Prove or disprove: Suppose $H$ is a bilinear form on a finite dimensional vector space $V$, with $\dim(V)>1$. Then for any $x\in V$ there always exists a non-zero $y\in V$ such that $H(x,y)=0$.
I feel the statement is correct, because the bilinear form $H(x,y)$ can be written as $H(x,y)=x^TAy$ where $A_{ij}=H(e_i,e_j)$ where $\{e_i:i=1,2,...,\dim(V)\}$ is a basis of $V$. We can always choose a $y$ such that $y$ is non-null and $H(x,y)=0$ because we are essentially finding solution to $(y_1,y_2,...,y_{\dim(V)})$ where $\sum_{i=1}^{\dim(V)}\sum_{j=1}^{\dim(V)}x_iA_{ij}y_j=0$ and we can of course find such a non-null $y$.
$\newcommand{\im}[0]{\mathrm{im}}$Given $x$, consider the linear map $f : V \to K$, where $K$ is the base field, given by $y \mapsto H(x, y)$. Since $\dim(V) > 1 = \dim(K)$, and $\dim(\ker(f)) + \dim(\im(f)) = \dim(V)$, the kernel will be different from $\{ 0 \}$.