TITLE: What is the geometric relationship between $A$ and $A^T$?
QUESTION [7 upvotes]: Posed a more specific way:
Let $A \in \mathbb R ^ {m \times n}$ and $S_k$ be the unit $k$-sphere.
What is the exact geometric relationship between $E_m = \{ A\vec x \mid \vec x \in S_m \}$ and $E_n = \{ A^T\vec y \mid \vec y \in S_n \}$?

REPLY [3 votes]: For $A \in \mathbb{R}^{n\times m}$ with $m>n$ define  the SVD as
$$A=U[\Sigma\;0]V^T$$
Since the sphere is unaffected by rotations, 
$$E_m =\{U [\Sigma\;0] \vec{x} \mid \vec{x} \in S_m\}= \{U \Sigma \vec{x} \mid \vec{x} \in B_n\} $$
where $B_n$ is the $n$-ball. Therefore $E_m$ is the $n$-ball  scaled by $\Sigma$ and rotated by $U$. For $A^T$
$$E_n =\left\{V \left[{\Sigma \atop 0}\right] \vec{x} \mid \vec{x} \in S_n\right\}$$
we have a scaling of the $n$ sphere by $\Sigma$ then rotated into $m$-dimensional space by $V$.
If we ignore rotation/embedding in space, both objects have the same bounding shape, but $E_m$ is filled while $E_n$ is not.