Angle preserving matrix Let $A$ be an invertible matrix of order $n$ such that for any nonzero vectors $u,v\in \mathbb{R}^n$,
the angle between $u$ and $v$ is always equal to the angle between $Au$ and $Av$.
Prove: $A=cP$ for some scalar $c$ and orthogonal matrix $P$.
By considering standard basis $E = \{e_1,e_2,\ldots,e_n\}$ and vectors $Ae_i$ for
$i = 1,\ldots n$, I have already shown that the columns of $A$ forms an orthogonal basis for $\mathbb{R}^n$. How can I proceed from here?
 A: Begin by looking at the definition of an orthogonal matrix:
\begin{equation}
Q Q^T = I
\end{equation}
Let's look at how to preserve the angle on two unit vectors. Consider $a$ and $b$ two unit vectors. There angle is $\cos^{-1}a^T\cdot b$. Lesson is we want to preserve $a^T b$. 
Multiplying our matrix $A$ gives \begin{equation}(Aa)^T (Ab) =a^T A^T A b \end{equation}
This immediately implies to us that if we want to preserve angle we need:
$A^T A = I$. This looks strangely familiar to the definition of an orthogonal matrix from wikipedia... 
A: let me see if it works in two dimensions. since $A$ is nonsingular and angle preserving, we can take $Ae_1 = d_1q_1, Ae_2 = d_2q_2$ where $e_1,e_2$ are the standard basis is two dimensions and $q_1, q_2$ are unit vectors orthogonal to each other. now we can write $A =(q_1,\  q_2)\pmatrix{d_1 & 0\\ 0 &d_2}= QD, $ where $Q$ is orthogonal and $D$ a diagonal matrix. to avoid $A$ being orthogonal we require that $D^\top D \neq I.$
we will see what the constraints on $D$ if $A$ were tp preserve angles. by construction, we know that $A$ preserves the angle between $e_1$ and $e_2.$ 
$A\pmatrix{\cos t \\ \sin t} = d_1\cos t \ q_1 + d_2 \sin t \ q_2, Ae_1 = d_1 q_1$ therefore the inner product of $A\pmatrix{\cos t \\ \sin t}$ and $Ae_1$ is $d_1^2 \cos t$ and the lengths are $\sqrt{d_1^2 \cos^2 t + d_2^2 \sin^2 t}, |d_1|$ the prservation of the angle between $A\pmatrix{\cos t \\ \sin t}$ and $Ae_1$ implies we must have $$d_1^2 \cos t = \cos t \ \sqrt{d_1^2 \cos^2 t + d_2^2 \sin^2 t}\  |d_1| \to d_1^2 = d_2^2 $$ 
that is $$A = \pmatrix{sgn(d_1)q_1,\ sgn(d_2)q_2}\pmatrix{|d_1|&0\\0&|d_2|} $$ this can also be written as $A = k. Q$ 
