, and closure under addition and multiplication is something i would want.
Attempts to reclose these "ternions" by writing ij = 1 makes j just be -i, and a little playing around with other initially plausible seeming guesses similarly fail.
In the end to make the ternions work we have to make ij be orthogonal to 1, i, j and then we have for dimensions not three.
Granted this is an intuitive and non rigorous plausibility argument. It's not a proof: it's saying that this is why it makes sense.
Once you mention dropping commutative multiplication, it occurred to me that you could skip 3D numbers and go to 4D in the form of 2x2 matrices. (I thought about bringing it back to 3D by putting 0 in the only space under the diagonal, but then you get 0 divisors.) ETA and then I realized that even the general 2x2's have 0 divisors when the second row is a multiple of the first.
Interestingly though, you can have a special subset of the complex 2×2 matrices (which is of dim 2 over C, 4 over R), that represents the quaternions. I think there is also a way to represent C as 2×2 real matrices.
@@quantspazar6731 try mapping a + ib to (a -b ; b a). Where the top row is a -b and the bottom row is b a.
@@terryendicott2939 Right after i commented I found that too. I knew that 1 was represented by the identity and that the matrix representing i would square to -I , so i picked a usual rotation matrix and everything followed
Btw, would it even be possible to have multiplication closed inside a 3D number system? At the end of the video, he had to define k = ij to make it possible.
@@terryendicott2939 there's even a couple more ways to restrict 2x2 real matrices to get an interesting associative algebra.
(a b; 0 a) gives us the dual numbers
and
(a b; b a) gives us the split-complex numbers (a composition algebra)
but either way you have zero divisors. these are the three most natural ways to constrain M2R.
perhaps (a b; 0 c) would give you an interesting 3D structure? i see something in it that squares to 1, and something else that squares to 0. they're both non-commutative
Great videoz the only step i did not understand is why the fact that x communtes with i implies it is a solution of a complex polynomial
I've wondered if the winding number is like the z-axis out of the complex plane. I think of it that way for multivalued functions
You can use the Clifford Algebra. It allows you to express 3d rotation through algebra and is commonly used in physics. The only downside is , that it is not 3d but 8d, and it is not a normed division algebra.
You should do a video on the octonians (Cayley numbers) sometime.
amazing thank you very interesting
Hey Michael, is it true that only the 2^(integer) dimensions are fully mathematically and self consistently complexify-able? I think I remember hearing this from somewhere, so I'm curious if there might be any exceptions to this rule, or does this rule hold true all the way out to infinite dimensions? Thanks.
Here's the Google food for you: the only division algebras over the reals have dimension 1, 2, 4, or 8. If you Google that claim, you might find more information on what you're looking for!
It surely depends on what properties you need them to have. You can make any system you want, including in 3 dimensions - the problem is having a useful system. There don't seem to be many of those.
Another way to think about this that if you have a sphere with fur on it, you cannot comb the fur so that it doesn't form a swirl or tuft. If you consider each hair a vector mapping to another point on the sphere, and you consider the numerical space consiting of infinite amount of spheres in layers (kind of like an infinitely large onion with infinitely thin layers), you will get an infinite number of swirls of tufts. And thus you would get an infinite amount of values that cannot be mapped in the numerical system.
If this is correct, I guess a geometrical explanation would be that you can rotate a 2D numerical plane by keeping multiplying with i so that a point on it will visit each quadrant. But you cannot do a similar thing to a 3D object, as when you keep rotating it along an axis, the points cannot visit all of the 8 sections. You would have to somehow twist the object, and then you would lose some of the algebraic features.
Michael: "You guessed it...."
Me: still trying to formulate a guess....
Thank you. Subbed.
Short answere from complex differential geometrie: if there exists something like an i with i^2 = -1 your three dimesional real vector space has an almost complex structure, meaning a map J: V -> V such that J^2 = -1, which is only possible in even dimensions. This is because an almost complex structure on a vector space induces a complex vector structure on the same space. Since every complex vector space is even real dimensional, the vector space is of even real dimension. Thus it is hopeless to look for three dimensional complex analogue.
