# False proof

## Main Question or Discussion Point

hi,
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

Related Linear and Abstract Algebra News on Phys.org
jbunniii
Homework Helper
Gold Member
hi,
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
Homework Helper
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
Homework Helper
Gold Member
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.