If $(Av,Au)=(v,u)$ then matrix $A$ is orthogonal 
Let $A\in M_{n \times n}(\Bbb R)$ and suppose that for every $u, v \in \Bbb R^{n}$ $$(Av,Au) = (v,u)$$ where $(\cdot,\cdot)$ is the standard inner product on $\Bbb R^{n}$. Prove $A$ is an orthogonal matrix.


I wasn't able to solve it, I got to the point
$\left(Av,Au\right)=\left(Av\right)^{T}Au=v^{T}A^{T}Au$ and $\left(v,u\right)=v^{T}u$
$\left(Av,Au\right)=\left(v,u\right)\ \ \ ➜\ \ \ v^{T}A^{T}Au\ =\ v^{T}u$
and now I'm stuck.. I don't know if I can conclude that $A^{T}A=I_{R^{n}}$ just by the last equation, and if not how to get to the point I can show it..
Now I have two questions, first the official solution is this

Let $u = e_{i}$ and $v = e_{j}$ . Then $(a_{i} , a_{j} ) = (Ae_{i} , Ae_{j} ) = (e_{i} , e_{j} ) = δ_{i,j}$ . Thus, the columns of $A$ are
orthonormal, so $A$ is orthogonal.

This is super unclear.. what is $a_{i},a_{j}$? what is $δ_{i,j}$?
I understand that $(e_{i},e_{j})=0$ if $i \ne j$ and 1 otherwise, but why can we choose $v,u$ if it's a for every claim? If someone can explain to be the logic behind the solution I'd be grateful.
the second question is if there is another way to solve it?

$e_i$ is an orthogonal basis. $a_i=Ae_i$ for each $i$. $\delta_{i,j}$ is called the kroneckerdelta function and is simply $1$ if $i=j$ and $0$ if $i\neq j$.
