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Norm and Rows of Complex Matrix

  1. Feb 12, 2013 #1
    1. The problem statement, all variables and given/known data
    I want to show that for an n x n matrix A with complex entries, if [itex]\left\|Ax\right\|=\left\|x\right\|[/itex] for any vector x in C^n, then the rows of A are an orthonormal basis of C^n.


    2. Relevant equations



    3. The attempt at a solution All I've managed to do so far is show that the columns of A all have length 1, which you can get by taking x to be e1. Is there a strategy of showing this for the rows and for showing orthogonality without having to write out the terms of the matrix multiplication Ax (I tried this and quickly became bogged down in notation)?
     
  2. jcsd
  3. Feb 12, 2013 #2

    haruspex

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    Can you think of interesting choices for the vector x based on the elements of A?
     
  4. Feb 13, 2013 #3
    I tried letting x be a column of the adjoint of A. This way when you multiply Ax, if your x is the ith column of the adjoint then the ith entry of the vector Ax is equivalent to the inner product <x,x>. But when you go to take the inner product <Ax,Ax> to find the norm of Ax I end up with a jumble of sums of entries of A being multiplied by each other so I'm not sure where to go from here. Am I on the right track?

    Thanks.
     
  5. Feb 13, 2013 #4
    Okay, suppose I'm setting x to be the first columns of the adjoint of A. Then when I take <Ax,Ax> the first term is <x,x>*<x,x> and the rest of the terms are the inner product of the first row of A with another term of A. Since we know that [itex]\left\|Ax\right\|=\left\|x\right\|[/itex] is it too fast for me to make the jump to saying that <x,x> must equal 1 in order for ||Ax|| to equal ||x|| and subsequently all the other terms must be 0? It seems like there should be more justification in between those steps, but I can't seem to put my finger on it.
     
  6. Feb 13, 2013 #5

    haruspex

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    Yes. If the row vectors of A are a1..an and x = a1T, you have <a1,a1>2+<a1,a2>2+... You know that <a1,a2>2 etc are non-negative, so...?
     
  7. Feb 13, 2013 #6
    Ah, yes. I was getting bogged down in notation and missing the bigger picture. Thanks.
     
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