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False proof

  1. May 27, 2009 #1
    hi,
    what is wrong about this proof?
    If [tex]x^TAx=0[/tex] then A is antisymetric matrix. True? false?
    P: False
    [tex]A=-A^T[/tex]
    [tex]x^TAx=-x^TA^Tx[/tex]
    [tex]x^TAx=-(Ax)^Tx[/tex]
    [tex]x^TAx=-\lambda\Vert x\Vert^2[/tex]
    If x^T.A.x is zero, then must be [tex]-\lambda\Vert x\Vert^2[/tex], but ||x|| is real nonzero number and lambda must be zero. But antisymetric matrix has imaginary eigenvalues [tex]b\mathrm{i}[/tex], and 0 is not in this form. So
     
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  3. May 27, 2009 #2

    jbunniii

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    You started the proof by writing down the false thing!?
     
  4. May 27, 2009 #3

    HallsofIvy

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    It's not exactly "writing down the false thing" but you start your proof asserting what you want to prove. It is an invalid proof.

    If your hypothesis is that [itex]x^TAx= 0[/itex] for some x, then the statement is certainly not true.
     
  5. May 27, 2009 #4

    jbunniii

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    What field are you working with? I'm assuming real scalars since you wrote [tex]x^T[/tex] instead of [tex]x^*[/tex].

    If

    [tex]x^T A x = 0[/tex] for every x in a real vector space, then it means that x and Ax must always be orthogonal.

    If

    [tex]x^* A x = 0[/tex] for every x in a complex vector space, then this actually implies that A = 0.
     
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