I have had this idea as well, but I haven't quite figured out where I land on the idea philosophically. On the one hand this makes sense as it would keep the "main channel" more consistent in terms of its level and niche. That being said, I really like how the popularity of this channel allows me to hack the system in order to get a larger number of eyes on advanced topics. Anyway, I would really like to hear everyone's thoughts about this....

@@MichaelPennMath You should make a community post or a poll on that matter. Maybe ask that in a video for the 200K subs 😛 But yeah, that’s what I would do : two channels for two different contents, for two different audiences.

The road to 200k has really slowed.... I should get better at self promotion CTAs. I will continue to mull this over.

@@MichaelPennMath You’re right. If you reserve this type of content for a second channel, then your just keeping this stuff hidden for only those who particularly are interested in it. On here, more people, including myself, get to see this stuff that I probably otherwise wouldn’t. I’d be down for a combination of your typical content and this where you could increase the frequency of this type of stuff.

Yeah, personally I disagree with splitting up the channels, thought it may sadly be better for KZfaq analytics...

I think variety is the greatest though. It's harder to get into videos that are part of a series

@mythic man sam Agreed

100% agreed. The contest videos are neat and seem to me like good ways to learn new and interesting methods for solving problems, but i much prefer these construction and exploration type videos

that is what "formal linear combination" means, so what's needed is a mention of that

I think that's a great thing to think about, and helps to get an intuition for what these objects are like, but it's a pain to write any examples, since they're all, by definition, awkward piecewise functions. Like {7 if x=1; -1 if x=7; 0 otherwise} is obviously not supposed to simplify to 0, but it's also not a nice notation to work with.

@@iabervon perhaps writing down specific vectors is slightly more cumbersome, but in exchange there is no ambiguity in any of the arithmetic operations which go into computing with the free vectors. It also makes dimensionality obvious: if S is a set with infinite cardinality, V(S) cannot be a finite dimensional space. A basis is also easy to construct: Use the kronecker delta. Free vector spaces as functions removes any requirement for "formal" sums and products. Instead we have literal sums and products. Much easier to compute when you know exactly where your + and * come from.

@@iabervon I agree that notation is cumbersome, but it's a separate question, first it's good to make the concept 100% clear and then use some convenient notation for it

We also did this. And then this helps justify the notation of R^n when you start with R^{1,2,3,...,n} which is V({1,2,3,...,n}) and if you abbreviate the set with numbers up to n by n_, then it's just R^n_ and then a small notational step is to leave out the _ at the end to end up with R^n. So for example, R^3 = R^{1,2,3} = V_R ({1,2,3}) with elements f: {1,2,3} -> R, 1 -> f_1, 2 -> f_2, 3 -> f_3 or in short f = (f_1, f_2, f_3)

I agree that this is a much better model. The use of "formal empty sum" from the video leave uncomfortable. I think to make that method work rigorously, you would need to toss in some extra set element to stand for this symbol. The approach you mention here doesn't run into these issues, and is much more elegant. Also more useful for proving the universal property.

Hey! On my channel I recently made a video that briefly goes over how tensors come up in physics, and where the definition comes from. I'm going to be expanding it into a fuller series where I give the gritty details about how to derive the transformation law, but the first video is more of a "bird's eye view".

unfortunately, "contest problems" videos gets much more views than these higher level maths :/ I hope you choose to do what you like most rather than worry about views :)

maybe its not new, but I enjoy a lot of iridescent colors on the board. aesthetically pleasing math is the way to go.

The usual way to construct the standard 3-dimensional vector space over R is to take the cartesian product RxRxR and define addition and scalar multiplication as follows: 1) (a, b, c) + (d, e, f) = (a + d, b + e, c + f) -- component wise addition 2) k (a, b, c) = (ka, kb, kc) where a, b, c, d, e, f in R and k in R. It is easy to see that this is a vector space satisfying the usual rules for being a vector space.

V(S) is isomorphic to R by sending ax to a. Show that this is a bijective linear map by checking all the requirements. V(S) is just one line because x is a chosen variable that cannot change. X doesn't even need to be a number.

@@somgesomgedus9313 I see. I didn't realize that x is not a variable. So we could take that example to be S={1} and let V(S) be all elements that consist of a formal finite sum of elements of S with coefficients from R. Then V(S) is isomorphic to R. Might have helped if we used c instead of x.

They can't be used to prove that every vector space has a basis, because that statement is actually equivalent to choice, which is independent of ZF, and the definition here only uses ZF. However, I do believe that you can use free vector spaces to prove if every vector space has a basis, then the axiom of choice holds.

You would use 0 as the coefficients for vectors not included. For example, let's say I have x, y, z as three of my elements in S. And I define v = x+3z and w = 2x-y. Then v = x+0y+3z and w = 2x-y+0z So v+w = 3x-y+3z

The confusion seems to be essentially self inflicted. We don't want to do mathematics formally, but to express things informally. Natural language and natural thought process are more appealing to us than the language of set theory and first order logic. I find the set theory definition of the free vector space over a set S (that is, maps from S to R) to be more understandable and easier to do proofs for than the informal definition of 'formal' sums of such and such. But, after doing the simpler proofs, the informal approach appears to be more suitable for understanding more advance results. It is amazing how much 'notation' like v * w allows us to proceed.

