False proof

  • Thread starter lukaszh
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  • #1
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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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  • #2
jbunniii
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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
You started the proof by writing down the false thing!?
 
  • #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.
 
  • #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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