If $x^{T}A^{T}Ax = x^Tx$ holds for every $x$, then $A^{T} A = I_n$ [closed]
Given $A \in \mathbb R^{n \times n}$, if $$\left( \forall x \in\mathbb R^n \right) \left(x^{T} A^{T} A x = x^T x \right)$$ how to conclude that $A^{T}A = I_n$?
I appreciate any help!
Solution 1:
Presumably the underlying field is real. It suffices to prove the stronger statement that if $S$ is a symmetric matrix and $x^TSx$ is identically zero, then $S=0$. To prove this, put $x=v+Sv$ for some arbitrary vector $v$.