Tensors for Beginners 13: Tensor Product vs Kronecker Product

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  • Опубліковано 20 лют 2018

КОМЕНТАРІ • 70

  • @danieldanieldadada
    @danieldanieldadada 5 років тому +114

    i need a joint

  • @zhiiology
    @zhiiology 5 років тому +28

    Hi eigenchris, just wanted to say that while the video is clear, the kronecker product is usually computed by shoving the tensor on the right into the tensor on the left, which is usually how it's done at least in quantum computing

    • @KleineInii
      @KleineInii 4 роки тому +3

      Yeah, I also found that confusing when I read a paper from Tamara Kolda (epubs.siam.org/doi/pdf/10.1137/07070111X, cited 5000 times). I think right to left is more common, but still it's explained brilliant in this video :)

    • @sathasivamk1708
      @sathasivamk1708 3 роки тому +2

      @@KleineInii There is math book called ABSTRACT ALGEBRA by DUMMIT AND FOOTE. Take a look, Kronecker product is just tensor product with respect to different ordering of basis

  • @anthonym316
    @anthonym316 3 роки тому +5

    My man this is exactly what I needed thank you

  • @dennisbrown5313
    @dennisbrown5313 5 років тому +3

    Just amazingly clear and good

  • @connorbeaton8375
    @connorbeaton8375 3 роки тому +1

    This channel is a goldmine

  • @IntegralMoon
    @IntegralMoon 6 років тому +7

    Awesome! Thanks for this :)

  • @zzzoldik8749
    @zzzoldik8749 4 роки тому +4

    tankyou, eigenchris, i never understand if i learn tensor before, you explain it from the star / beginner. these really helpful

  • @underratedPie
    @underratedPie Рік тому +1

    I told my colleague to study the Tensor calculus on this channel only

  • @RaoBlackWellizedArman
    @RaoBlackWellizedArman 4 роки тому +1

    Thanks a million!

  • @Timelaser001
    @Timelaser001 6 років тому +7

    Hi, I'd just like to clarify something. On 2.21, is the summation sign implied for v(j)e(i)? I understand Einstein's notation is used for letters which are the same on top and the bottom, but in this case the letters at the top and bottom are different.

    • @A.Arbabinezhad
      @A.Arbabinezhad 4 роки тому +1

      Tensor product(e_i, epsilon^j) is a stack of vectors all parallel to e_i and the result of acting of this linear map on a vector is linear combination of these parallel vectors. So the result is a vector prallel to e_i (wich is v^j e_i)

    • @abstractnonsense3253
      @abstractnonsense3253 2 роки тому

      It's not a sum. v^j e_i is the *number* v^j multiplying the *vector* e_i. The *number* v^j is the jth component of the *vector* v.

  • @trendytrenessh462
    @trendytrenessh462 2 роки тому

    Same thing, different context
    Got it, cheers

  • @alessandrogardini5469
    @alessandrogardini5469 3 роки тому +6

    Hi! It seems to me that at 2:49 the new vector components should be represented by a latin letter and upper indices (like [w^1 over w^2] rather than [omega_1 over omega_2]). Thank you again for the videos.

  • @AJ-et3vf
    @AJ-et3vf Рік тому

    great video. thank you

  • @jacobshin4279
    @jacobshin4279 Рік тому +1

    Would the array at 3:02 be a rank 3 tensor?
    Am I correct in guessing that the rank = m+n, where m is the # of covariants and n is the number of contravariants?

  • @jamescook5617
    @jamescook5617 4 роки тому +1

    Hold a second... now a row vector is an array. Good.

  • @geneyi8703
    @geneyi8703 4 роки тому +1

    I didn't get it that what is the definition of Kronecker product? "We just distribute the array on the left into the array on the right."

  • @mathgeek43
    @mathgeek43 2 роки тому

    Something seems off at 1:36. It is my understanding that the tensor product takes a (p,q)-tensor and (a,b)-tensor and produces a (p+a,q+b) tensor. In the example shown, we have the vector e_i which is a (1,0) tensor and eps^j which is a (0,1) tensor. This means that the tensor product between e_i and eps^j is a (1,1) tensor. What's confusing is about 1:36 is that the tensor resulting from the tensor product of e_i and eps^j should have a covector and a vector as an input. This will make sure that the output of the tensor is a scalar and not a vector.

