False proof

[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.