In this post norm denotes a matrix norm, i.e. it is sub-multiplicative. All matrices are real. $A$ is of size $n \times k$ with independent columns ($k \leq n$). $B$ is of size $k \times n$.
Let $\| \cdot \|_\square$ be some arbitrary matrix norm. Assume $\| AB\|_\square \leq 1$. Then we know $\rho(AB) \leq 1$. Then we know from this post that $\rho(BA) \leq 1$.
Edit: thanks to @loupblanc I realized this was incorrect.
From this we know that there must be a matrix norm $\| \cdot \|_\triangle$ such that $\| BA\|_\triangle \leq 1$. . In fact we only know that there is a norm which is arbitrarily close to the spectral radius.
I am interested in whether there is an easy way of defining $\| \cdot \|_\triangle$ from $\| \cdot \|_\square$. For example, is it true that $\| \cdot \|_\triangle = \|F (\cdot) Q\|_\square$ for some choice of $F$ and $Q$?
Edit: when I say the norm $\| \cdot \|_\square$ is sub-multiplicative I mean it is sub-multiplicative in the space of $n \times n$ matrices. I do not mean $\| A B \|_\square \leq \|A\|_\square \| B \|_\square$, which indeed doesn't make sense. What I mean that if, say, one decomposed $AB$ into a product of square matrices, i.e. $AB = C_1C_2$, where $C_1,C_2$ are square, then we would have $\| A B\|_\square = \| C_1 C_2 \|_\square \leq \| C_1 \|_\square \| C_2 \|_\square $.