  • @user-je4xw6tx3k
    @user-je4xw6tx3k 4 роки тому +2

    If we can download all the slides, it would be perfect, since we can print it out and study it without electronic devices, also we can drop down notes for deeper understandings.

    • @eigenchris
      @eigenchris  4 роки тому +8

      I have the slides uploaded here: github.com/eigenchris/MathNotes/tree/master/TensorsForBeginners

  • @vassillenchizhov290
    @vassillenchizhov290 8 місяців тому

    Your notation with the array of arrays suggests that the Kronecker product produces arrays of higher dimension that two. However the Kronecker product of a matrix is always a matrix, for instance if I take $u\oplus v \oplus w$ this will not produce an object having three indices, it will still be a matrix. Of course you can relate this to the object made from arrays of arrays, but technically they are not the same thing.

  • @PaulWintz
    @PaulWintz 4 роки тому +1

    At 1:56, when the j-th basis covector of V* acts on the k-th basis vector of V to become the Kronecker delta, why does the tensor product operator disappear? Is it a valid operation to take the tensor product between a vector and a scaler? And does it merely equal the scalar times the vector?

    • @eigenchris
      @eigenchris  4 роки тому +1

      It probably should have disappeared in the 2nd line, since epsilon(v) is a scalar, not a vector.

  • @auvski5903
    @auvski5903 3 роки тому +1

    I'm actually pretty sure the Kronecker product as used in this video (and the previous one) is backwards. Both in my university's coursework and on the Wikipedia page, it seems like you're actually supposed to distribute the array on the right over the one on the left, that is, the left array is meant to be used as a "template" and the right array is copied for each block. You can take a look here:
    en.wikipedia.org/wiki/Kronecker_product

    • @eigenchris
      @eigenchris  3 роки тому +1

      Yeah, I noticed that. I'm not sure if it's just two different conventions or if the wikipedia one is fundamentally correct.

  • @jasonbroadway8027
    @jasonbroadway8027 4 роки тому +2

    I am hazy now. I followed you until this video and the previous video.

    • @eigenchris
      @eigenchris  4 роки тому +1

      Which part is confusing you?

    • @jasonbroadway8027
      @jasonbroadway8027 3 роки тому +2

      @@eigenchris I get it. Thank you, Eigenchris.

  • @ahmadfaraz9279
    @ahmadfaraz9279 3 роки тому

    Is dot product is one of the tensor product?

  • @mithsaradasanayake3211
    @mithsaradasanayake3211 2 роки тому +1

    Thanks eigen chris

  • @paradigmshift03
    @paradigmshift03 3 роки тому

    Hey eigenchris, at 2:04, I'm a bit confused about what v^j e_i actually is. Should this not result in a vector? Therefore, shouldn't the top and bottom indices end up being the same?

    • @jonathanchippett4036
      @jonathanchippett4036 3 роки тому

      v^j e_i is a vector. It is the ith basis vector of V scaled by the jth component of v.
      In regards to your last question, the top and bottom indices should not be the same because we are not summing over anything. We do not need to do an implicit sum to get a vector.

    • @paradigmshift03
      @paradigmshift03 3 роки тому

      @@jonathanchippett4036 Ahh ok, I now understand :) Thank you!

  • @thedorantor
    @thedorantor 2 роки тому

    1:57 so I can just as well put a co-vector in as argument, the epsilon of the co-vector "eats" the e_i of the (1,1) tensor, and the result would be a co-vector?

  • @jasonbroadway8027
    @jasonbroadway8027 4 роки тому +1

    Linear map...how?

  • @chenlecong9938
    @chenlecong9938 Рік тому

    Hey as far as i know kronecker product is a branch of tensor product?

  • @minecrafthowtodude
    @minecrafthowtodude 2 роки тому

    so.... I understood a little bit of that :)

  • @dirrelito
    @dirrelito 4 роки тому +3

    Typically, you also flatten when using the Kronecker product. Tensor product increases tensor order, but kronecker product does not. This is a quite imporant distinction both in theory and practise. The Kronecker product as you explain it here is not how it works in e.g. Numpy or PyTorch.

