Is separability a property of the homomorphism or just the matrix that represents it? I've been trying to learn about separable matrices in the context of the Singular Value Decomposition. I suspect one of the following is true but don't know which:

*

*If a homomorphism is separable, it remains so no matter what bases are chosen for the domain and codomain; or

*Any homomorphism can be written as a separable matrix with the right choice of bases.

Is either of these true? Suggested references for this sort of thing would also be greatly appreciated.
 A: This is a little bit difficult to google for so for the sake of any readers who, like me, had not seen this term before: a separable matrix $M$ is a matrix that can be written as an outer product $v \otimes u^T$ of two vectors. Here is the relevant result:

Claim: Let $U, V$ be vector spaces with bases $B, C$, let $T : U \to V$ be a linear transformation, and let $M$ be the matrix of $T$ with respect to the given bases. Then $M$ is separable iff $T$ has rank at most $1$, and in particular this condition does not depend on the choice of bases.

$U$ and $V$ don't need to be finite-dimensional here but we can assume that condition for simplicity.
Proof. If $M = v \otimes u^T$ is separable then every column of $M$ is a scalar multiple of $v$, so $M$ (and hence $T$) has rank at most $1$.
Conversely, if $T$ has rank at most $1$, then either $T = 0$ (in which case the desired property is trivially satisfied) or any matrix $M$ of $T$ with respect to any bases has the property that, if we pick some nonzero column $v$, then every column is a scalar multiple of this column. These scalars can then be assembled into a vector $u$ such that $M = v \otimes u^T$ as desired. $\Box$
To say all this in a coordinate-independent way, a linear map $T : U \to V$ has rank at most $1$ iff it can be written $T = v \otimes f$ where $f \in U^{\ast}$ is an element of the dual space. This generalizes to any finite rank as follows: a linear map has rank at most $k$ iff it can be written $T = \sum_{i=1}^k v_i \otimes f_i$ where $v_i \in V, f_i \in U^{\ast}$. Again $U$ and $V$ don't need to be finite-dimensional.
