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?


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.


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