Background & Motivation
The geometric and intuitive nature of matrix transpose is well explained (e.g. What is the geometric interpretation of the transpose? and Truly intuitive geometric interpretation for the transpose of a square matrix) as the same stretch but the inverse twist. To better understand this, I formulated this problem:
Definitions
Let $A$ be a real $n \times n$ matrix, and consider $x \in \mathbb R^n$ such that $y = Ax$. If $y \neq 0$, define $Q(A,x)$ to be the unique orthogonal $n \times n$ matrix with determinant 1, and $\lambda(A,x)$ the unique positive real number, such that $y = \lambda(A,x) Q(A,x)x$. And for $y = 0$, define $Q(A,x) = I$ and $\lambda(A,x) = 0$.
Thus, $Q$ captures the twisting action of $A$ on $x$ and $\lambda$ the stretching.
Then define $\mathcal T(A)$ to be the unique real $n \times n$ matrix such that for all $x, \lambda(\mathcal T(A), x) = \lambda(A, x)$ and $Q(\mathcal T(A), x) = Q^{-1}(A,x)$.
Problem
Problem: Show that for any such $A$, $T(A)$ exists, is unique, and equals the transpose of $A$. Do this for both definitions of transpose:
- $T(A)_{ij} = A_{ji}$
- For all $x, y, Ax \cdot y = x \cdot \mathcal T(A)y$
Update
The comments have pointed out some gaps in how I set up this problem: namely in defining $Q$ to be unique I'll accept an answer which provides a good way to define $Q$ uniquely.
As the comment says, the uniqueness cannot be guaranteed. Thus it may be better to drop the restrictions on the uniqueness. The following is a method to define $Q$ uniquely.
If $x,y$ are colinear, i.e. $y=\lambda x$, then it natural to set $\lambda(A,x)=\lambda\in\mathbb R$ and $Q(A,x)=I_n$. Hence in the following we consider the case where $x,y$ are not colinear. Moreover, only $Q$ needs to be considered, since we have $|Ax|=|\lambda||Qx|=\lambda|x|$ and therefore $\lambda=|Ax|/|x|$.
Let $x_0=\frac{x}{|x|}$ and $y_0=\frac{Ax}{|Ax|}$. Suppose $U,V$ are the orthogonal matrix whose determinants are both $1$ and first columns are $y_0,x_0$ respectively. We set $Q(A,x)=UV^\mathsf T$, then we have $\det Q=1$ and $$ Qx_0=UV^\mathsf Tx_0=U(1,0,\dots,0)^\mathsf T=y_0. $$ The inverse of an orthogonal matrix is just the transpose, so $Q^\mathsf {-1}(A,x)=VU^\mathsf T$. However you should note that $Q(A^\mathsf T,x)\neq Q^{-1}(A,x)$ in this case. Instead we have $Q(A^+,y)=Q^{-1}(A,x)$, where $A^+$ is the Moore–Penrose inverse of $A$.