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Math Help - Simple linear algebra justification - clarification

  1. #1
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    Simple linear algebra justification - clarification

    Good day to all.

    I have been reading on transposes and came across the following question:

    If A is a n x n matrix where A is non zero can AA^T = 0

    I thought about it and came to the conclusion that the answer is no with the following justification:

    Since A is a non zero matrix then at the very least some (A)i,j element is <> 0.

    By definition of the transpose this implies that the (A)j,i element is <> 0 and (A)j,i = (A)i,j

    Therefore the cross product of the ith row of A with the ith column of A^T will have a minimum value of ((A)i,j)^2

    Does this conclusion make sense or have I missed something along the way.

    Thanks for your help
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  2. #2
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    Quote Originally Posted by gate13 View Post
    Good day to all.

    I have been reading on transposes and came across the following question:

    If A is a n x n matrix where A is non zero can AA^T = 0

    I thought about it and came to the conclusion that the answer is no with the following justification:

    Since A is a non zero matrix then at the very least some (A)i,j element is <> 0.

    By definition of the transpose this implies that the (A)j,i element is <> 0 and (A)j,i = (A)i,j

    Therefore the cross product of the ith row of A with the ith column of A^T will have a minimum value of ((A)i,j)^2

    Does this conclusion make sense or have I missed something along the way.

    Thanks for your help

    Again, looks fine to me.

    Tonio
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  3. #3
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    Thanks tonio for your input. I have become somewhat paranoid with my linear algebra TA as he tends to not be very accepting of answers that deviate slightly from his. At least this way, I know that I have not misunderstood the material and I can attempt to vigorously defend my point of view. Again many thanks for your input.
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  4. #4
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    Attached Thumbnails Attached Thumbnails Simple linear algebra justification - clarification-app57.jpg  
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    Last edited by math2009; February 4th 2010 at 05:20 AM.
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  5. #5
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    Thanks math2009for the attached document. It will be very useful, although right now we have not covered images yet. That portion of the proof I will look into on my own so that I may understand it. Finally I apologize for my tardy reply. I have been wrapped up in mid-term studies. Again many thanks
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