These sums are finite by definition. So just take the finitely many elements you're going to add, and name index them 1,,,,n. You have to change your indices for different sums, otherwise as you correctly point out we wouldn't have enough indices.

If S = {0}, we get the 1 dimensional vector space (not the 0 dimensional vector space). Michael Penn used the notation v sub x. One could also use the notation e sub x (which might be more familiar). Thus, if S = {1, 2, 3} , then one wants to form all formal sums of e1, e2, e3 over the real numbers R. That is, one wants to consider all the following as vectors: a1 * e1 + a2 * e2 + a3 * e3 with a1, a2, a3 in R (where addition + is commutative and scalar multiplication * follows the normal rules for vectors). If S = {0}, then one wants to form all formal sums of e0 over the real numbers R: a0 * e0 where a0 in R. Of course, it would be clearer if one started with the set {e1, e2, e3} to begin with rather than with the set {1, 2, 3} since we associate much more structure to 1, 2, and 3 than we do to e1, e2, e3. However, we really want to be able to form the free vector space over any set S (since we are going to later cheat and make possible use of the additional structure that the elements of S have). As has been pointed out in other replies, it may be easier to think of the free vector space over S to be just the maps from S to R that are zero almost everywhere. This formal approach is not used in general since we want to be informal. Compare with the way a polynomial in x is sometimes defined: informally a polynomial in x is like a0 + a1 * x + a2 * x^2 + ... an * x^n. But what is that x? The usual formal definition of a polynomial in x is to take x to be equal to (0, 1, 0, 0, ...) and polynomials to be (a0, a1, a2, ...) where almost all the a's are 0 and where one defines sum in the usual way and the product of two such polynomial sequences using the "convolutional" type product so that x * x = (0, 0, 1, 0, 0, ...) etc.

It is just juxtaposition. It is similar to the subring of all polynomials in x, y, z, ... where there are no constant terms and no quadratic or higher terms (that is, only linear terms in the variables). For example, consider the polynomial 3 x - 5 y + 7 z added to 10 x - 1 z to get 13 x - 5y + 6z. I suspect that you are not asking how x is being multiplied by 3 in this case.

@@billh17 No... because a polynomial comes from R[x] and thus multiplying two elements is well defined because any real and also 'x' are in ring. At least that's my understanding. What I was asking was that we are taking a (sub i) to be defined as a member of the reals... and a set S and then defining V(S) to be a sum over the ordinality of S of products of what? Am I thinking too hard on this. For example if S is (apple, banana, orange) then I should just be happy with 5 apple + 10 banana and not worry about the 'product' of 5 and apple being well defined?

And 'No' was meant as an answer to your question, not meant as disagreement

​@@aoehler1 asked " I should just be happy with 5 apple + 10 banana and not worry about the 'product' of 5 and apple being well defined?" Yes. But, the product of 5 and apple is well defined: it is the juxtaposition of 5 before apple. The 'sum' + is also well defined: it is the concatenation of juxtaposed terms with other juxtaposed terms separated by the + sign. One is trying to define the elements of the set V(S). This must be done before we can define addition and scalar multiplication on V(S) to make it into a vector space. The + sign and the scalar multiplication sign * in the expression " 5 * apple + 10 * banana " are just constant symbols without meaning or interpretation at the beginning stage of just defining the set of elements that make up V(S). When we get to defining a vector structure on V(S), then the * will "transform" into the tensor product which is a real product.

Thanks. Makes sense now

@Joshua Grumski asked "In order to use the convention V(S), must all elements be linearly independent?" S is just a set. There is no vector space structure on S. Even if S is the underlying set of vectors for a vector space structure imposed upon S, that extra structure is forgotten and ignored. Since S is just a set, elements of S being linearly independent has no meaning.

@@billh17 alright, makes sense. So from my understanding, V(S) is an object that provides a vector space structure for the set S? And then the vectors corresponding to the elements of S, are they linearly independent, or do we “forgive and forget” that too? In other words, do the vectors corresponding to the elements of S have to form a basis for V(S)?

​@@joshuagrumski7459 Note: S is not made into a vector space (the elements of S become linearly independent elements of a vector space V(S) that contains S where V(S) is the 'smallest' such vector space). So, yes the vectors corresponding to the elements of S form a basis for V(S), but they don't form a basis for the vector space they originally belong to (if S was a vector space to begin with). @Joshua Grumski asked "do we forgive and forget" the underlying vector space for S if S is a vector space? Yes and no (this is what I mean by cheat). First, note that there are two vector spaces V(S) and S if S is a vector space. They are not related (except the set S is a subset of the underlying set of vectors in V(S)). But, you don't want to think too much about S (the underlying set of vectors) being a subset of V(S). This is not going to lead to too much. The cheating will come in when we take the quotient of V * W by I where we make use of how vectors in V and W can be expressed in terms of other vectors in V and W. But, V(S) doesn't know about this (maybe "I" sort of knows about it). The same thing is done when forming the group algebra over the complex numbers C for a group G. One first forms V(G) to make a vector space. Note that this is a vector space even though G is a group. Now, technically V(G) doesn't know that G is a group: it knows only that G is a set. But, then we make use of the group structure of G within V(G) to make V(G) an algebra that has a multiplication.

