ABA-product of matrices and length of chains of principal inner ideals

Question

Let $k$ be a field, $p,q$ positive integers, and let $R$ be the space of $(p \times q)$-matrices over $k$, and $S$ be the space of $(q \times p)$-matrices over $k$. For every matrix $A \in R$, we define a so-called principal inner ideal

$$[A] := \{ ABA \mid B \in S \} \subseteq R.$$

Define the principal rank of $A$ as the length $n$ of a maximal chain of principal inner ideals

$$[0] \subsetneq [A_1] \subsetneq [A_2] \subsetneq \dots \subsetneq [A_{n-1}] \subsetneq [A].$$

Is it true that the principal rank of a matrix coincides with its usual rank?

Motivation: The pair $(R, S)$ is an example of a so-called Jordan pair, introduced by O. Loos, and the inner ideals play an important rôle in the theory of Jordan pairs. Loos seems to indicate that for matrix pairs, these two rank notions coincide, but he doesn't provide any details.

Answer 1

This is a nice exercise in linear algebra. It's more elegant to work without coordinates here, that is, let $R= \hom(U,V)$ and $S=\hom(V,U)$, where $U$ and $V$ are vector spaces over $k$. For $\alpha\colon U\to V$, show that $$ [\alpha]:=\{ \alpha\beta\alpha \mid \beta\in S \} = \{ \phi\colon U\to V \mid \ker\alpha \leq \ker \phi \quad\text{and}\quad U\phi \leq U\alpha \}. $$ Thus $[\alpha] \mapsto (\ker \alpha, U\alpha)$ yields a well defined bijection between principal inner ideals and pairs $(K,B)$ of subspaces $K\leq U$, $B\leq V$, with $\dim K + \dim B = \dim U$, and $[\alpha_1] \subseteq [\alpha_2]$ iff $\ker\alpha_2 \leq \ker\alpha_1$ and $U\alpha_1 \leq U\alpha_2$. Thus, a maximal chain of principal inner ideals in $[\alpha]$ has length $\dim(U\alpha)$.