>>>Why are there no 3 dimensional "complex numbers"?
Just R, C and H? But what about the octonions (O), the sedenions (S) and the trigintaduonions (T) [where dim(O)=8, dim(S)=16, dim(T)=32]? And other (2^n)-ions in general?
Those are all non-associastive. Without associativity, the argument involving linear algebra won’t work.
In the field of electromagnetic theory, wave propagation is described in terms of phasors. Phasor notation allows description of electromagnetic wave propagation in 3D space. Phasor is a complex entity. This is a routine work of microwave and antenna engineers. You perhaps need to double check...
Thanks!
so, to dumb it down:
if your field is commutative -> isomorphic to complex plane/real number line -> at most 2 dimensions
if non-commutative elements are added -> implies different right and left multiplication/inverses -> 1 extra dimension needed for each -> 4 dimensions
am I understanding this right?
i stg, this is the type of thought i have high, and i think, "i'm a genius!" so it's really cool to see other people ask these types of questions
Incredible to have this on TH-cam
Here’s the TL;DW: The dimensional expansion from real to complex to quaternion to octonion does follow an incremental progression (0,1, 2, 3). It’s just that the increase in spatial dimensions is an exponential function (2^n) rather than a linear function (1+n). This exponential function produces a progression of (1, 2, 4, 8) spatial dimensions for respectively real, complex, quaternion, and octonion algebras.
It seems to me the proof can be simplified a lot by noting that for any x in A\R, the subalgebra R(x) is commutative and thus a field, and of finite degree over R, and thus isomorphic to C.
This is one of those videos where after watching it, i feel simultaneously very smart, and very stupid. I think that means you did a good job.
Is there an name for the transformation f(x) = (a^-1)(x)(a), I recall that that operation is used for the Normal Subgroup test, but I do not recall if it had a name.
In German I know this operation as "konjugieren" which would translate to conjugate in the sense of conjugating a verb, so maybe the transformation is called conjugation? This is just a wild guess but since German and English math vocabulary is often translated one to one there is a chance it is called that
@@flyni7249 Conjugate is the term I was looking for, my mind was in coset, but that is something completely different. Goes quickly to my book. Yep, there it is, Conjugacy Class of a in the section of Sylow Theorems. TYVM
7:55 Wait, linear maps A -> A with trivial nullspace needn’t be surjective: consider [0,1] -> [0,1], x |-> x/2.
[0,1] is not a vector space.
Love your t-shirt
Hey Michael, loved the video, but I do not understand why x being outside of R implies it is a root of a real polynomial. It probably has to do with the fact that this space is finite dimensional and has inverses, but I feel it is not a trivial step.
Quaternions,octonions ?
The higher-dimensional algebras defined by the Cayley-Dickson construction (starting with the sedenions) all fail to satisfy this property. They all have zero divisors.
So the conjecture makes me wonder if you could represent 0 division as a symbolic quantity where the numerator is the multiple argument with signage determined by that. It'd be nice to see if this yields any useful properties.
It's not really clear to me why would an algebra with dimension > 2 contain "i" because most of the proof relies on the fact that C is contained in A (which then makes it that the dim of A is a multiple of 2)
Good question! I think the video could have made this a bit clearer, but it goes back to the argument Michael made about if dim A = 2, then A is isomorphic to C. Only an absolutely tiny bit of the argument actually relied on dim A = 2. Essentially, most of the argument is spent proving the following statement: if x is an element of A (where A is a finite-dimensional R-algebra without zero-divisors) and x is not an element of R, then R(x) is isomorphic to C. The only place where dim A = 2 is used is to conclude that R(x) is _all_ of A.
So the brunt of the dim A = 2 step of the proof is really showing that if dim A > 1, then A contains at least one copy of C.
