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Math Help - Identity Matrix - Show That AI=IA=A For All Square Matrices

  1. #1
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    Identity Matrix - Show That AI=IA=A For All Square Matrices

    Hey all,

    I've got a question that I'm having a little trouble with. I understand what the identity matrix is and how it works, but am unsure how to clearly show this. The question is as follows.

    --
    Let I be a n x n matrix whose entry in row i and column j is

    { 1 if i = j
    { 0 if i != j
    *This is an identity matrix

    Show that AI = IA = A for every n x n matrix A.
    --

    Once again I know that this statement is true as I know what the identity matrix is and when I use the identity matrix I understand why this works, but I am unsure how I would clearly show this. Any help is greatly appreciated. Thank you very much!
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by BraydenW View Post
    Hey all,

    I've got a question that I'm having a little trouble with. I understand what the identity matrix is and how it works, but am unsure how to clearly show this. The question is as follows.

    --
    Let I be a n x n matrix whose entry in row i and column j is

    { 1 if i = j
    { 0 if i != j
    *This is an identity matrix
    We call this \delta_{ij} for short, the so called "Kronecker delta".

    Show that AI = IA = A for every n x n matrix A.

    Once again I know that this statement is true as I know what the identity matrix is and when I use the identity matrix I understand why this works, but I am unsure how I would clearly show this. Any help is greatly appreciated. Thank you very much!
    Let's just calculate what the element in the i-th row and k-th column will be for the two cases:

    (AI)_{ik} = \sum_j a_{ij}\delta_{jk}= a_{ik} by definition of \delta_{jk}, since for all j\neq k we have \delta_{jk}=0.

    and, similarly,

    (IA)_{ik} = \sum_j \delta_{ij} a_{jk}= a_{ik}, by definition of \delta_{ij}

    So what we have shown that AI=A and IA=A, and thus AI=IA.
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