its a well made video but unfortunately there is no substitute for spending hours struggling with a concept with a pen and a book. these videos should be used as revisions or supplements to your understanding only. having a good supply of analoogies for understanding the same thing is always helpful when studying mathematics.

But you don’t learn anything because what we explained in the video is too simple for you or because you think you need to spend hours with pen and a book in order to learn?

@@dibeos what I meant to say was the video can't possibly serve as an introduction for someone who's never studied any linear algebra, but its a good video for reinforcing concepts and providing visual analogies. Normally it would take a person 15 minutes to fully internalize the concept of something like 'linear span' on its own for example. Hence, sitting with a book and working through it yourself is important. I loved the video but I can also empathize with people who feel like they come away with no greater understanding, since they don't have the background necessary to digest the information at the pace its provided. Also, could you explain how distributions form a vector space?

@ Thanks for letting us know. Now I understand your point better. We are improving our videos more and more every week. If you compare our videos now with those 6 months or a year ago, you will see a huge difference. What I mean is that the videos are far from being perfect, but we guarantee that they will become even better over time and will be more useful for people who want to learn math or solidify concepts. But we still really appreciate constructive feedback. About distributions that form vector spaces, I think we were not clear enough in the video and that's why some people were confused. What we meant is that in certain conditions probability distributions can be treated as vectors in a vector space, for example in the context of functional spaces like L^1 and L^2. But again, just in specific cases where linear combinations of densities preserve normalization and non-negativity

Yes, you are right. We just did not want to make the video tooooo long, but I agree that this would have been a nice to mention 😎

Glaze

@joeybasile545 You try it

No

Huh??? Hahahaha. Your comment makes no sense.

@@samueldeandrade8535 I agree 😂😂🤝

I believe he was talking about linear independence, in which case all coefficients must be zero, for the liniar combination to yield the zero vector. in 2D case, say we got c1*v + c2*w = 0 and if c1 is 0 then c2*w is the zero vector, and since w is not the zero vector, this implies that c2 must be zero as well. Unless i'm mistaken and you meant something else.

@@stefan-danielwagner6597listen to the part in referring to, he’s talking about linear dependence there. In that case any nonzero coefficient leads your vector to not be linearly independent. What you’re saying is correct, it’s just not what I’m referring to

@@Happy_Abe I get what you mean now. if any subspace is linearly dependent, then the entire vector space is linearly dependent. c1*u+ c2*w + c3*w = 0; you say for example c2 can be zero and c1 and c3 nonzero means (u,v,w) are linearly dependent. But in this 2D case, if c1*v + c2*w = 0 and c1 != 0 then c1*v = -c2*w in which case if c2 is 0 then c1*v = 0 with c1 != 0 and v != 0 which means that c2 has to be non-zero as well. For R^2 it holds, but for R^n if any subspace is linearly dependent, the rank will be < n and the image will not be R^n.

@@stefan-danielwagner6597 yeah but in theory v or w can be 0 vectors in which case, say v is 0, that you trivially can have a nonzero c1 and a zero c2 and still have c1v+c2w=0 and {v,w} are linearly dependent even though not both of the coefficients c1 and c2 were 0. My point being any nonzero coefficient will make the set of vectors dependent.

@@Happy_Abe Agree. I enjoyed the comment exchange. Have a lovely day

Yeah I’m confused by the density statement as well

I think they meant Z/5Z is not a vector space over R. Yes it’s a vector space over itself, it seems they were just defining vector spaces here as real vector spaces not vector spaces over a general field

Hehehe. Yep, that doesn't make sense at all, does it? About the Z modulo 5 (considerer over R), it was a counterexample, so it is not a R-vector space. This video is not that good. But this is a nice channel.

@@samueldeandrade8535 it’s just not the best counterexample since the scalar multiplication they were doing wouldn’t work in a vector space anyway because the distributive property fails. A more basic example would be the integers or even the rational numbers which fail to be a vector space over R.

What I meant is that Z/5Z is defined by a quotient of Z. The operation 0.5x3 with 3 in Z/5Z and 0.5 in R does not even exist. But you are right, maybe they meant {0,1,2,3,4} in the video. This would have been a correct and less complicated choice to name this counter-example.

Yes you do get a vector space structure on Z/5Z by what you’re saying but that vector space is over Z/5Z not R(real numbers). In general, a vector space is defined over a field. In this video it’s just treating that field to be R in which case Z/5Z is not a vector space. But using the field Z/5Z, Z/5Z is a vector space. Every field is actually a vector space over itself. But what you said about a vector space needing both the vector elements and what you’re scaling by to be from the same set to scale them is not correct. As long as you properly define a multiplication operation on a set by field elements that respect the vector space definition it will be fine. More generally, if the “scaling” set is not a field but is instead a ring(more general algebraic structure than a field) we call this a module over that ring. Scaling in this sense means how the ring/field “acts” on the underlying group that is then made into a module/vector space.

He didn’t multiply two vectors in Z. A vector space is defined over a field. What he did was scalar multiplication using a member of the field R ( 0.5) and a vector from Z. But the result was no longer in the presumed vector space Z, hence Z is not a vector space over R (the real numbers). To make Z a vector space would require choosing a different field, or having Z become a “module” by choosing a “commutative ring “ instead of a field. These terms are, loosely speaking , “an easy going vector space over a laid back field” where certain restrictions are relaxed. Bottom line: Results of scalar multiplication cannot leave the vector space!

Indeed, you didn't get it. 0.5 is not, as a real number, in Z modulo 5. But they didn't assume that. A vector space V is considered over a field of scalars K. In their counterexample, Z modulo 5 is the candidate for vector space R is the candidate for the field of scalars The expression "0.5*3" in the video should be interpreted as (scalar in the candidate for field)*(vector in the candidate for vector space) You are confusing the sets, the candidate for vector space with the candidate for field of scalar.

"Practical reasons"? In Math or in the video?

Yes, eigenvectors are practical.

Eigenvectors and their corresponding Eigenvalues characterize the operator (System).

@@geraltofrivia9424 thanks for the nice words!!! We will publish more videos about Differential Geometry and Tangent Spaces just as you asked. 😎

@@rahulmandal1920 yessss 😎 love from Italy ♥️

@BCarli1395 thanks for the nice words! We really appreciate it. We are working on a longer video about the core of Linear Algebra. Hopefully we will post it this weekend - stay tuned 😎

@@hamzamohamed7935 awesome! 😎

yeah, many people asked us for Clifford Algebra. we will make a video about it. Thanks for letting us know!

@willyh.r.1216 we really appreciate your support, Willy! Just because you asked us we will keep going… 😉 The next video will be about the Core of Linear Algebra. But we added your suggestions to our list of ideas 😎