The strange utility of this mutant matrix.

[@MichaelPennMath]

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[@TheEternalVortex42]

We can't have a normal matrix representation because matrix multiplication is associative.

[@nathanisbored]

its cool that these mutant matricies are kind of like non-associative matricies. i wonder how many things from the study of matrices can still be applied to them, with only minor tweaks. If they dont correspond to linear maps, what do they correspond to in general?

[@SimonClarkstone]

The definition of matrix multiplication using cross product inside entries like that has a nice property: if you replace the R^3 vector with an R (scalar), you get normal matrix multiplication because cross-product on scalars is always 0, so they vanish from the sum. (I deduce that the cross-product of scalars is 0 from something mentioned in an earlier video: 0, 1, 3, and 7 dimensional cross-products are the same as multiplication of the reals, complex, quaternions, and octonions respectively, if you *only* use the imaginary parts and keep setting the real parts to 0. Specifically, scalar cross-product is related to the multiplication of imaginary parts of complex numbers, which always results in a 0 imaginary part.)

[@nerdatmath]

If the split octonions have non-associative multiplication, then they can't be represented by normal matrices, for which multiplication is associative.

[@Happy_Abe]

Is it because the octonians aren’t associative and matrix multiplication is?

[@franzlyonheart4362]

10:59, ... "plus ZERO times (1,0,0)", that zero was missing here but is used later on when cancelling. All good.

[@robshaw2639]

it's a little unexpected that the basis is split into 3 elements that square to -1, and 5 elements that square to 1... Does 1 "not count", so the 3/4 split is the most even you can get?

[@TheEternalVortex42]

Well, you could consider the split-complex numbers which have 1^2 = 1 and j^2 = 1 as basis elements. Removing those we see that the remaining elements do produce an even split (the same is true of split quaternions, etc.). Of course if we had started with 1^2 = 1 and i^2 = -1 then we would get the normal complex/quaternions/octonions.

[@mairc9228]

Loving this mini series on real algebras!

[@VaradMahashabde]

I think there is some utility in writing the the representation in terms of Pauli matrices. e1,e2,e3 use i*s_y and the rest use s_z or s_x. However we need multiple copies of s_x and s_z with defined multiplication between them to encapsulate the vector part, and we are kinda back at the split octonions

[@xizar0rg]

Is the mapping going to be a bijection? It seems like going from the resulting summed matrix back to the "regular" scalar-ish representation should be unique.

[@ivandebiasi6657]

Yes, it is. You have for any couple of coefficient the sum and the differencein the matrix, and so you can easily solve and prove uniqueness

[@franzlyonheart4362]

Yes. The Octonians are obviously 8-dimensional over |R, hence their name. For the number-vector matrix representation, if you count the "number entries" in those whacky matrices, there are the 2 regular numbers on the main diagonal, plus one 3-dimensional vector each on the two off-diagonal entries, which gives you the remaining 3+3 = 6 dimensions for the total of dim = 2+6 = 8. So the dimension is already correct, and it's pretty easy to pick numbers (1s and 0s should suffice) using that map which will give you 8 linearly independent vector-matrices. So the dimensionality works out which proves the bijection. More concretely, ignoring the matrix shape and just using standard linear algebra for the 8 coefficients, it will even be easy enough to write up the inverse map. On terminology, I guess the term "representation" ought to always imply an isomorphism (my memory may be a bit rusty here), and an isomorphism is certainly always bijective (that's part of its definition).

[@DrYankeeDoodle]

@@franzlyonheart4362 Thank you for this clarification, and also a line from a user above, about it having sum and difference incorporated, so it's possible to extract coefficient for e's.

[@GastroenterologyPINNs]

The channel uploaded video in 2160p so that audience could understand the problem clearer

[@jimdotz]

The next step up would be to do the "split sedenions" (if such a structure exists). My hunch is that the "mutant matrix" representation of the "split sedenions" would be a 2x2 matrix with all four components having the form of 3x1 column vectors... but I could be wrong.

[@esajpsasipes2822]

If i go off other comments - sedenions have 16 degrees of freedom, while the 2x2 matrix would have only 12 diffirent degrees of freedom.

[@pierreabbat6157]

Is it possible for a zero divisor to have an inverse?

[@robshaw2639]

is there typo in the diagram? not all of the outside edges have the same orientation: e4->e3->e7 goes against the other two edges.

[@Curmesh]

No, this is the structure of split-octonions, otherwise it would be just normal octonians

[@assassin01620]

This is "split-octions." If they have the same orientation, then it would be regular octonians.

[@robshaw2639]

@@assassin01620 in regular octonions, all the e's square to -1. The change is some now square to 1 instead - I thought that was the only change... I'm sure the diagram should change

[@robshaw2639]

oops meant to say "not sure" the diagram should change

[@robshaw2639]

Actually, we are both wrong - I just looked it up. That is the correct diagram for regular octonions - not sure why it has that wart. So, the diagram doesn't change for split octonions, just what squares to -1.

[@Alan-zf2tt]

As a second guess: for computational algebra. Either for numerical analysis such as working out the values, plotting charts, graphs interpolating/extrapolating OR for algebraic computational math retaining surds and powers and stuff like that. A limitation on numerical analysis might be error bounds either by percentage or by powers of epsilon e.g. accuracy up to epsilon squared. Basis for conjecture: computers do it quicker, cheaper and within parameters set at the time - forming an algebra incorporating mutant matrix lays groundwork to defining a computer algebra to make it all happen. Randomized tangents: projective geometries and (guessing here) injective geometries - a wonderful world of application potentials?

