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Kronecker Delta use

  1. Feb 12, 2009 #1
    Kronecker Delta expression

    Please, give me an example of this identity using a 3 dimensional matrix R (maybe representing a rotation). My difficulty lies in the indices manipulation.

    R_{ii'}R{jj'}\delta_{i'j'} = \delta_{ij}

    I know it is obvious, but I'm really stuck in my self-teaching. Thank you.
    Last edited: Feb 12, 2009
  2. jcsd
  3. Feb 12, 2009 #2


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    Are you summing over repeated indices, e.g.
    [tex]R_{ii'} R_{jj'} \delta_{i'j'} \text{ means } \sum_{i', j'} R_{ii'} R_{jj'} \delta_{i'j'}?[/tex]
    By the definition of matrix multiplication:
    [tex]R_{ij} R_{kl} \delta_{jl}= R_{ij} R_{kj} = R_{ij} R_{jk}^T = R_{ij} R_{jk}^T = (R R^T)_{ik}[/tex]
    This is important, you should convince yourself of each step. If you must, write it out in matrices and compare each step to the index notation, seeing how operations like multiplying two matrices or taking a transpose translate into operations on the indices.

    If this is equal to [itex]\delta_{ik}[/itex] that means that [itex]R R^T[/itex] is the identity matrix. An example of such a matrix is, such as
    [tex]R = \begin{pmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}[/tex]

    If you want you can explicitly write out all the 9 components of the left hand side (let i and j run and sum over k), and show that you are really just multiplying R by R^T. Or if you are not sadistic, take a 2x2 matrix :)

    Sorry if I am too general, but I think it is best if you work through the manipulations yourself.
  4. Feb 13, 2009 #3
    Thank you for the answer. That's exactly what I expected. :)
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