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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:

  1. $T(A)_{ij} = A_{ji}$
  2. 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.

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    $\begingroup$ I expect that the matrix $Q(A,x)$ as you define it, can be not unique. For instance, if $n=3$, $y=x=(0,0,1)$ then $y=Qx$, for any matrix $Q$ such that $Q=\begin{pmatrix} \cos\varphi & \sin\varphi & 0\\ -\sin\varphi & \cos\varphi & 0\\ 0 & 0 & 1 \end{pmatrix} $ for some real $\varphi$. $\endgroup$ Commented Dec 13, 2023 at 7:33
  • $\begingroup$ If $x$ and $y$ are not colinear, you should add the condition that $Q$ coincides with the $I$ in the orthogonal of the plane spanned by $x$ and $y$. But if $x$ and $y$ are negatively colinear, there is no canonical way to choose $Q$ if you want that $\det Q = 1$. $\endgroup$ Commented Dec 13, 2023 at 19:20
  • $\begingroup$ @ChristopheLeuridan What is "negatively colinear"? Also: I chose $\det Q = 1$ to avoid having ambiguous $\lambda$ (otherwise, if $\lambda$ works, then $-\lambda$ would too). But if you have a better way to solve that, it's fine to drop the $\det Q = 1$ requirement. $\endgroup$ Commented Dec 13, 2023 at 19:36
  • $\begingroup$ I mean $y = \lambda x$ with $\lambda < 0$, my english is probably not good. $\endgroup$ Commented Dec 13, 2023 at 20:55

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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 $A$.

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