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Kronecker Delta

  1. Sep 1, 2007 #1
    1. The problem statement, all variables and given/known data
    Show by matrix multiplication, [tex]\delta_{ij}v_j = v_i[/tex]

    3. The attempt at a solution
    I'm having trouble understanding how to do this, because I'm under the impression that [tex]v_j[/tex] is a row vector, which can't be multiplied by a 3x3 matrix which [tex]\delta_{ij}[/tex] is; or am I horribly wrong here?
     
  2. jcsd
  3. Sep 1, 2007 #2

    learningphysics

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    I believe the Kronecker delta is just the identity matrix... if it's a 3x3 matrix, then [tex]v_j[/tex] is 3x1 (3 rows, 1 column)
     
  4. Sep 1, 2007 #3

    dextercioby

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    Row vectors (n,1) can be multiplied by matrices (n,n) only if they stay at the right of the matrix, which is the case here.
     
  5. Sep 1, 2007 #4
    Then shouldn't the unity matrix give another row vector as an answer? I'm trying to understand how [tex]v_i = v_j[/tex]
     
  6. Sep 1, 2007 #5

    dextercioby

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    But the unit matrix 0 nondiagonal elements, so that v_{i}=v_{j} only for i=j.
     
  7. Sep 1, 2007 #6

    learningphysics

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    Aren't we talking about column vectors here... 3x1 is a column vector... and the result of the multiplication gives the same column vector back...

    [tex]\delta_{ij}v_j[/tex] denotes the sum over all j... for a particular i... ie: it is analogous to the multiplying the ith row of the matrix by the column vector [tex]v[/tex]... and the result is [tex]v_i[/tex]
     
    Last edited: Sep 1, 2007
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