Let $\varphi:V\times V\to k$ is bilinear function over field $\mathbb{k}$ where $\operatorname{char}\mathbb{k}\neq 2$ and let $b:V\to \mathbb{k}$ defined as $b(x)=\varphi(x,x)$ is quadratic form.
Let $B$ be the matrix of quadratic form $b$, i.e. matrix with elements $b_{ij}=\varphi(e_i,e_j)$.
Suppose quadratic form $b$ has the property: for all $x\in V$ we have $b(x)=0$.
What we can say about the matrix $B$?
After some thoughts I came up with the following conclusion: first of all one can show that $b_{ii}=0$ and $b_{ij}+b_{ji}=0$.
Is my conclusion correct?
Would be very thankful for any comments!
Let $A=(B+B^T)/2.$ Then $A$ is smmetric and for any $n-$ place column-vector $x,$ wehave $$x^TBx=x^TAx$$. Thus $x^TBx=0$ for all $x$ iff $x^TAx=0$ for all $x$. Since $A$ is symmetric and the characteristic of $k$ is not 2, there exists a non-singular $n \times n$matrix $P$ such that $P^TAP$ is diagonal. Let $x=Py$. Then $$x^TAx=y^TP^TAPy$$ The conditions $x^TAx=0$ for all $x$ and $y^TP^TAPy=0$ for all $y$ are equivalent. Let $y$ be the column-vector whose $i-$th entry is 1 and whose other entries are all 0. Then $$y^TP^TAPy=0$$ = $i-$ th diaagonal entry of $P^TAP$. So $P^TAP=0$ and hence $A=0$. Thus $B+B^T=0.$ We conclude that $x^TBx=0$ for all $x$ iff $B^T=-B$, i.e. iff B is skew-symmetric.