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!
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!!
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.
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.
Sir. I want to talk U. Can I get Ur personal contact no
5 лет назад+14
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.
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.
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.
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.
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.
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".
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)
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.
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!
Very useful video. Students often get bogged down in the checking and the proofs, yet this insight is so important.
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!!
That’s awesome hahaha
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.
Love this guys energy
Nice simplicity you put to the subject. Thank you for your lectures Dr.
Well u just saved me so I highly appreciate this
amazing video clears up all my questions! thank you!
Always love your videos
🙂
Merci Dr peyam pour ce cours, je vais vous présenter à mes camarades qui ont des problèmes avec l’anglais et l’algèbre avec un grand A :)
De rien 🥰
I wish you were my lecturer xD. Lectures would never be boring!
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.
U are the best teacher ever I saw. I love your mathod. Sir keep it up. Love from Pakistan
Sir. I want to talk U. Can I get Ur personal contact no
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.
Eigenchris has these content ❤️✌️
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.
underrated teacher
This explains a lot. Thanks
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.
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.
Literally studying this rn thank you!
👍👍👍
So if the first three properties hold, all the others in the list automatically hold?
Pls make a video about basis vectors in that abstract spaces
Of course! It’s already on my playlist
@@drpeyam I remember that you have a vid about orthonormal basis aka Fourier Series but are other ways to do it?
God bless you sir!
excellent, thanks
17:40 unfortunately you lose AB = BA, part of the ten important features. 1 * 4 = 4, 4 * 1 = 1
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.
Omg all subjects I struggle with u explain I Want to meet you in real life and hug you ❤️❤️❤️
Reals as a rational vector space.
That’s a nice one too
Dr Peyam, {0, water, syrup, drink} is still a vector space even though syrup+drink=drink ? aren't there two Os ? syrup+drink=drink=0+drink -> syrup=0
How do you get so much enthusiasm?
Chocolate
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".
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.
Yeah
at 20:20 you say that this is a subspace of a non-vector space (r^2) but isn't r a vector space
I mostly clicked on this because I live in Friedberg (in Bavaria (there are several...))
LMAO 😂😂😂😂
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)
Functions and mappings are the same thing
@@drpeyam what is it called when some function has multiple possible outputs? like x=y^2, f(x)->y, functions are a subset of that.
Multivalued function
Kyle Foregard?
Damn!
Wow Blackpenredpen is here !!!
good shit
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...
@@NAMEhzj oh, I think you may be right. I initially thought the distubutice law does not hold, but I checked it out and it is fine
That doesn't mean there aren't any discrete vector spaces with scalars in {0,1} with 1+1=0, 1+1+1=1, -1=1. But syrup can't be the zero vector.
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?
I want to start linear algebra, can i start it from here ?
It’s best to start from the Systems of Equations playlist!
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
The null vector space is still a vector space
@@piguyalamode164 Yep, never said anything else, I just pointed out that he used the example twice (11 and 16)
Killing 25 flies with 1 stone🤣lame joke but funniest ever
I was taught an extra axiom: 1 ≠ 0. Any thoughts on this?
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.
That's for a field. There's usually no concept of "1" for a vector space (just for the underlying field).
@@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.
I still don't know how to reach you😂
Idk your e mail so...
Ptabrizi (at)uci.edu
betichod 0 vector=1 how dude?
Thank you!