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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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Please rephrase your question so it doesn't sound like you are assigning us homework. Also, if you have gotten anywhere thinking about the problem, it might help to let people in on your work. –  Gerry Myerson Aug 26 '11 at 12:43
    
What did you try? Can you express $q_A(x)$ with $x$ and $A$? –  Davide Giraudo Aug 26 '11 at 12:43
    
@Gerry Myerson I think it's suffice to show x*Px=0 iff P=0, but I don't know whether that's correct. –  BVFanZ Aug 26 '11 at 12:45
    
Related: math.stackexchange.com/questions/12172/… –  Hans Lundmark Aug 26 '11 at 16:24
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2 Answers 2

up vote 3 down vote accepted

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. –  BVFanZ Aug 26 '11 at 12:51
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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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brilliant! But no need to introduce $S=A+A^T$. The below is a simpler proof. Since $x^TAx=0, \forall x$, $e_i^TAe_i=0$, so $A_{ii}=0$; $(e_i+e_j)^TA(e_i+e_j)=0$, so $A_{ij}+A_{ji}=0$. Hence $A=-A^T$. –  Shiyu Aug 26 '11 at 15:08
    
@Shiyu: you're right. I agree we don't need to introduce the matrix $S$, we it show that if a symmetric matrix is such that $x^tSx=0$ for all $x$ then $S=0$. I think it's an interesting property in itself, especially when we study quadratic forms. –  Davide Giraudo Aug 26 '11 at 15:16
    
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. –  Shiyu Aug 27 '11 at 1:31
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