# Can we prove $A \hat x = r \hat y$ implies $A^T \hat y = r \hat x$?

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:

1. Manipulations of transpose, such as $$A^TA\hat x = A^T r \hat y = r A^T \hat y$$.
2. Definition of $$A^T_{ij} = A_{ji}$$. I didn't see how to use this.
3. 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?

• This fails if $A$ is the zero matrix and $x$ is a non-zero vector. Commented Dec 6, 2023 at 21:12
• @timon92 If $A$ is the zero matrix, then $r = 0$, and so $A^Tv = rw$ for all $v,w$. Commented Dec 6, 2023 at 21:18
• Not really. If $A$ is the zero matrix and $y$ is the zero vector, then $r$ can be arbitrary. Commented Dec 6, 2023 at 21:25
• I'll add another counterexample to the answer below. Suppose $A$ is an invertible square matrix that is not symmetric. Consider any of its eigenvectors $v$, where the relationship $Av = \lambda v$ holds. Because $A$ is not symmetric, $A\neq A^T \implies A^Tv \neq \lambda v$
– Sam
Commented Dec 6, 2023 at 21:28

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.
• Thanks. I added context to the OP. I think my error is in thinking that diagonal matrix $P$ can't change a vectors direction. $P$ has a basis of real eigenvectors, but since they have different eigenvalues, $P$ can change direction of a linear combination of these eigenvectors. Commented Dec 6, 2023 at 21:33