Show that $A \in M_n$ and $B \in M_n$ need not be equal if $x^TAx = x^TBx$ for all $x \in \mathbb{C}^n$.

My attempt: $x^TAx - x^TBx = x^T(A-B)x = x^TCx = 0$, where define $C = A-B$, then either $A=B$, or the eigen values of $C$ correspond to zero. Does that proof make sense? or do I need more information to corroborate the statement?

Thank you in advance


So you need $C\not=0$ such that $x^TCx = 0$ for all $x$. For $2\times2$-matrices this may be achieved with $C=C_2=\pmatrix{0&1\\-1&0}$. For $n>2$ you may use $\pmatrix{C_2&0\\0&0}$.

  • $\begingroup$ Thank you so much. $\endgroup$ – user550103 Apr 14 '18 at 13:48

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