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Simple Tensor proof-Need help

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that [tex] A \cdot A^{T} [/tex] is symmetric (A is a 2nd order tensor)


    2. Relevant equations



    3. The attempt at a solution

    So I got down to this and i can see that it will be symmetric however when I try taking the transpose of the solution I cant seem to make it equal the starting equation.

    [tex]
    \sum_{ij}A_{ij}\delta _{i}\delta _{j}\cdot \sum_{kl}A_{lk}\delta_{k} \delta_{l}=\sum_{ijl}A_{ij}A_{lj}\delta _{i}\delta _{l}
    [/tex]

    Is there anyway to manipulate the the transpose to make it equal to the original equation or is what I got as far as I can go?

    Transpose:

    [tex] \sum_{ikl}A_{jl}A_{ji}\delta _{i}\delta _{l} [/tex]


    Thanks
     
  2. jcsd
  3. Sep 30, 2011 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    What is a delta with a single index supposed to mean? I've never seen that before. The matrix product of A and B is (AB)_ik=sum over j A_ij*B_jk. Your notation here is pretty unconventional.
     
  4. Oct 1, 2011 #3
    the deltas represent the unit dydas (according to the book Im using) and I agree with your matrix product however the other A is transpose so I would think it would be the sum over A_ij*A_kj which is essentially what I wrote in my original post.
     
  5. Oct 2, 2011 #4
    Anyone else have an idea?
     
  6. Oct 2, 2011 #5
    Ah. You should be careful using the deltas like that though, since they are usually reserved for a very special kind of tensor. Be sure to explain your notation also. I will also assume that you're not working with covariant/contravariant notation, otherwise your tensors don't make sense.

    That being said, evaluate the product. You'll need to be a bit careful when you do because your notation leaves something to be desired. When you combine the [itex] \delta_i \delta_j^T [/itex] you will be able to say something something about this product. In particular, you will run into a conflict of notation because you'll need to use a Dirac delta here.
     
  7. Oct 2, 2011 #6
    I guess I will clear up some of my notation since I dont think a Dirac delta comes into place here. The e in this case represents the unit dydas, and delta is a standard Kronecker delta.

    So form the start:
    [itex]\bar{}[/itex]
    [itex]\sum_{ij} A_{ij}e_{i}e_{j}\cdot {(\sum_{kl}A_{kl}e_{k}e_{l}})^{T}=\sum_{ijkl}A_{ij}A_{lk}e_{i}e_{l}\delta_{jk}[/itex]

    I reduced this down to:
    [itex] \sum_{ijl}A_{ij}A_{lj}e_{i}e_{l}[/itex]

    Taking the transpose of the above gives:
    [itex]\sum_{ijl}A_{jl}A_{ji}e_{i}e_{l}[/itex]

    from here i need to show that
    [itex] \sum_{ijl}A_{ij}A_{lj}e_{i}e_{l}=\sum_{ijl}A_{jl}A_{ji}e_{i}e_{l}[/itex]

    I know I can do it by writing out all the terms but Im having trouble finding a clever way of proving it without taking the grunt work way
     
  8. Oct 2, 2011 #7
    Yes sorry. I meant Kronecker, not Dirac.

    Now again, the notation you're using is quite inconsistent. Since you're not using co/contra-variant indices, the [itex] A_{ij} [/itex] are just coefficients while the [itex] e_i, e_j [/itex] are tensors. Hence they are not affected by the transposition. Futhermore, [itex] e_i^T \neq e_i [/itex] but you haven't included this either. Have you been given some sort of consistent notation with which to work?
     
  9. Oct 2, 2011 #8
    My understanding for my transport book is that coefficient is effected by transposition, which is what I showed. The notation I am using corresponds pretty similarly to what is presented here: http://www.foamcfd.org/Nabla/guides/ProgrammersGuidese3.html

    With the e's representing the tensor directions. From my understanding in what i learned in class and what my book says up to solving before the transpose everything should be correct.
     
