Proving Properties of orthogonal Matrix

Question

I was given a task whereby its defined that a $n \times n$ matrix, A, is orthogonal if $\langle A\vec{u},A\vec{v}\rangle$ = $\langle \vec{u}$,$\vec{v}\rangle$ and i have also been given the property that

$\langle A\vec{u},\vec{v}\rangle$ = $\langle \vec{u},A^T\vec{v}\rangle$ .

Now if $A$ is an orthogonal matrice how do i prove that $A^TA = I_n$

I know how to prove it using the property that an orthogonal matrices has column vectors that are orthonormal. However we were told to prove this property in the later part so I don't think we should use that in this question. So if not using this property, I'm completely lost in proving this and I would appreciate any help at all.

Answer 1

If $A$ is orthogonal, then, for any two vectors $\vec u$ and $\vec v$, you have $\left\langle A\vec u,A\vec v\right\rangle=\left\langle\vec u,\vec v\right\rangle$. But you also have $\left\langle A\vec u,A\vec v\right\rangle=\left\langle\vec u,A^\top A\vec v\right\rangle$. So, for any two vectors $\vec u$ and $\vec v$,$$\left\langle\vec u,A^\top A\vec v\right\rangle=\left\langle\vec u,\vec v\right\rangle.$$In other words,$$\left\langle\vec u,\left(A^\top A-\operatorname{Id}\right)\vec v\right\rangle=0.$$In particular,$$\left\|\left(A^\top A-\operatorname{Id}\right)\vec v\right\|^2=\left\langle\left(A^\top A-\operatorname{Id}\right)\vec v,\left(A^\top A-\operatorname{Id}\right)\vec v\right\rangle=0.$$So, for every vector $\vec v$, $\left(A^\top A-\operatorname{Id}\right)\vec v=\vec0$; in other words, $A^\top A\vec v=\vec v$.

Answer 2

For some matrix $C$, we know that if $u^T C v = 0$ for all vectors $u$ and $v$, then $C = O$. $$ \left<A u, Av\right> = u^T A^T A v = u^T v $$ or $$ u^T (A^T A - I) v = 0 $$ for all vectors $u$, $v$.