TITLE: Skew-symmetric matrix and quadratic form
QUESTION [6 upvotes]: Show that the quadratic form of a matrix is $0$ if and only if the matrix is skew–
symmetric, i.e., show that $q_A(x) = 0$ for all $x$ iff $A^t = −A$.
Thanks a lot!

REPLY [5 votes]: If $A^t=-A$ then for $x\in\mathbb R^n$ we have $-x^tAx=x^tA^tx=(x^tAx)^t=x^tAx$ since $x^TAx$ is a real number hence $x^tAx=0$.
For the controverse, let $S:=A^t+A$. Since for all $x\in\mathbb R^n$ we have $q_A(x)=x^tAx=0=-x^tAx$, we get $x^tSx=0$ for all $x$. We denote by $e_j$ the vector whose entries are $0$, except the $j$ which is $1$, and $s_{ij}$ the entries of $S$. Since $e_j^tSe_j=0$ for all $j$ we get $s_{jj}=0$ and for $i\neq j$ we have $$0=(e_i+e_j)^tS(e_i+e_j)=e_i^tSe_i+e_j^tSe_i+e_i^tSe_j+e_j^tSe_j=e_j^tSe_i+e_i^tSe_j=s_{ji}+s_{ij}.$$
Since $S$ is symmetric, we have $s_{ji}=-s_{ij}$ hence $s_{ij}=s_{ji}=0$.