Matrices with equal entries are super interesting! 

Yes! 6 is the neutral element of ({2,4,6,8}, multiplication under mod 10) 6*2=12=2 mod 10 6*4=24=4 mod 10 6*6=36=6 mod 10 6*8=48=8 mod 10 Multiplication is commutative, so multiplication under mod 10 is commutative "for free". And I don't need to check for the reverse multiplication 2 and 8 are inverses 2*8=16=6 mod 10 4 and 4 are inverses 4*4=16=6 mod 10 6 is the identity.

I think your constraint needs more constraints. Consider elements eae and ebe. Multiplying them gives eaebe under these rules, but this only works if aeb is an a-type term, which is not guaranteed for all choices of e and a.

The identity of a subgroup must be the same with the identity of its parent group. Notice that G defined at 5:50 is not a subset of GL2 and G is not a subgroup of GL2. Hence they can have different identity.

@@yuan-jiafan9998 oh thanks! Hadn't noticed that. So why is the identity destined to be the same in a subgroup?

@@atreidesson because for all elements in the subgroup we know that it has the identity of its parent group and that it is unique so therefore it must be the same identity in the subgroup

@@wynautvideos4263 Well, we prove that an identity A is identical by applying it to an other potential identity B and receiving 1) B, which means B is not an identity because it doesn't preserve B*A = A, 2) A, which means, our A was not an identity because it doesn't preserve A*B = B. So we use the fact that the two identities are contained in our operating set, and so they must preserve each other. But the point is, we can abandon the original identity when choosing subgroup elements, but our subgroup will have some element which is "locally identity" for that subgroup

Another commenter pointed out that whenever you have any element e of a ring where e*e=e, the elements e*a*e (where a is another element of the ring) form a subgroup with identity e over the same operation. This element doesn't have to be the identity e.g. 6 over multiplication mod 10.

And even a commutative field :-) . And yes, isomorphic to (R, +, x) .

​@@ericbischoff9444 in other words the matrices behave exactly the same as their traces

A field is a field

I think the span of any single vector can be taken as a field isomorphic to the one used to define the vector space

Sounds pretty related. The set of means is a group.

@@MrRyanroberson1 I'm not sure what you mean by that. Can you elaborate on that?

I was wondering about that myself -- it seemed like it'd be equivalent to multiplication of the real numbers if you relabeled the matrices the right way, but I couldn't quite figure out how to prove whether the internal connections between elements actually corresponded or not. Using the trace is a simple and elegant way to prove the isomorphism!

But isn't it just a coincidence that trace gives the correct scaling factor?

@@adayah2933 What do you mean by that?

@@Convergant It's obvious that the isomorphism should be defined so that the matrix with all entries equal to x maps to s*x, where s is some scaling factor. With a little calculations it can be found that s=2 is the right factor (or s=n in n dimensions). Trace happens to give the same factor. But is there any direct reason why trace would be the correct isomorphism, other than randomly producing the same constant? For example the identity tr(AB) = tr(A) tr(B) could be such a reason, but it is not true in general (although it is true for our matrices - but then again, it is not just a fluke?).

​@@adayah2933 I wouldn't say it's a coincidence, but rather, a special case. In general, you can write a function of the form f(A) = sgn(a) * k^(log2(|a|)+1) as the isomorphism, for k>0, k=/=1. Using tr(A) = 2a is the special k=2 case. The trace having this property is a necessary part of the construction of this group, which is why I think it's more than a coincidence. To check this really works: f(A)f(B) = sgn(a)sgn(b) * k^(log2(|a|)+log2(|b|)+2) =sgn(ab) * k^(log2(|ab|)+log2(2)+1) =sgn(2ab) * k^(log2(|2ab|+1) = f(AB) f being a bijection is left as an exercise to the reader. f is well-defined from G to R*, and that is left as a much more trivial exercise for the reader. Thus, f is an isomorphism from G to R*. I suspect (although I haven't verified) that you can pepper the function with discontinuities without stopping it from being an isomorphism. The function I used still has to hold for integers values of x, though.

Look up doubly stochastic matrices

@@literallyjustayoutubecomme1591 Yup, just glancing at the Wikipedia article that is exactly the type of thing I had in mind. :) I find one thing interesting, it was noted that multiplication is closed, however the inverse of a DS matrix need not be a DS matrix. (Relevant for this video.)

So it actually has some real world applications?

Wait but when you use matrices for modeling quantum states, the entries will not usually be the same

@redpepper74 This will be used only as a subgroup(component).

