Debunking the False Proof of x^TAx=0 and A Being Antisymmetric Matrix

[lukaszh]

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

 

[jbunniii]

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

You started the proof by writing down the false thing!?

 

[HallsofIvy]

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 x^TAx= 0 for some x, then the statement is certainly not true.

 

[jbunniii]

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

If

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

If

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