If $x^{T}A^{T}Ax = x^Tx$ holds for every $x$, then $A^{T} A = I_n$

Question

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 $A^{T}A = I_n$?

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I appreciate any help!

Answer 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 $v$.

Answer 2

Slightly different approach (in the final stage of proof).
We have $$\left( \forall x \in\mathbb R^n \right) \left(x^{T} (A^{T} A -I) x =0 \right)$$

$S=A^{T} A -I$ is obviously symmetric hence it is diagonalizable.

Equation above must be hold also for any eigenvector $v$ of $S$. i.e. $v^TSv=0 =v^T\lambda v$,
hence every eigenvalue $\lambda$ of $S$ is zero and it's diagonal form must be zero matrix what means $S$ is zero matrix itself.