[@mathechne]

Wonderful! Is there any links with Markov matrix?

[@BrianDominy]

Why is b4 the coefficient in the diagonal entries? b4, b5, b6, b7 look symmetric in the original representation.

[@iabervon]

The definition of multiplication having those cross products is kind of unexpected. Is there a good explanation as to why that's the natural definition, beyond that it's in the split octonian multiplication formula and nobody wanted this notation for anything else?

[@TheEternalVortex42]

I don't really think this definition is "natural" in any sense, it's not like we use these matrix-vectors elsewhere. And cross product only existing in 3/7 dimensions means it's not very generalizable (though the 7 dimensional one is roughly giving you the octonions already). I imagine Zorn was just messing around with how to add non-associative operations on matrices and found this.

[@franzlyonheart4362]

These octonians and split version are interesting to me. I was doing all that group and ring theory stuff, and that Lie algebra stuff, myself earlier in my life. I work in a totally different field now, but it's good fun to watch such quirky new things like "Octonians" on KZbin. I hadn't even come across those back in my mathematician former life (everyone learns about quaternions in the first semester, but not everyone hears the stuff beyond). Having said this, I totally forgot how the cross product works. Haha, perhaps it's too simple!

[@SimonClarkstone]

The definition of matrix multiplication using cross product inside entries like that has a nice property: if you replace the R^3 vector with an R (scalar), you get normal matrix multiplication because cross-product on scalars is always 0, so they vanish from the sum. (I deduce that the cross-product of scalars is 0 from something mentioned in an earlier video: 0, 1, 3, and 7 dimensional cross-products are the same as multiplication of the reals, complex, quaternions, and octonions respectively, if you *only* use the imaginary parts and keep setting the real parts to 0. Specifically, scalar cross-product is related to the multiplication of imaginary parts of complex numbers, which always results in a 0 imaginary part.)

[@patrickstrasser-mikhail6873]

How does this compare to a Clifford Algebra (Geometric Algebra) of Cl(3,5)?

[@BongoFerno]

What is the dimension of that matrix? Is an integer number or a rational? Von Neuman defined matrices with non integer dimension, but it goes over my head. All I wanted to know is what is a matrix with non integer dimension.

[@wilderuhl3450]

Oh jeez. This is super cool, but what motivations led someone to come up with this?

[@TheLiuzp]

Why does this product of matrices needs the cross product? Wouldn't be enoutgh to have just the linear combination of the vectors?

[@siquod]

Why did you draw the diagram in such an asymmetric way? e4 should be in the middle.

[@Axacqk]

Where does "so on and so forth" go further in the direction octonions -> quaternions -> C -> R?

[@muhammadariffjazlanbinjoha9031]

one way is to say R -> Q -> Z -> N, but in algebraic geometry some would consider the tropical numbers as the 'next step' after C and R

[@mostly_mental]

This is a really cool result, and it's not hard to prove it works once you've seen it. But how on earth would anyone come up with this representation?

[@schweinmachtbree1013]

I imagine by first writing split-octonion multiplication in terms of dot products and cross products (you can do the same thing for quaternion multiplication) and then noticing that the resulting formula can be written in terms of a "matrix" product if you squint enough.

[@MathsMadeSimple101]

Are we living in a matrix

[@goodplacetostop2973]

19:24

[@yairman4]

As a graduated EE all i can say is WTF mathematicians do weird things

[@MDNQ-ud1ty]

Mathemagicians run the world.

[@jimdotz]

We pure mathematicians wear that WTF badge proudly! It's what we're all about.

[@MDNQ-ud1ty]

@@jimdotz They laugh at us now, they always do... but we get the last laugh because we know about limits.

[@someheree6362]

Having block matrices simplifies calculations with nice properties in big matrices. Vector entries is a way to represent tensors to people who cannot process tensors

[@Alan-zf2tt]

Well in one way the matrix operation may provide easier ways to find solutions based on isomorphisms. Is it correct to say if we have an isomorphism then we have 2 isomorphisms or is that a step too far? My simplistic thought: an isomorphism can be helpful in finding an easier solution event space for things that have a bothersome problem in source event space. Interim conclusion: it depends. Just like all isomorphisms? Plus! If there exists a one-to-one correspondence then really we have to 2 two-to-two too? Dah-dah de daaaah 🙂

[@trueriver1950]

Hmmm. Not only is associativity an issue for this mutant matrix, but I'm unclear how is take it's determinant, or figure out it's inverse (if any).

[@robshaw2639]

Is this a group? If so, then there is at least some representation using normal matrices, but maybe the dimension is inconvenient...

[@iabervon]

It's not a group, because it's not associative.

[@TheEternalVortex42]

No, since it contains elements without inverses. Indeed since matrix multiplication is associative we know that no matrix representation can exist.

[@franzlyonheart4362]

No … I presume you are attempting to answer his "homework question" why no (isomorphic) Matrix representation exists. The reason is the fact that matrix multiplication is always associative, but Octonian multiplication is non-associative.

[@JosBergervoet]

ℂ𝕆𝕆L !!