Hey what is the group operation on this? I know they’re defined by cohomology classes, but I’m not immediately seeing what the operation between classes is

​@@noahtaulHey Noah, thanks for the question! Say you got two Extensions 0 -> N -(f1)> E1 -(g1)> M -> 0 and 0 -> N -(f2)> E2 -(g2)> M -> 0. For clarity: by N -f> M I just mean f: M -> N. The following construction is also known as the Baer sum of E1 and E2. First we define the pullback (of g2 and g1) over M: E' = {(e1,e2) in E1×E2 | g1(e1) = g2(e2)} Then we set F = E'/{(f1(n),-f2(n)) | n in N}. Now we get an exact sequence (hence an extension of M by N): 0 -> N -f> F -g> M -> 0 where f(n) = [(f(n),0)], where [•] denotes the equiv. class and g([e1,e2]) = g1(e1) = g2(e2). You still have to proof the maps I gave are well defined and the sequence is exact :).

I think the group is best understood as the semidirect product of its subgroups: (normal) D consisting of diagonal matrices and S consisting of two matrices, the identity and the "read coordinates backwards" matrix (aka 1s on the "antidiagonal", 0s elsewhere). That's where parity comes from. Also it can be easily generalized by taking S to be the set of matrices corresponding to an arbitrary group of permutations of coordinates.

Just noticing that, for any n, it's also Abelian, because all its elements are scalar multiples of one another.

It can be relatively easily seen to be isomorphic to R because ab + a + b = (a + 1)(b + 1) - 1, so it’s like shifting them by 1 and doing the multiplication and then shifting them back again

I live in Poland and can't confirm this. At the University of Warsaw linear algebra is a 1st year undergraduate course white abstract algebra is a 2nd year undergraduate course. In linear algebra the only thing related to abstract algebra I had was that linear algebras are defined over fields but they didn't even explain properly what fields are at that point.

@Maciek300 So it apparently depends on the university. Over a decade ago, Automatics and Robotics at AGH had an Algebra course, starting with abstract algebra basics. The same is the case for Physics at UJ nowadays.

@@Macieks300 Same on University of Wrocław, except they do explain properly what fields are. ;P

@@adayah2933 How much time was spent on explaining fields in your linear algebra class?

I don't know about Poland, but when I was a student vector spaces were seen after groups in France. Which kind of makes sense because it's more complex structures (and vector addition defines a group).

Yeah, this is the clearest way to see what's going on. All that's needed to go between the two pictures is to note that the square of the matrix (1 1; 1 1) is twice itself.

I guess it is the same but it just shows how the same group can be defined in seemingly different ways.

Not really fake complexity. It's a representation, like yours. I will add a very nice one, but that is subjective. Also, there are no arbitrary rules, it was totally natural. You are suggesting a path that requires defining a*b=2ab for real numbers, even when R-{0} is already a group with the "natural" operation.

What makes complexity "fake"? Isomorphisms are what make math interesting! :)

It is very much a fake complexity. Just being another representation doesn't mean that they are equally simple or complex. The moment you realize that it's something as simple as 2ab (by which I mean that you can prove homomorphism) , the matrix farce along with its complexity can be dropped.

Yup. That would be a much easier way to demonstrate that this set forms a group

It looks like you don't need the square root or the sign; you can just take f(M_a) = 2a. This is obviously a bijection, and has the property f(M_a) * f(M_b) = f(M_a * M_b) because f(M_a) * f(M_b) = 2a * 2b = 4ab, and f(M_a * M_b) = f(M_(2ab)) = 2*2ab = 4ab, so it is an isomorphism. Similarly to the isomorphism presented in my original comment, you can extend it to the nxn repeated matrices by changing f(M_a) to na. I'll change the original comment and show my original isomorphism here: f(M_a) = sign(a) * sqrt(2|a|), where M_a is the 2x2 matrix with a as all its entries.