@@MuffinsAPlentythanks I kind of see it now, the confusing part is that all the steps for searching for i suppose that it is commutative, but on the other hand I think it's not possible to have a 2d non commutative algebra on R which then will kinda solve the problem. I was thinking of like a n-vector algebra where R does not commute with anything (unlike H where it commutes with i)
@@dariofagotto4047 I see what you're saying! The commutativity is a subtle point.
I think it is due to one of the underlying assumptions of an "algebra" - that the ground field R commutes with everything in the algebra. (This is either forced by the definition of an algebra or taken as part of the definition.) From here, R(x) is necessarily commutative (even if x is not dimension 2) because x commutes with everything in R, powers of x commutes with powers of x (since we have associativity), and since multiplication distributes over addition.
Using those facts, you can prove that given any two elements p and q of R(x) that pq = qp.
@@MuffinsAPlenty And since R(x) is commutative and every element has an inverse, it must be a field over R of finite degree, so it must be C. If we accept as given that C is algebraically closed over R, it seems to me this can greatly simplify the proof. (At least for those familiar with field extensions, which is perhaps the tricky point here.)
@@ronald3836 It seems that Michael _did_ use that C is algebraically closed in the proof, in order to say that the only irreducible polynomials over R are of degree 1 or 2.
What always makes me nervous about proofs like this is 9:05 where we are proving left-right multiplicative identity, and somehow we use a factoring technique to say that a(b-eb)=0. The assumption being that factoring "works" as a technique, but you have to prove that it does without assuming multiplicative identity. Makes me a little itchy.
How so? It s just the distribution of multiplication over addition
I think distribution was implied as an assumption when he said A is an n-dimensional algebra
Right, this is simply what it means to have a distributive rule. It may be possible to create some object that doesn't have the distributive property but that means it isn't even a vector space, so all sorts of rules are out the window. Might be fun to explore though.🎉🎉
@@BridgeBum the algebra of wheels: en.wikipedia.org/wiki/Wheel_theory weakens distributivity to (x+y)z+0z=xz+yz in order to allow "division" by zero.
I feel like a small but important detail has been left out. Around 9:40, a statement about distributivity should be mentioned. I guess it has been implicit through out the video that we are trying to construct a "field"-like structure; so distributivity has been assumed. However, if we go to the next level, and try to construct an 8-dimension number, we can only do it if we are willing to give away distributivity.
Isn't distributivity a property of algebras in general?
So by assuming A is an algebra, we automatically get distributivity I believe
@@fedem8229 Yes, I agree. But I would say that for someone like myself, these fine details are always easy to miss... and better call them out explicitly.
Very happy to see this... finally.
I watched another excellent video of yours a few months ago in which you showed how to construct a 2^n dimensional extension of the real numbers for an arbitrary positive integer n. This was very enlightening, but what I don't get is how as n increases, the resulting extension keeps losing more and more structure. For instance, the complex numbers aren't ordered, the quaternions aren't commutative, the octonions aren't associative, and the sedenions have zero divisors. Does this loss of structure continue ad infinitum, or does it stop somewhere?
Clearest explanation I've seen.
Amazing!
Great video! I'm wondering why at 22:15 you can say that i exists in A. You proved it for when A is 2 dimensional over R, but I'm not sure why this should apply for the case when the dimension is greater than 2.
The dim A = 2 part of the argument actually is a lot more general than just showing "if dim A = 2, then A is isomorphic to C". Most of his argument is showing
*If x is an element in A but not in R, then R(x) is isomorphic to C.*
Nothing about this argument relied on dim A being _equal_ to 2. It only relied on dim A being _at least_ 2 to guarantee that there is an element x in A but not in R.
@@MuffinsAPlenty Thank you! I missed that, and also him saying an irreducible polynomial over R has degree as most 2, so I assumed he was making the polynomial quadratic or linear from the dimension of A.
