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Proof that AI_n = A

  1. Oct 3, 2013 #1
    I'm trying to prove AI_n = A for my linear algebra class:

    ## A = (a_{st}) ##
    where ## 1 \leq s \leq m ## and ## 1 \leq t \leq n ##

    then ## (AI_n)_{sk} = \displaystyle\sum_{t=1}^n a_{st}(I_n)_{tk} = \displaystyle\sum_{t=1}^n a_{st}\delta_{tk} ##

    we know that most of ## \delta_{tk} = 0 ## from the definition,

    therefore

    ##\displaystyle\sum_{t=1}^n a_{st}\delta_{tk} = a_{sk}\delta_{kk} + \displaystyle\sum_{t\not=k} a_{st}\delta_{tk} = a_{sk}(1) + 0 = a_{sk} ##
    I don't understand the final bit of the proof, we start with defining A = a_{st}, and we end up with a_{sk}, how is this equivalent?
     
  2. jcsd
  3. Oct 3, 2013 #2

    D H

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    Which part don't you understand?
    1. ##\sum_t a_{st}\delta_{tk} = a_{sk}\delta_{kk} + \sum_{t\ne k} a_{st}\delta_{tk}##
    2. ##a_{sk}\delta_{kk} + \sum_{t\ne k} a_{st}\delta_{tk} = a_{sk}(1) + 0##
    3. ##a_{sk}(1) + 0 = a_{sk}##
     
  4. Oct 3, 2013 #3
    the second one, i.e. how you get from 1. to ##a_{sk}\delta_{kk} ## how does the t change to a k? for a and delta, am I righ tin thinking it's because of the deifnition of matrix multiplication?
     
  5. Oct 3, 2013 #4

    D H

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    The first step is just taking one specific term out of the sum, the term for which t=k. This results in ##\sum_t a_{st}\delta_{tk} = a_{sk}\delta_{kk} + \sum_{t\ne k} a_{st}\delta_{tk}##.

    The second step merely replaces the value of the Kronecker delta with its value. The value of ##\delta_{kk}## in first term, ##a_{sk}\delta_{kk}##, is just one. The second term is the sum ##\sum_{t\ne k} a_{st}\delta_{tk}##, and here ##\delta_{tk}## is identically zero since ##\delta_{tk}=0## for all t ≠ k.
     
  6. Oct 3, 2013 #5
  7. Oct 4, 2013 #6
    anyone?
     
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