so has nice geometric interpretation. i consider (rows x columns) matrix to be 1-nested row-ary column-digit number system. Our 2x2 matrix is interior (edge) of 2-vertex simplex ( binary 2-digit number system). 2 vertices and exterior are taken away because a0 (only "11" of "00 01 10 11" is valid). What is impossible, first order shape is broken. So , by means of self-closure, interior(edge) became point(exterior and interior not counted), wich is valid shape. Matrix operator "*" has no carry. Digits connected other way: 1)each row of at right matrix is a face-vector of row-shape. 2)each column of at left matrix is a total_face-vector of column-shape, i-element of this vector is total amount of faces of (i-1)-dimensional column-shape(if exterior counted i starts at 1 else at 0). For example. 1. if row-shape is square(2D 3-shape,3-ary 2-digit number system) then : (0^0 0^1 0^2)*(4 4 1)=(0+2)^2; column-shape is 0-shape , negative shape wich change kind of direction,gives -1 (see formula at the end), -1-ary 0-digit number system. (1^0 1^1 1^2)*(4 4 1)=(1+2)^2; column-shape is 1-shape, kind of neutral shape wich only convert face-vector into total amount of faces of same shape(see formula at the end), point exterior and interior not counted(donut_1, 0D sphere), unary 1-digit number system. (2^0 2^1 2^2)*(4 4 1)=(2+2)^2; column-shape is 2-shape, point exterior xor interior not counted (1-vertex simplex), binary 1-digit number system. (3^0 3^1 3^2)*(4 4 1)=(3+2)^2; column-shape is 3-shape, point exterior and interior counted or 1-chain exterior not counted or 3-ary 1-digit number system. (4^0 4^1 4^2)*(4 4 1)=(4+2)^2; column-shape is 4-shape : 1) 2-vertex simplex exterior and interior counted, binary 2-digit number system. 2) Donut_4 (2 vertices connected by 2 edges or 2-ring ) exterior and interior not counted, 4-ary 1-digit number system. ((odd)^0 (odd)^1 (odd)^2)*(4 4 1)=((odd)+2)^2; column-shape is odd-shape , cube_odd exterior not counted, odd-ary 2-digit number system. ((even)^0 (even)^1 (even)^2)*(4 4 1)=((even)+2)^2; column-shape is even-shape , donut_even exterior and interior not counted , even-ary 2-digit number system. 2. if row-shape is cube(3D 3-shape,3-ary 3-digit number system) then: (0^0 0^1 0^2 0^3)*(8 12 6 1)=(0+2)^3 (1^0 1^1 1^2 1^3)*(8 12 6 1)=(1+2)^3 (2^0 2^1 2^2 2^3)*(8 12 6 1)=(2+2)^3 3.if row-shape is 4-cube(4D 3-shape or 3-ary 4-digit number system) then: (0^0 0^1 0^2 0^3 0^4)*(16 32 24 8 1)=(0+2)^4 (1^0 1^1 1^2 1^3 1^4)*(16 32 24 8 1)=(1+2)^4 (2^0 2^1 2^2 2^3 2^4)*(16 32 24 8 1)=(2+2)^4 So you see total_face-vector of column-shape * face-vector of row-shape = total amount of faces of (row+column-1)-shape. Complexity to all above stuff gives point. Using rule "those place where located Counter is not counted by him" we receive 3 types of point. 1) "exterior xor interior not counted". Counter located in exterior xor interior , he counts cube-like shapes and 1-vertex simplex, wich is charge monopole in physics to my mind,. This type gives 2 subtypes + and -, or signature of curvature of cube-like shapes, wich are inherited by all possible cube-like shapes, you can not change it except by operator like NOT, wich inverses dark (interior not counted) and not dark(exterior not counted) cube-like shapes . 1-vertex simplex can be "+"("0" exterior "1" face) and "-" ("0" face "1" interior),. 2) "exterior and interior not counted". Counter located in exterior and interior , he counts donut-like and sphere shapes 3) "exterior and interior counted". Counter dislocated , he counts simplex--like shapes except 1-vertex simplex., wich is cube-like and can be dark and non-dark and to Wich can be applied operator like NOT. It is useless to apply operator NOT to other simplexes and typ 2 shapes because it does not change anything,, they have no signature.All what is happens is reverse of face-vector(curvature-vector), what gives nothing new.

Here’s my idea: The Cayley-Hamilton theorem says that a matrix A satisfies its characteristic polynomial, the constant term of which is a sign times the determinant of the matrix (which is nonzero if A is invertible). If we move that constant term over and multiply by A^-1, and divide by the constant, we find that A^-1 is just a polynomial in A. So then if A is a linear combination of the matrices, then we can show that all nonnegative powers of A are too, since the base set of matrices is closed under multiplication. Then any polynomial in A is also a linear combination of the base matrices, and hence A^-1 is too.

@@noahtaul That's pretty nifty! I have to think about some of the details but this is the simplest solution my non-algebra brain can understand. Thanks!👍

You're almost there. Take (a a a a) -> 2a as your isomorphism, and identity is mapped to the identity :-) .

They are not a group as part of a bigger set, but are a group on their own. Like soup is an appetiser if you eat it as the first course of a meal, but it is not if it forms the complete meal.

It is a different group. Even though you are using matrix multiplication for both the calculations it is a different group operator. In the first it behaves as you expect, but in the second it is behaving as normal multiplication of real numbers, just with all the numbers relabled to funny matrices.

A group is defined by its elements, operations AND identity. This is a different group from std mat mul because its elements are a subset and its identity element is different. And, of course, all the group conditions hold.

So you find numbers worthless to study then, since isomorphisms are symmetric? I guess the two groups do not require to be studied separately, beyond their isomorphism, is what you mean? 😉

How many students will think at the isomorphism approach ? Also, this might be presented even before you introduce isomorphisms.

@@ericbischoff9444 Many, I bet. Not sure if they would attach the label "isomorphic", bit many would intuit that these matrices behave just like scalar numbers then.

@@landsgevaer Maybe. I don't know.