Is the "Its transpose is its inverse" definition of an orthogonal matrix equivalent to the "It preserves the dot product" definition?

Question

A long time ago, I was taught that a real $n\times n$ matrix $A$ is called orthogonal if $AA^t=I$. But recently I learned from a DG book that $A$ is said to be orthogonal if it preserves the dot product: $$(Ax)\cdot(Ay)=x\cdot y\quad\text{for all }x,y\in\mathbb{R}^n.$$ Are these two definitions equivalent? It is easy to see that the former implies the latter: $$(Ax)\cdot(Ay)=(Ay)^t(Ax)=(y^t A^t)(Ax)=y^t x=x\cdot y$$ But I have a hard time going from the latter to the former. Is it possible? Thank you for your time.

Answer 1

Suppose $Ax\cdot Ay=x\cdot y$ for all $y$.

Recall that $Ax\cdot Ay=A^{T}Ax\cdot y$.

So $(A^{T}Ax-x)\cdot y=0$.

Put $y=A^{T}Ax-x$ to see that $A^{T}Ax=x \ \forall x$.

Answer 2

Notice that if $$ (Ax)\cdot (Ay)=x\cdot y$$ for every $x,y\in \mathbb{R}^n$ then $$ (x)\cdot(A^tAy)=x\cdot y.$$ Using bilinearity of the dot product, we have $$(x)\cdot ([A^tA-I]y)=0. $$ By non-degeneracy of the dot product, there holds $$(A^tA-I)y=0, \;\;\forall y\in\mathbb{R}^n. $$ Therefore $$ A^tA=I,$$ and we are done.