Thank you! I failed linear algebra last time I took it and I think when we started talking about vector spaces was when everything started to go way over my head, but now I understand it a lot better!
I think one of the coolest things is that you can always equip the set of all homomorphisms from one vector space to another with an addition and scalar multiplication such that they form a vector space.
My cousin sent me this saying you looked like the male version of me, at first I was upset as recently someone said I looked like a brown snape, but after skimming your video your mannerisms made me feel nice and I'm sure make vector spaces more digestible. Thanks for the good content twin!!
Quantum physics got its vector space mentioned (or close to it, Hilbert space), but General Relativity got left out, as usual. Simple example is the tangent spaces (tangent bundle) to a circle, one vector space (the tangent line) for each point of the circle. For a sphere, the spaces are the tangent planes. Each vector defines a directional derivative for functions of points on the sphere, and the vectors work just like differentials. See Robert Wald's text. And Tensors are defined as linear maps from vectors to reals, etc.
For a while the only kind of vectors I knew about were in some sense functions. For instance, a vector in R^3 is a function from {1,2,3} to R. An interesting vector space where the vectors are not functions is a quotient space: let V be a vector space and U a subspace. For every v in V we define the "coset" [v] := {v + u | u in U} We define the obvious addition and scalar multiplication on cosets: [v] + [w] := [v + w] a*[v] := [a*v] Under these operations the set of cosets is a vector space, denoted V/U. Its vectors are sets of vectors in V.
amazing video clears up all my questions! thank you!
5 лет назад
Tensor calculus, please :) I would like to learn (as a hobby ...) in that direction, but the topic is not always clear for me (I've tried several youtube lectures on it, maybe I have some idea now, but the Peyam style can boost the learning curve, I bet ....).
Yes. Quantum physics got its vector space mentioned (or close to it, Hilbert space), but General Relativity got left out, as usual. Simple example is the tangent spaces (tangent bundle) to a circle, one vector space (the tangent line) for each point of the circle. For a sphere, the spaces are the tangent planes. Each vector defines a directional derivative for functions of points on the sphere, and the vectors work just like differentials.
There is an issue with the vector space of {0, water, syrup, drink} . What is water-drink? They aren't in the set itself, thus it breaks the first/third rule of vector space.
Mon 1er professeur d'algèbre linéaire nous a introduit les applications linéaires avec un marché de tomates, carottes, et aussi avec des lapins et des chapeaux de magicien, pour le côté abstrait.
That was not one of the ten if you meant the list of axioms at 1:05. In number ix c and d are scalars so you are not multiplying, say, matrices by each other. Or in other words AB= BA is not a requisite of a vector field.
Very interesting. So for P2[x] must be regarded as a subset of P2[x,y] and we cannot specifically imply 'in 2 variables' when subsets in P2[x] 'in 1 variable' are acceptable (to establish or demonstrate a vector space for P2[x,y]). For example, let me modify your example: (x^2 + y^2 +xy) + (x^2 - y^2 -xy) = 2x^2 an element of set P2[x] which must now be regarded as a subset of P2[x.y] if P2[x.y] is to meet the test (axiom) and be a vector space. Likewise in general Pn[v] must be a subset of any arbitrary dimensional Pn[v1, v2, v3,...vm] if they are to be considered vector spaces.
are functions just a subset of mappings? f(x) = 3 and 2 is not a function because its output isn't one value (perhaps if the set of all values were collected into a vector it could be called a function)
Isn't every field (Körper) also a vector space? e.g. the real numbers. The ten commandments for vector spaces strongly resemble the axioms of addition and multiplication those must obey
Yeah it is, a vector space over itself, but that’s not interesting, it’s like studying R as a vector space over R. What’s more interesting is studying things like R as a vector space over Q
@@drpeyam What does it mean to say "___ is a vector space OVER ____"? so far in your video I've only understood what it means to say "___ is a a vector space" so I do not understand "over".
11 is not close to being a vector space. And 13 is not one either EDIT: Namehzd has informed in in the comments that 13 does seem to work, so my bad. But we can all agree that 11 is not close to being a vector space
Elaborating: #11 is not a vector space since in a vector space, a+b=b if and only if a=0 (since the set of vectors with addition is a commutative group), however here we have a case where a+b=b, and a+c≠c, which is a contradiction. -#13 is not a vector space since there are scalars c and d and vectors u such that (cd)u≠c(du), for example, take c=2, d=3, u=1-
@@Demki I think 13 is correct, the example c=2, d=3, u=1 works: c*d (in R) is 6, (cd)u = 6 times 1 which is 6 + (1 - 6) = 1. And du = 3 + (1 -3) = 1 and c(du) = 2 times 1 which is 2 + (1 -2) =1. The space is just R (as a real vector space) but shifted to the right by 1. So you define x +' y = ((x - 1) + (y - 1)) + 1 and c *' x = (c * (x - 1)) + 1. When looking at it this way it seems petty clear that it checks out. You're right with #11 though...
Vector space {0,a,b,c,a+b,a+c,b+c,a+b+c}, scalars {0,1}, 1=-1, a=-a, a+a=0. I believe all the axioms are satisfied. a+b=b+a, k(a+b)=ka+kb, (i+j)a = ia + ja. What is missing?
You only showed us 24 or even 23 vector spaces :( 11) is clearly equal to 16 as water = syrup = drink = 0 so 11 is the null vector space. Also P2[x] is contained in Pn[x] so it is trivial. Well basically all of the examples are sort of trivial so I was not expecting anything really
That axiom is only to remove {0} from the set of vector spaces, which iirc simplifies some things so you don't have to specifically exclude the trivial vector space. It's up to you whether it makes things better or not.
@@Demki Oh yeah. I was thinking of rings. Sorry. I completely missed the big difference between properties (iii) and (x) at the start of this video, i.e.: vector *0* and scalar 1.
