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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!

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    Related: http://math.stackexchange.com/questions/12172/does-there-exist-a-nonzero-linear-transformation-t-such-that-alphat-t-a/2011-08-26

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Here is the if part, $x^TA x = x^TA^Tx= - x^TA x \in \mathbb{R}$ Only if part does not hold, counter-example $ \begin{pmatrix} 1\\ 0\end{pmatrix}^T\begin{pmatrix} 0&1\\1& 0\end{pmatrix}\begin{pmatrix} 1\\ 0\end{pmatrix}=0 $ unless you are missing "for all $x$" before the "iff"

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    I mean for all x. Sorry for that.2011-08-26
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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 $s_{ij}=s_{ji}=0$.

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    Yes, it is useful that symmetric matrix $A$ will be zero if $x^TAx=0,\forall x$. But I think it is merely a special case because a matrix is zero if it is both symmetric and skew-symmetric.2011-08-27