If a real matrix induces an isometry, is the matrix orthogonal? Where can I find a proof? Math people:
This is a pretty simple question, but I am having trouble finding an answer.  I did not 
find an answer in the Similar Questions, and I apologize ahead of time if this is a duplicate.
Here is my question: if $A$ is a real square matrix, and $\|A\mathbf{x}\| = \|\mathbf{x}\|$ for all real column vectors $\mathbf{x}$ of the right length, is $A^T A = I$, that is, is $A$ an orthogonal matrix?  (the converse is true, of course).  I am pretty sure the answer is "yes", but it is surprisingly hard to verify this, and I would like at least a link to a proof.
EDIT: This is in fact true, and it appears as Proposition 5.9 in the document at http://www.cis.upenn.edu/~eas205/eas205-12-sl5.pdf .  This is in the thirty-third page of a forty-two page document, and it is not apparent whether there is an elementary proof. 
Stefan (STack Exchange FAN)
 A: All you need to do to show that $A$ is orthogonal is to show that each column of $A$ has length 1, and any that two columns of $A$ are orthogonal.
Let $e_i$ be the vector with $1$ in the $i$th component, and $0$ in the other components. Then the $i$th column of $A$ is $A e_i$, and $||A e_i || = ||e_i|| = 1$.
Now, consider the vector $v = e_i + e_j$. Since $A$ preserves lengths, $Av$ must have length $2$. But if the columns of $A$ are $c_i$,  the length of this is $2 + 2c_i\cdot c_j$, which shows that $c_i \cdot c_j = 0$. 
A: Hint: $||Ax||^2=(Ax)^T(Ax)=x^T(A^TA)x$. (EDIT: I no longer think this hint is actually useful)
Alternative hint: $A$ is orthogonal iff the columns of $A$ are orthogonal. The norm being preserved means that the inner product is as well. (i.e. $\langle x,y\rangle = \langle Ax,Ay\rangle.)$  The columns of $A$ are the images of the usual basis under the transformation that $A$ represents, so since inner products are preserved, the columns are orthonormal also.
A: You could start with $<A^TAx,x>=<Ax,Ax>=||Ax||^2=||x||^2=<x,x>$. So $<Ix,x>=<Ax,x>$ for all x. ($<>$ is the inner/dot product.) The by observe that 
$0=<A^TAx,x>-<Ix,x> = <(A^TA-I)x,x>$ for all x and conclude that $A^TA-I$ must be the zero matrix.
A: You say (in a comment) that you would like to give a proof to "non math majors". Speaking as quite the opposite of a non math major, I find that the SVD (singular value decomposition) is very important. Maybe you can use it:
$$ A = USV^*$$
$$ ||Ax|| = ||USV^*x|| = ||x|| \quad \forall x \Rightarrow \exists X \quad \text{s.t.}\quad X^*X=I \wedge ||AX|| = ||USV^*X|| = ||X|| $$
and by the property of unitary $X$ (and unitary $U$ and $V$) that may be converted to
$$||USV^*|| = ||I||=||S||$$
Which gives $S$ with only unitary elements on the diagonal (otherwise if some element on the diagonal of $S$ is not unitary then some $x$ gives $||Ax|| \ne ||x||$), so $S^*S = I$ and
$$A^*A = (VS^*U^*)(USV^*) = VS^*SV^* = VV^* = I$$
and of course as you seem to be using reals, use $\star^\top $ instead of $\star^*$.
