0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.