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Math Help - Prove A . In = A = In . A

  1. #1
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    Prove A . In = A = In . A

    How do you prove this:

    The matrix multipication definition I have is:

    Given A = (aij) : m x n matrix
    B = (bij) : n x p matrix

    We define the product C = AB where C has general element cij = n sigma k=1 aikbkj
    =ai1bij + ai2b2j + ai3b3 + .. + ainbnj
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  2. #2
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    Is that Prove A times the identity is the identity times A?
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    Is that Prove A times the identity is the identity times A?
    No. Mainly prove that A times the Identity is equal to the identity...

    Can you help?
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  4. #4
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    When you multiple anything times the identity what happens? You will have a sum of a_{ij}s but all but 1 will be zero per location (unless that a_{ij} in A is zero then the sum will add up to zero).
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  5. #5
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    That is only true if A is the identity.

    The identity matrix is a square matrix that acts like the number 1 from the reals.
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    When you multiple anything times the identity what happens? You will have a sum of a_{ij}s but all but 1 will be zero per location.
    I understand that AI = I through many examples, I am just struggling to prove it using the matrix multiplication definitions! I don't quite understand your comment..
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  7. #7
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    This is a sheer exercise in notation.
    Do you know about the Kronecker’s delta notation:  \[<br />
\delta _{jk}  = \left\{ {\begin{array}{*{20}c}   {1,} & {j = k}  \\   {0,} & {j \ne k}  \\ \end{array} } \right~?
    It is used to define I_n=(\delta_{jk}):~n\times n.

    We must also assume that A=(a_{jk}):~n\times n.

    So A \cdot I_n  = \sum\limits_{j = 1}^n {\left( {\sum\limits_{k = 1}^n{a_{jk}  \cdot \delta _{kj} } } \right)} .

    Now ask yourself, ‘If 1\le J\le n~\&~1\le K\le n is it true that
    a_{JK}  = \sum\limits_{j = 1}^n {\left( {\sum\limits_{k = 1}^n {a_{Jk}  \cdot \delta _{kK} } } \right)}~?

    If so then you are done.
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  8. #8
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    Quote Originally Posted by Plato View Post
    This is a sheer exercise in notation.
    Do you know about the Kronecker’s delta notation:  \[<br />
\delta _{jk}  = \left\{ {\begin{array}{*{20}c}   {1,} & {j = k}  \\   {0,} & {j \ne k}  \\ \end{array} } \right~?
    It is used to define I_n=(\delta_{jk}):~n\times n.

    We must also assume that A=(a_{jk}):~n\times n.

    So A \cdot I_n  = \sum\limits_{j = 1}^n {\left( {\sum\limits_{k = 1}^n{a_{jk}  \cdot \delta _{kj} } } \right)} .

    Now ask yourself, ‘If 1\le J\le n~\&~1\le K\le n is it true that
    a_{JK}  = \sum\limits_{j = 1}^n {\left( {\sum\limits_{k = 1}^n {a_{Jk}  \cdot \delta _{kK} } } \right)}~?

    If so then you are done.
    Thanks for that well written explanation. I have heard of the kronecker delta notation but not actually come across it yet in my course.

    I don't quite understand the final line though. I understand everything up to that, but not quite how you get to the final part of the proof ..
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  9. #9
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    Quote Originally Posted by chr91 View Post
    I don't quite understand the final line though. I understand everything up to that, but not quite how you get to the final part of the proof ..
    I must tell you that I find that statement hard to believe.
    Either the notation is clear or it is not clear.
    Note that \delta_{KK}=1 and it is zero else where.
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