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Suppose that $B, C \in M_3(R) $ are three-by-three matrices over $R$. Suppose that there is an open set $U \subset R^3$ so that for all vectors $u, v \in U$ we have $u^TBv = u^TCv$. Prove that $B = C$.

This is part of a homework assignment, so any hint will help. Thanks.

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  • $\begingroup$ HINT: Try to consider the difference $u^T(B-C)v$. $\endgroup$ – Crostul Nov 30 '15 at 12:34
  • $\begingroup$ I considered that but hit a dead end. $\endgroup$ – J Simmons Nov 30 '15 at 12:37
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Given $u^TBv = u^TCv$ $\forall u,v\in U\subset\mathbb{R}^3$

Consider $A=(B-C)$, let $v\in U$, then $$\|Av\|_2^2 = (Av,Av)=v^tA^TAv=0$$ we got that $\|Av\|_2=0,\forall v\in U$, therefore $A=0$ and hence $B=C$.

This could only be true for an open subset (of $\mathbb{R}^3$ in this case), because otherwise $U$ could be a discrete set, e.g. subset of the kernel of $A$, for which the last argument above won't work.

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