Conclude that $A^TA=I$, so $A^{-1}=A^T$. A matrix $A$ is said to be orthogonal if for every vector $\textbf{x}$ we have 
$||A\textbf{x}||_2=||\textbf{x}||_2$. 
Show that such a matrix will necessarily have the property that 
$\left<A\textbf{x},A\textbf{y}\right>=\left<\textbf{x},\textbf{y}\right>$. 
From here, conclude that $A^TA=I$, so $A^{-1}=A^T$. 
$\textbf{Hint:}$ Make use of the identity $\left<\textbf{x},\textbf{y}\right>=\frac{1}{4}||\textbf{x}+\textbf{y}||^2_2 - \frac{1}{4}||\textbf{x}-\textbf{y}||^2_2$.

I am not sure how to even begin this question. If someone could at the least provide definitions of the notation. How do you read the first line? Is it read, "The norm of the matrix A with vector x in $\mathbb{R}^2$"? Then what does the hint say? Why is there a 2 as a superscript and subscript?
 A: Hint
$$\left<Ax,Ay\right>=\left<x,y\right>
\text{ }\text{ }\Rightarrow\text{ }\text{ }
\left<A^TAx,y\right>=\left<x,y\right>
\text{ }\text{ }\Rightarrow\text{ }\text{ }
\left<(A^TA-I)x,y\right>=0$$
A: To try to decipher this for you:
Here $A \in \mathbb{R}^{n \times n}$ is some matrix and $x,y \in \mathbb{R}^n$ are vectors. Then $Ax = y \in \mathbb{R}^n$ and
$$
\|Ax\|_2 = \|y\|_2 = \sqrt{\sum_{k=1}^n y_k^2}
$$
where $y = (y_1, y_2, \ldots, y_n)$.
So the first line claims that $A$ has the property that
$$
\|Ax\|_2 = \|x\|_2.
$$
A: The 2 in the superscript means squared. The 2 in the subscript means the L-2 norm, aka $\Vert x \Vert_2=\sqrt {\sum_{i=1}^n x_i^2}$. Usually the subscript is dropped when you know it’s L-2 because it’s most common. Other norms include 1 norm, infinity norm, etc. Notice that after squaring the norm there’s no overarching square root.
So
$$
\begin{split}
(Ax)^T(Ay)
 &= 1/4(Ax+Ay)^T(Ax+Ay)-1/4(Ax-Ay)^T(Ax-Ay)\\
 &= 1/4(A(x+y))^T(A(x+y))-1/4(A(x-y))^T(A(x-y)) \\
 &= 1/4(x+y)^T(x+y)-1/4(x-y)^T(x-y) \\
 &= x^Ty
\end{split}
$$
Using the notation you’re given,
$$
\begin{split}
\langle Ax, Ay\rangle
 &= 1/4\Vert Ax+Ay\Vert ^2-1/4\Vert Ax-Ay \Vert^2 \\
 &= 1/4\Vert A(x+y)\Vert^2 -1/4\Vert A(x-y) \Vert^2 \\
 &= 1/4\Vert x+y\Vert^2 -1/4\Vert x-y\Vert^2 \\
 &= \langle x, y\rangle \ldots
\end{split}
$$
For the second part, You showed already that $(Ax)^T(Ay)=x^Ty$. But $(Ax)^T(Ay)=x^TA^TAy$. Looking at both these, It’s sandwiched in between, so it must be that $A^TA=I$.
Because $AA^{-1}=I$, A^T is the inverse. Learn the moves involving matrix distribution, inner products, norms squared, and transposing a matrix-vector product and you’re good to go.