@@JPJ280 To be fair, I think it is the presenter that missed a few steps ;-)
I have thought about it multiple times
Is this correct? If aa’=1, then a’(aa’)=a’ so (a’a)a’=a’ so a’a=1 since the identity is unique so every right inverse is also a left inverse.
If there are no zero-divisors, this is correct. Otherwise you might not be able to cancel that a'.
(a'a)a' = a' means (a'a - 1)a' = 0. If there are zero divisors, we could have a'a-1 and a' both non-zero.
Hey ! I'm trying to find a number system well-suited to descibe space-time (and all the relativity stuff). But is space-time, the "distance" is x^2 + y^2 + z^2 - t^2 (so not euclidean) and the fact that all this is 4D makes it impossible to graph ; is there a 3D number system with x^2 + y^2 - t^2 as "distance" ? I ask that because in this case, there should be zero divisors (light-like cone) and so the Frobenius theorem does not hold.... Thanks for the answers ^^
And really cool video, I had already seen explainations, but never as clear as this one, thanks !!
Number systems such as complex numbers and quaternions do not come into play here.
What you are talking about is called a _Minkowski metric._ Spacetime (at least, flat spacetime, as in special relativity) is an example of what is called a _Minkowski space,_ specifically it is the 4-dimensional kind. There are in fact lower- and higher-dimensional analogues of spacetime; these are the other Minkowski spaces, whose dimension can be any integer 2 or greater.
The concept of the _(Riemannian) metric_ at a point in space comes from the theory of _manifolds_ (potentially curved spaces), which is known as (pseudo-)Riemannian geometry. The metric is the object that encodes the notions of distance and angles at each point in the manifold. In an n-dimensional manifold, the metric is an n-by-n matrix whose specific value depends on your location in the manifold. At any given point in the manifold, the off-diagonal components of that matrix describe the curvature of the space at that point. In a Minkowski space, the metric is constant (i.e. its value actually doesn't depend on your location), and its value is (-1, 1, 1, 1, 1, 1, ...) along the diagonal, and zero in the off-diagonal cells. A metric that is a diagonal matrix describes a "flat" space, so the Minkwoski metric is an example of a flat space. In particular, in the flat 4-dimensional spacetime of special relativity, its value is (-1, 1, 1, 1) along the diagonal, and zero everywhere else.
Once you take general relativity into account, spacetime is actually curved, not flat, and the value of the metric is given by the Einstein field equations, which take into account the distribution of mass/energy throughout spacetime.
If you'd like to learn more about all this, I highly recommend Alex Flournoy's General Relativity lectures, which you can find on here on TH-cam. (I'm not providing a link here because that will cause TH-cam to flag my comment as spam.)
@@JivanPal Thanks for the answer !
I've heard and read a bit about metric space and Minkowski space, and in fact I learned that split-compex numbers is a number system that naturaly encodes 1+1 Minkowski space ("flat" 1+1 space-time), or other exotic number system like split-quaternion, biquaternions, ... Then I read trough a blog called "Introduction to 3D complex numbers" which is very interresting (the style of the blog is... also quite interresting ^^) and seen a video called "Let's invent triplex number".
In fact, I've been playing around myself a bit with this, espacially trying to unroll i^3= +/-1 (I've heard in the video called "So tiny its square is zero!?" at 5:30 that the dimention of a quotient ring equals the degree of the polynomial with respest to which one quotients, I don't know the name of the theorem ^^'), and there are some interresting stuff (the subspace of norm 0 is a cone for example !). The hard part seems to find a good définition of norm...
All this to say that I'm still wondering if a well suited number system could be find to make the relativity calculation easier/more intuitive in 2+1 space-time (of even 3+1 space-time, unfortunately split-quaternion don't do the job...)
