4
$\begingroup$

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!

  • 1
    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.2011-08-26
  • 0
    What did you try? Can you express $q_A(x)$ with $x$ and $A$?2011-08-26
  • 0
    @Gerry Myerson I think it's suffice to show x*Px=0 iff P=0, but I don't know whether that's correct.2011-08-26
  • 0
    Related: http://math.stackexchange.com/questions/12172/does-there-exist-a-nonzero-linear-transformation-t-such-that-alphat-t-a/2011-08-26

2 Answers 2

4

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"

  • 0
    I mean for all x. Sorry for that.2011-08-26
3

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 $s_{ij}=s_{ji}=0$.

  • 2
    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 $A=-A^T$.2011-08-26
  • 0
    @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.2011-08-26
  • 0
    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