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Math Help - Elementary row operation

  1. #1
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    Elementary row operation

    Let
    e be an elementary row operation,
    A,B,C be a n \times n matrices
    By e(A) we mean the resultant matrix after we perform e on A

    Does the the following hold true -
    e(A)B = Ae(B) ?

    Also, we know the e(A)=e(I_n)A. One way to prove this is for each kind of elementary row operation establish this fact by performing the computation on the matrix elements.

    I was wondering if there is a more conceptual/abstract argument, logic, reason behind this to work.

    Thanks
    Last edited by aman_cc; October 5th 2009 at 12:39 AM. Reason: mis-spelling
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    Let
    e be an elementary row operation,
    A,B,C be a n \times n matrices
    By e(A) we mean the resultant matrix after we perform e on A

    Does the the following hold true -
    e(A)B = Ae(B) ?
    no! for example in M_n(\mathbb{C}), choose e=e_{11} + 2e_{22}, \ A=e_{12}, \ B=e_{21}. then: (eA)B=e_{11} \neq 2e_{11}=A(eB).
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    no! for example in M_n(\mathbb{C}), choose e=e_{11} + 2e_{22}, \ A=e_{12}, \ B=e_{21}. then: (eA)B=e_{11} \neq 2e_{11}=A(eB).
    Thanks NonCommAlg.

    Though I am fairly confident of the following still would check this:
    if C=AB then e(C) = e(A)B

    Also is there another conceptual-argument (based on linear transformation etc) for e(A)=e(I_n)A to be true? The proof I have at my disposal is based on actual computation.
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  4. #4
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    Quote Originally Posted by aman_cc View Post
    Thanks NonCommAlg.

    Though I am fairly confident of the following still would check this:
    if C=AB then e(C) = e(A)B
    this is true because, (at least) over a field, "matrix multiplication" is associative. so e(AB)=(eA)B. you're making your job hard by looking at e as a map rather than just a matrix.


    Also is there another conceptual-argument (based on linear transformation etc) for e(A)=e(I_n)A to be true? The proof I have at my disposal is based on actual computation.
    i don't know what kind of "concept" or even "computation" you're talking about here? I_n is the identity matrix and so eI_n=e, which gives us e(I_n)A=(eI_n)A=eA=e(A).
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    i don't know what kind of "concept" or even "computation" you're talking about here? I_n is the identity matrix and so eI_n=e, which gives us e(I_n)A=(eI_n)A=eA=e(A).
    Thanks for confirming the part (1). After you said it - I realized I was asking a very trivial question.

    For part(2) - Let me try to explain.
    When I say 'e' is an elementary row operation. I look at it as some kind of a function on A. e(A) in my way of thinking is image of A under e.

    Now it so happens that e(A) = e.A
    In R.H.S. I guess e is treated like a matrix ( e=e(I_n)). And, e.A is a matrix multiplication.

    I guess it is this multiplication that you refer when you write eI_n=e

    My question is why should e(A) = e.A? Is there a logic to it - or this is a fact which just happens to be true based on how we define
    1. Elementary Row Operation
    2. Matrix Multiplication

    Please ignore this question if it is really not relevant. Maybe I need to get more clear in my head.
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  6. #6
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    Quote Originally Posted by aman_cc View Post

    My question is why should e(A) = e.A? Is there a logic to it - or this is a fact which just happens to be true based on how we define
    1. Elementary Row Operation
    2. Matrix Multiplication
    well, the only way to prove that is direct calculation, i.e. to check that eA really does the required row operation.

    here's a question for you to think about: can you write an elementary matrix in terms of e_{ij}?

    ( e_{ij}, as usual, is the square matrix with 1 in the i-th row and j-th column and 0 everywhere else)
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  7. #7
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    Quote Originally Posted by NonCommAlg View Post
    well, the only way to prove that is direct calculation, i.e. to check that eA really does the required row operation.

    here's a question for you to think about: can you write an elementary matrix in terms of e_{ij}?

    ( e_{ij}, as usual, is the square matrix with 1 in the i-th row and j-th column and 0 everywhere else)

    Let me try

    I_n = \sum e_{ii} , where  i \in [1,n]
    In the equations below I have used the above mentioned expansion.

    Interchange of row - Row_a \leftrightharpoons Row_b
    e = I_n - e_{aa} - e_{bb} + e_{ab} + e_{ba}

    Multiplication by a scalar Row_a = x \times Row_a
    e = I_n - e_{aa} + x \times e_{aa}

    Adding a scalar multiple of a row to another row Row_a = Row_a + x \times Row_b
    e = I_n  + x \times e_{ab}

    Am I anywhere close to what you wanted me to try? If yes, any specific importance of this method?

    Thanks very much !!
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  8. #8
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    Quote Originally Posted by aman_cc View Post
    Let me try

    I_n = \sum e_{ii} , where  i \in [1,n]
    In the equations below I have used the above mentioned expansion.

    Interchange of row - Row_a \leftrightharpoons Row_b
    e = I_n - e_{aa} - e_{bb} + e_{ab} + e_{ba}

    Multiplication by a scalar Row_a = x \times Row_a
    e = I_n - e_{aa} + x \times e_{aa}

    Adding a scalar multiple of a row to another row Row_a = Row_a + x \times Row_b
    e = I_n + x \times e_{ab}

    Am I anywhere close to what you wanted me to try? If yes, any specific importance of this method?

    Thanks very much !!
    right! of course, "multiplying by a scalar" and "adding a scalar multiple of a row to another row" have the same form, which is I + ce_{ij}. so we basically have two distinct forms of matrices.

    the reason that i gave you this question is that using these forms makes the proof e(A)=eA easier. that's because we can write the matrix A=[a_{ij}] in the form A=\sum_{i,j}a_{ij}e_{ij} and then

    we can use this fact that e_{ij}e_{rs}=\delta_{jr}e_{is}, where \delta_{jr} is, as usual, the Kronecker's delta.
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  9. #9
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    I get the idea. Thanks very much !!
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