Thanks :)
cant you just loop around to the next dimension? x's i points to y, y's i point to z and z's i point back to x? or at a minimum define a vectorspace with those conditions
So xy=z yz=x and zx=y or something like that? Then what is 1? If 1 is x, y or z then you just defined the other two equal to each other. If 1 is something else then it's 4 dimensional.
So actually the Quaternions can be interpreted as a polynomial ring C[j] and k is just an unnecessary symbol.
So all Quaternions can be written as (a+bi)+(c+di)j.
This makes it easier for me to understand why only number systems with 2^n dimensions make sense.
Q(2^(1/3)) has dimension 3 over Q :-)
The dimensionalities of these number systems seem to be sequential in powers of 2:
2^0 == 1 dimensional R
2^1 == 2 dimensional C
2^2 == 4 dimensional H
2^3 == 8 dimensional O.
Why?
Something about the conjugation-style operations used to generate each algebra from the preceding one (the Cayley-Dickson construction) makes the successive algebra into a direct sum of the previous one and a sort of "conjugate" to it.
That construction can be done *ad infinitum* but past the Sedenions (16-dimensional), no more properties are lost: They're all power-associative (meaning x(xx)=(xx)x for all x), and they don't even have special names.
Although it is true that R, C, H, and O are the only real division algebras, it is not true that those algebras and subsequent Cayley-Dickson algebras are the only power-associative real division algebras: In particular, the n×n real matrices form associative division algebras of dimension n², and some interesting sub-algebras can be defined, along with ways to model the complex numbers as 2×2 real matrices and the quaternions as 4×4 real matrices.
@@JamesLewis2 Thanks for the insights!
28:06 When you defined “y”, did you assume without proof that A is the union of B and the Complex numbers? How did you get from “y is in A” to “y = 1/2 (y - sigma(y)) + 1/2 (y + sigma(y))”?
Edit: the right hand side simplifies to y via distributivity
i thought about this once, but my thought ended at sqrt(i) is in C since ((1+i)/sqrt(2))^2 = i
What systems open up if you allow zero divisors? what functionality do these systems lose over quaternions / complex / reals?
I learned so many new words today.
7:56 - I'm not sure why L having trivial mull space and L(x) = ax will show that L is surjective
So L is a linear transformation from A to A, and A is a finite-dimensional vector space.
By the rank-nullity theorem, nullity L + rank L = dim A. Since nullity L = 0, we get that rank L = dim A. So the range of L is a subspace of A having the same dimension as A.
A general fact from linear algebra is that if V is a finite-dimensional vector sapce, then whenever U is a subspace of a vector space V and dim U = dim V, then U = V. So using this fact, we get that the range of L is all of A.
@@MuffinsAPlenty thank you!
Hey, i don"t fully understand why the transform must be surjective based on the size of the null space. Thanks for the help !
When it comes to linear transformation, the injectivity is equivalent to having a null space equal to {0}. Here, the transformation goes from A to A, and since A has a finite dimension the injectivity of such a linear transformation is equivalent to its surjectivity.
At 7:55 is an interesting claim: that ax is surjective on A. It's intuitively sensible but it is not immediately obvious why it rigorously must be true. Then again, a lot's deliberately elided. Perhaps the notion that linear polynomials have one root is part of making A an algebra?
The surjectivity claim follows from the combination of two useful facts from linear algebra:
1. The Rank-Nullity Theorem
2. Assume V is a finite-dimensional vector space and U is a subspace of V. If dim U = dim V, then U = V.
I think it is interesting to ask when linear polynomials might have more than one root, though. Suppose ax+b is a linear polynomial with more than one root. Then there are c != d with ac+b = 0 = ad+b. If we are allowed to subtract b from both sides, we get ac = ad. From there, we can do some algebra to get a(c-d) = 0. In order to be a linear polynomial, a can't be 0, so by the no zero-divisor rule, c-d = 0, giving c = d (by adding d to both sides).
So it seems like linear polynomials have only one zero because every element has an additive inverse (subtracting b, subtracting ad, adding d) and because we do not allow zero-divisors.