  10. Oct 2, 2011 #9
    Here is an important difference between the link you sent and the notation you are using. In the link you sent, the notation [itex] A_{ij} [/itex] represents the entire tensor. There is no summation, no unit vectors, nothing of the sort. On the other hand, when you write [itex] \sum_{i,j} A_{ij} e_i e_j [/itex] you are saying that [itex] e_i, e_j [/itex] are the tensors and [itex] A_{ij} [/itex] is the coefficient corresponding to the tensors. This is an important difference.

    With proper notation, this question is very simple and results from simply applying definitions. However, I believe you are confusing different notational conventions. However, I may simply have never seen the notation you are using. If that is the case, I apologize for misleading you and would encourage you to continue to try to find the answer.
     
  11. Oct 2, 2011 #10
    correct, based off my notation ei and ej represent the tensors and Aij represents the coeffiecent of the tensor. The problem lies I feel in what the coefficent should be when you take the transpose. I get AljAij as my transpose coeffiecent, but it should be equivalent to AijAlj, and I am not sure how to show that without expanding everything out.
     
  12. Oct 2, 2011 #11
    Here's what I'm saying though. If when you take the transpose, you flip ALL the indices, then you haven't changed anything. That's not A transpose, it's just A where you've changed the dummy variables. Think of it like this: we can think of [itex] A_{ij} e_i e_j [/itex] as the number [itex] A_{ij} [/itex] in the (i,j) spot (indicated by the order of the tensors). Similarly, [itex] A_{ji} e_j e_i [/itex] is still just [itex] A_{ji} [/itex] in the (j,i) spot. You haven't changed anything. Okay, so what is the transpose? Well, we want [itex] A_{ij} [/itex] to be the (j,i) spot, that means
    [tex] (A_{ij} e_i e_j )^T = A_{ij} e_j e_i [/tex]
    so you do not flip the indices on the coefficient.
     
  13. Oct 2, 2011 #12
    While I do not disagree with you, and using your method does seem to give me the correct answer my book explicitly states that given a tensor [itex]\tau[/itex]:

    [itex]\tau=\sum_{ij}\tau_{ij}e_{i}e_{j}[/itex]

    then [itex]\tau^{T}=\sum_{ij}\tau_{ji}e_{i}e_{j}[/itex]

    I am assuming this means that the transpose would be when you only switch one of the indices for one of the A terms, such as Aij*Alj -> Aij*Ajl?
     
    Last edited: Oct 2, 2011
  14. Oct 2, 2011 #13
    Ah, this is the exact same thing! I chose not to flip the coefficients, whereas your book chose to flip the coefficients, but not to flip the order of the tensors. They are equivalent up to renaming the indices.
     
  15. Oct 2, 2011 #14
    So would that mean that the transpose of AijBlj= BijAlj, but since they are both A in this case you get AijAjl?
     
  16. Oct 2, 2011 #15
    I finally see what you're doing, though I have to say this notational weirdness is throwing me off.

    Note that your sum is as follows
    [tex] \sum_{i,\ell} \left[ \sum_j A_{ij} A_{\ell j} \right] [/tex]
    That is, you can define a new tensors [itex] B_{i\ell} = \sum_j A_{ij} A_{\ell j} [/itex] so that you can write this as
    [tex] \sum_{i,\ell} \left[ \sum_j A_{ij} A_{\ell j} \right]=\sum_{i,\ell} B_{i\ell} [/tex]

    Now, are you sure that when you take the transose, that [itex] B_{\ell i} = \sum_j A_{ji} A_{j\ell} [/itex]? I don't think this is correct. It seems to me that if we're just flipping i and l then
    [tex] B_{\ell i} = \sum_j A_{\ell j} A_{i j} [/tex]
    wouldn't you agree?
     
  17. Oct 2, 2011 #16
    I think I agree with your second statement:

    [tex]B_{li}=\sum_{j}A_{lj}A_{ij}[/tex]

    In my books notation that would be:

    [tex]\sum_{ilj}A_{lj}A_{ij}e_{i}e_{l}=\sum_{ilj}A_{ij}A_{lj}e_{i}e_{l}=A \cdot A^{T}[/tex]

    Since [tex]A_{ij}[/tex] is just a constant it can be rearranged.

    correct?
     
  18. Oct 2, 2011 #17
    Correct.
     
  19. Oct 2, 2011 #18
    Cool, thanks so much for your help
     
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