@@billh17 ah, ok. So, from my understanding, V(S) defines an addition and a multiplication for the elements of S, and those definitions along with the elements of S form V(S). Is it correct to say that now? Or am I still missing something?

Maybe this might be a better explanation. It looks like you are not assuming that S is a vector space (so things I said about that assumption don't apply). @Joshua Grumski asked "In order to use the convention V(S), must all elements be linearly independent?" The convention V(S) will define a vector space (it must be prove though). After one proves V(S) is a vector space, then one can prove that S is a linearly independent set of vectors in V(S). But the proof that V(S) is a vector space doesn't depend upon proving the elements of S are linearly independent or form a basis for V(S) (which can also be proved). @Joshua Grumski said "Because in the given definition, all elements of S were linearly independent, yet it was never in the definition that this was necessary." In the given definition, all elements of S were not claimed to be linearly independent in V(S) (hard to prove since V(S) has not been defined until it is defined). Yes, the definition is set up so that the elements of S will be linearly independent, but that is just because that is the intention of the definition and not a requirement of the definition. Since "formal" sum is not quite defined sufficiently, it might appear that one is imposing that the elements of S are linearly independent, but with a more explicit definition of "formal" sum you would see that the elements of S being linearly independent is not a requirement to form such sums. @Joshua Grumski asked "would the free vector space just be the span of S?" Yes, but this would require a proof (easy).

Yeah, technically we have a pair of functors between the category of sets and the category of vector spaces.

yup! people usually say this as "every vector space is free" (which is no longer true when we generalise vector spaces to modules)

Free objects typically have a universal mapping property. For vector spaces, that universal mapping property is as follows: Let F be a field. For a set S, let V_F(S) denote the free F-vector space with respect to S, and let ι : S → V_F(S) be the natural map with ι(s) = 1s. Given any F-vector space V, any set S, and any function f : S → V, there exists a unique F-linear transformation g : V_F(S) → V with g∘ι = f. This is a generalization of the following theorem: Let V and W be F-vector spaces. Given any basis β of W and any function f : β → V, there exists a unique F-linear transformation g : W → V so that, for each x in β, f(x) = g(x). Essentially, you can define a unique morphism out of your free object by choosing where the "basis" of the free object gets sent. And you are free to choose where the basis gets sent however you like. (The concept of a free object can be generalized to any concrete category.) This universal mapping property would fail if we allowed for infinite formal sums in V_F(S). As an example, let R be the field of real numbers and let S = {1, x, x^2, ...}. I could choose V = R[x] (the polynomial ring with coefficients in R, but viewed as an R-vector space) and the function f : S → V to be f(x^i) = x^i. If V_F(R) were to be R[[x]] (the formal power series ring with coefficients in R, but viewed as an R-vector space), there is not a unique R-linear transformation g : R[[x]] → R[x] so that g(x^i) = f(x^i) = x^i, since S is not a basis for R[[x]]. S is, indeed, linearly independent in R[[x]], so it can be extended to a basis, but the fact that it can be extended (and therefore can be extended in infinitely many ways) shows that there are infinitely many such R-linear transformations. But you could see this as pushing the problem to only allowing finite sums generally in vector spaces. If we allowed infinite sums in vector spaces, then there would be a unique map there. But algebra typically cannot handle infinite sums except formally. If you consider R as an R-vector space, you would have to contend with questions like 1+1+1+1+... and things like 1+(-1)+1+(-1)+... To handle infinite sums in vector spaces like this, you need a way to determine which infinite sums are "good". I don't know much about Banach and Hilbert spaces. I think perhaps those are the typical ways to do this. Maybe if one has a topological vector space over a topological field, that might work too. Since I'm not very experienced with these topics, I can't tell you if your idea of allowing infinite sums would represent a "free" Banach space or Hilbert space or topological vector space, but those would be the first sort of settings I would look investigate to see if it can be a notion of freeness anywhere.

Seems an answer à 7.46: the operation is not an addition (as in common sens) and what is called 0 is not a real.

But there is still an issure here: + is a formal operation between some amount of elements of S. I understand I can "add" nXi, i.e 3Xi=Xi "+" Xi "+ Xi, but how do you define aiXi where ai is a real number. If S is the set of fruits, how can I define 3.5 banana ?

With formal addition, it's always: a_i v_i + b_i v_i simplifies to (a_i+b_i) v_i, but nothing else simplifies. That is, you write a formal sum as terms with "+" between them, where each term is a number next to an element of S, and you add these formal sums to each other by combining like terms and not combining unlike terms, which means you only need to understand equality for the elements of S.

no, its not.

@@Euler108 huh? It's absolutely true that every vector space is free, assuming the axiom of choice.

which cutlery set u want?

@@user-wl3ts5rd1i fork please

But unfortunately not the last.

@@matthiasbergner8911 XD