3
$\begingroup$

I am wondering the following: if $f:\;V\to \mathbb{R}$ is a quadratic form that is neither positive or negative definite, must its kernel be non-trivial?

Here quadratic form means $f(v)=g(v,v)$ where $g:\;V\times V\to\mathbb{C}$ is a Hermitian form.

I can prove that this is true when $V$ finite dimensional, through diagonalisation; but I am stuck on the more general case. Help?

$\endgroup$
1
$\begingroup$

Choose $x,y\in V$ such that $f(x)<0<f(y)$. Then $f$ is a quadratic form on $V'=\text{span}(x,y)$. Now you are back in the finite dimensional case.

If you do not like diagonalisation choose $x,y$ as before, and look at the function $$F\colon [0,1] \to \mathbb{R}, t \mapsto f(tx +(1-t)y). $$ Since $f$ is quadratic $x$ and $y$ can not be linearly dependent and since $F$ is continuous by intermediate value theorem it has a zero. Thus the kernel is non-trivial.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.