    • @sathasivamk1708
      @sathasivamk1708 3 роки тому

      Kronecker product is just tensor product with respect to different ordering of basis

  • @srtghfnbfg
    @srtghfnbfg 3 роки тому

    Since the kronecker product can act on nested arrays, is there an index notation formula that i can find anywhere ? I've scoured google and can't find anything related to this specifically

    • @eigenchris
      @eigenchris  3 роки тому

      I think the Kronecker product, as seen in most math classes, doesn't use "nested arrays". If you have a 2x2 matrix krocker-product'ed with a 3x1 matrix, you will just get a 6x2 matrix, without any "nesting". The "nested arrays" thing is something that I came up with myself. I was desperately trying to make "array multiplication" work for arrays beyond 2D matrices. But my view now is to "give up" trying to multiply arrays bigger than 2D and just use the tensor index notation instead.

  • @no-one-in-particular
    @no-one-in-particular 7 місяців тому

    2:07 The "circle times" shouldn't be there from the second row onwards as epsilon acting on v is a number

  • @quantabot1165
    @quantabot1165 3 роки тому

    If i want to do a kronecker product between a 1-column array and a nxn matrix. Do I distribute the column to each row?

    • @eigenchris
      @eigenchris  3 роки тому +1

      You distribute the column to each element of the matrix. If the column is m elements tall, you'll end up with an nxnxm array.

    • @quantabot1165
      @quantabot1165 3 роки тому

      @@eigenchris I see, Thank You eigenchris!

    • @sathasivamk1708
      @sathasivamk1708 3 роки тому

      Nothing fancy here.
      Kronecker product is just tensor product with respect to different ordering of basis

  • @shenyuan9810
    @shenyuan9810 4 роки тому

    Aren’t they just the same thing but in different forms?

    • @sathasivamk1708
      @sathasivamk1708 3 роки тому

      Kronecker product is just tensor product with respect to different ordering of basis, You are right

  • @mythic3187
    @mythic3187 6 років тому

    Hey Eigenchris, can we do some problems of some of these.. I'm kinda getting a grasp of what tensors are..what are these good for besides computer programs and/or GR

    • @eigenchris
      @eigenchris  6 років тому +2

      I'd like to get another 2-3 videos out of the way just to finish the series, but after that I can make a video or two on applications. In addition to GR, there are applications in quantum mechanics, electricity & magnetism, and continuum mechanics. Unfortunately I am less experienced with the physics compared to the pure math, so I don't have a ton to say on those topics right now.

    • @mythic3187
      @mythic3187 6 років тому

      Thank you very much

  • @sdu28
    @sdu28 3 роки тому

    Is a kronecker product between 2 same-size square matrices valid?

    • @eigenchris
      @eigenchris  3 роки тому +1

      Yes. You can do the Kronecker product of any two matrices.

    • @sdu28
      @sdu28 3 роки тому

      @@eigenchris thanks

  • @kimchi_taco
    @kimchi_taco 9 місяців тому

    It looks like
    Tensot product is operation for vector and covector
    Kroneker product is operation for vector and covector components

  • @rs-tarxvfz
    @rs-tarxvfz Рік тому

    This is so confusing. When you did first Kronecker you distributed the RHS into the LHS but when you did second Kronecker you distributed LHS into RHS. This is like WTF !!

  • @debendragurung3033
    @debendragurung3033 2 роки тому +2

    Kinda lost here.

  • @sathasivamk1708
    @sathasivamk1708 3 роки тому

    I guess it's probably intended for physicist, or engineer . As a mathematician, I think it's not good way to learn it.
    Kronecker product is just tensor product with respect to different ordering of basis

  • @farriskft4847
    @farriskft4847 3 роки тому +2

    Same as prevoius: when why how can this information be used...? Start with some practical approach pls.

  • @MrSidTaylor
    @MrSidTaylor 4 роки тому +1

    This is hopeless - it's supposed to be for beginners!!

    • @eigenchris
      @eigenchris  4 роки тому

      Which parts are you struggling with? Have you watched the rest of the series up until this point?

  • @shone7064
    @shone7064 Рік тому

    I think your definition of Kronecker product is reversed.
    The definition on Wiki says that A⊗B distributes B into A, but you just distribute A into B