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Tensor multiplication

  1. Nov 20, 2011 #1
    This book says that if [itex]W^aX_a=0[/itex] and [itex]X_a[/itex] is arbitrary, then I should be able to prove that [itex]W^a=0[/itex]. I don't see how this is possible. This is the equivalent of the vector dot product, so if, say, [itex]X_a=(1,0,0,0)[/itex], then [itex]W^a[/itex] could be (0,1,1,1), and the dot product would be [itex]1*0+0*1+0*1+0*1=0[/itex]. Why would [itex]W^a[/itex] have to be 0?
     
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  3. Nov 20, 2011 #2

    lurflurf

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    It means for any Xa not for some Xa. So in your vector example the only vector orthogonal to all vectors is the zero vector.
     
  4. Nov 21, 2011 #3

    HallsofIvy

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    If [itex]X_\alpha[/itex] can be any vector, it can be [itex]W_\alpha[/itex]. If [itex]W^\alpha X_\alpha= 0[/itex] for [itex]X+_\alpha[/itex] any vector then [itex]W^\alpha W_\alpha= 0[/itex] which immediately gives [itex]W^\alpha= 0[/itex]
     
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