# Prove the geometric nature of matrix transpose: same stretch, inverse twist

## 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.

• 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 $\varphi$. Commented Dec 13, 2023 at 7:33
• 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 $\det Q = 1$. Commented Dec 13, 2023 at 19:20
• @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. Commented Dec 13, 2023 at 19:36
• I mean $y = \lambda x$ with $\lambda < 0$, my english is probably not good. Commented Dec 13, 2023 at 20:55

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 $$A$$.