@@MuffinsAPlenty Proof by contradiction may be fun, but simple calculation suffices: Given ax=b, then x=b/a makes a root when division works as we expect. Regardless, the formula relies on division to work properly -- i.e. it relies on the existence of a multiplicative inverse, which is to say it relies on multiplication being surjective, which is the original question. Obviously it works in R, but there's no reason a three-dimensional vector-space would have to have surjective multiplication. It's an assumption. MP showed eventually that R^2 has to be the complex numbers but yet has used the dual numbers in his own vids; evidently these lack some property relied upon to be an algebra of the sort we seek. But they're a thing with interesting properties of their own, so the question becomes how much familiarity must we sacrifice to get a meaningful R^3?
@@IanKjos "Obviously it works in R, but there's no reason a three-dimensional vector-space would have to have surjective multiplication. It's an assumption."
I gave you the key steps of the proof in my original comment. Perhaps I should spell it out more. Since ℓₐ is a linear transformation, by the Rank-Nullity Theorem, we have that nullity ℓₐ + rank ℓₐ = dim A. Here, the no zero-divisor property gives that nullity ℓₐ = 0, which in turn allows us to conclude that 0 + rank ℓₐ = dim A, or more simply, rank ℓₐ = dim A. Recall that rank ℓₐ = dim(range ℓₐ). Hence, we have that range ℓₐ is a subspace of A which has the same dimension as A. Since A is finite-dimensional, this implies that range ℓₐ = A (fact 2 that I stated above), which means that ℓₐ is surjective.
This does not rely on R specifically. You can replace R with any field F and A with any finite-dimensional algebra (vector space with a bilinear "vector multiplication") over F.
"MP showed eventually that R^2 has to be the complex numbers but yet has used the dual numbers in his own vids; evidently these lack some property relied upon to be an algebra of the sort we seek."
Yes, the dual numbers have zero-divisors. In particular, ε ≠ 0, but ε^2 = 0, so ε is a zero-divisor. Note that the "multiplication by ε" function would have a nontrivial null space, since ε would be an element of the null space.
"so the question becomes how much familiarity must we sacrifice to get a meaningful R^3?"
You can get something like that by giving up the no zero-divisor rule. You could, for instance, take R[X]/(X^3) for a variable X. This would be a lot like the dual numbers except that instead of having something that squares to 0, you get something that cubes to 0. You could do this for any power of X to get any dimension you want.
Alternatively, if you don't want to bring in the sledgehammer that is the rank-nullity theorem, you can prove the following statements.
Let L : V → W be a linear transformation of vector spaces.
1. null(L) = 0 if and only if the image of every linearly independent subset of V is a linearly independent subset of W.
(i.e., null(L) = 0 if and only if whenever S is a linearly independent set of V, we have that L(S) = {L(x) : x is in S} is a linearly independent subset of W.)
2. L is surjective if and only if a spanning set of W is contained in the range of L.
These are much easier statements to prove than the rank-nullity theorem, since essentially you only need to use the definitions of things like null(L), linearly independent set, spanning set, surjective, and linear transformation to prove these.
Now, since ℓₐ : A → A has trivial nullspace, we know that every linearly independent set of A (domain of ℓₐ) is mapped to a linearly independent set of A (codomain of ℓₐ). In particular, every basis of A (domain of ℓₐ) is mapped to a linearly independent set of A (codomain of ℓₐ). But this linearly independent set of A (codomain of ℓₐ) has the "right number of elements" (dim A) to be a basis of A (codomain of A). Since it's linearly independent and "the right size" (dim A), it is a basis of A (codomain of ℓₐ), and hence, is a spanning set of A (codomain of ℓₐ). But this means that a spanning set of A (codomain of ℓₐ) is contained in the range of ℓₐ, so ℓₐ is surjective.
The thumbnails keep getting better and better!
2 robux. 1 for you, another for your graphics designer.