Proof of A is orthogonal $\Leftrightarrow \|Ax\|=\|x\|$ I want to proof this property.
A is orthogonal $\Leftrightarrow \|Ax\| = \|x\|$    
I tried to elaborate from this, but cannot see how to get any further:
$\|A\| \times \sqrt{<x,x>}$    
I really appreciate your answer!!!
 A: Clues:


*

*$\|x\|^2=x^tx$

*A orthogonal: $A^tA=I$
A: You should add to your equivalence
$$
A \quad \text{orthogonal} \qquad \Longleftrightarrow \qquad || Ax|| = ||x|| \quad \text{for all} \quad x \ .
$$
Said this, one implication is clear: if $A$ is orthogonal, then
$$
||Ax||^2 = (Ax)^t(Ax) = x^t(A^tA)x = x^tx = ||x||^2 \ , \quad \text{for all} \quad x \ .
$$
And, since on both sides you have positive numbers, you can delete both squares.
For the other implication, you can take advantage of the computations already done: $||Ax|| = ||x||$ is the same as $x^t(A^tA)x = x^tIx$. Of course, this wouldn't mean that, necessarily $A^tA = I$, if the equality was true just for one or some $x$: you can't just cancel $x$'s on both sides. But, if it's true for all $x$, then you can say more:


*

*First, for any matrix $B$, what's $e_i^tBe_j$? (Here $e_i^t = (0, \dots, 1, \dots, 0)$: just one $1$ in the $i$th coordinate, and the rest are zeros.)

*Then you should try to prove this equality (probably you already know it):


$$
x^ty =\frac{1}{2} \left( ||x+y||^2 - ||x||^2 - ||y||^2  \right) \ .
$$
