Can $A \cdot \vec{x}=\vec{b}$ be re-written as $\vec{x}=A^T \cdot \vec{b}$?

Question

An orthogonal matrix $Q \in \mathbb{R^{n,n}}$ is a matrix such that $Q \cdot Q^T=I_n=Q^TQ$, which means that $Q^{-1}=Q^T$. If we have an orthogonal matrix $A \in \mathbb{R^{m,m}}$, $\vec{x} \in \mathbb{R^m}$ and $\vec{b} \in \mathbb{R^p}$ and we want to solve the equation $A \cdot \vec{x}=\vec{b}$, can the equation be re-written as $\vec{x}=A^T \cdot \vec{b}$?

Answer 1

Assuming you mean $\vec b \in \mathbb{R}^m$ instead of $\vec b \in \mathbb{R}^p$, the following equalities hold:

$A\cdot\vec x = \vec b$

$A^{-1}A\cdot\vec x = A^{-1}b$

$I_m\cdot\vec x = A^{-1}b$

$\vec x = A^{\top}b$

So this is indeed true.

Answer 2

Yes if $p=m$, in fact $$A \vec{x}=\vec{b} \Leftrightarrow A^T A \vec{x}=A^T \vec{b} \Leftrightarrow \vec{x}=A^T \vec{b}, $$ since $A^T A=I$.