Let $A$ be a real matrix. If $A \hat x = r \hat y$, does $A^T \hat y = r \hat x$?
I believe it does, but am struggling to prove it.
Geometrically, I believe this is true, because $r$ captures the change of magnitude of $x$ under $A$, which should be shared by $A^T$, and $\hat y$ captures the change of direction of $x$ under $A$, which should be inverted under $A^T$. (Thinking of the polar decomposition of $A = UP$, the transpose of $A$ is often described as inverting the orthogonal matrix $U$ while preserving the diagonal matrix $P$.)
To prove this, I've experimented with the following approaches:
- Manipulations of transpose, such as $A^TA\hat x = A^T r \hat y = r A^T \hat y$.
- Definition of $A^T_{ij} = A_{ji}$. I didn't see how to use this.
- Finding the image of a basis under $A$. I didn't see how to go from $A\hat x$ to a full basis.
Is my conjecture true? If yes: How do I prove it? If not: What is wrong with my geometric argument?
The conjecture is false: here $A = \begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix}$, $\hat x = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \hat y$
$$\begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = 3\begin{pmatrix} 1 \\ 0 \end{pmatrix}$$ $$\begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix}^T \begin{pmatrix} 1 \\ 0 \end{pmatrix} =\begin{pmatrix} 3 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \neq 3 \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$
The flaw in your geometric argument is that the "change in direction of $x$ under $A$ is not simply reversed by transposing $A$, and the "change in magnitude" is not simply preserved either. It's not entirely clear what you mean by this precisely, so I can't be more specific as to how it's flawed.
Rather than the transpose, you would want some operation that's more akin to the inverse of a matrix, but one that generalizes to non-square and non-invertible matrices.