I've always said that things thought to be difficult are very often simply poorly explained. This video is a confirmation: it makes things considered difficult easy. Thank you.
Consider the vector space V=RxR and a bilinear map f from V to R. Fix the standard basis e_i so that f is represented by the matrix A with element a_ij =f(e_i,e_j). This way f(x,y)= in the standard way being the standard scalar product. So basically your correspondence takes A to f and is a bijection. The tensor product is then the space the linear map represented by A acts uniquely. This work for finite dimensional vector spaces or even finitely generated R modules over a commutative ring R. Do you agree?
Yes, the converse holds. In other words, for every linear map g on V⊗W there exists a unique bilinear map f on V×W such that g ∘ τ = f. Uniqueness is obvious from the equation g ∘ τ = f because this specifies that, as functions, f is equal to g ∘ τ. We can write the map as f(v,w) = g(τ(v,w)) = g(v⊗w). Checking that this map is bilinear follows straightforwardly from the construction of the tensor product as a quotient space shown in this video and the fact that g is assumed to be linear.
Remember that * is not denoting ordinary multiplication in this case. v*w just means "the basis vector associated with the pair (v,w)". Since each distinct pair has its own basis vector, v*w and 2v*2w are distinct basis vectors, hence not multiples of each other.
I've always said that things thought to be difficult are very often simply poorly explained. This video is a confirmation: it makes things considered difficult easy. Thank you.
This has got to be the best video on tensor products on UA-cam. I hope you make more videos on this stuff!
This is fire thank you. Perfectly rigorous with great examples.
Very well done! I really liked how you discussed all details 100% concisely.
Excellent! Thanks a lot!
Fantastic explanation ❤️
Always as clear as crystal🎉🎉🎉
Great skills of explaining!
Thank you for this enlightening lecture
amazing video.
Great video! ❤
Consider the vector space V=RxR and a bilinear map f from V to R. Fix the standard basis e_i so that f is represented by the matrix A with element a_ij =f(e_i,e_j). This way f(x,y)= in the standard way being the standard scalar product. So basically your correspondence takes A to f and is a bijection. The tensor product is then the space the linear map represented by A acts uniquely. This work for finite dimensional vector spaces or even finitely generated R modules over a commutative ring R. Do you agree?
I thank you so much for this video, helped me a lot!
But does the converse hold? This is, every linear well-defined map on the tensor product comes from a bilinear map on the cartesian product?
Yes, the converse holds. In other words, for every linear map g on V⊗W there exists a unique bilinear map f on V×W such that g ∘ τ = f.
Uniqueness is obvious from the equation g ∘ τ = f because this specifies that, as functions, f is equal to g ∘ τ. We can write the map as
f(v,w) = g(τ(v,w)) = g(v⊗w).
Checking that this map is bilinear follows straightforwardly from the construction of the tensor product as a quotient space shown in this video and the fact that g is assumed to be linear.
@@MuPrimeMath I see. Thanks a lot. Keep up the good work!
4:40 Is v*w not a multiple of 2v*2w?
Remember that * is not denoting ordinary multiplication in this case. v*w just means "the basis vector associated with the pair (v,w)". Since each distinct pair has its own basis vector, v*w and 2v*2w are distinct basis vectors, hence not multiples of each other.
Hi sorry what text are you working off of?
This video wasn't working off of any particular text.
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Awesome video!!! Thank you so much!!!