# Standard terminology for the relation between $A$ and $B$ if $B= Q^t A P$?

Let $A,B$ be two rectangular $m\times n$ matrices related by $$B= Q^t A P$$ with $P$ an $n\times n$ and $Q$ an $m\times m$ matrix.

Is there a standard terminolgy for this relation? If instead of the transposed $Q^t$ one takes the inverse $Q^{-1}$ above, they are just called "equivalent" according to http://en.wikipedia.org/wiki/Matrix_equivalence

I know that when $m=n$ (Edit 1: and P=Q) one uses "congruent" (transposed case) and "similar" (inverse case).

Edit 2: $P$ and $Q$ are assumed both to be invertible (sorry for forgetting to write it). As Marc van Leeuwen pointed out, there is no point in distinguishing among the cases $Q$, $Q^t$ and $Q^{-1}$ since $Q$ is arbitrary ( and invertible). It only makes sense when interpreting $Q$ as coordinate change matrix and $A$ as a linear operator ( -> $Q^{-1}$) or bilinear form (-> $Q^t$) (see explanations of Paul Garrett).

Thanks to everybody who contributed to clarify my confused question.

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The term "similar" is used when $P=Q$. –  M Turgeon Aug 18 '12 at 0:02
@M Turgeon. Yes, thanks, I forgot to write it. –  Hans Aug 18 '12 at 0:05
What is the point of transposing $Q$? If $Q$ is an arbitrary $m\times m$ matrix, then so is $Q^t$. As you state it, the relation means "$B$ has the same shape as $A$ and $\mathrm{rk}\,B\leq \mathrm{rk}\,A$". –  Marc van Leeuwen Aug 18 '12 at 8:22
@Marc van Leeuwen. You are right, I edited the question. –  Hans Aug 18 '12 at 13:47

If both $P$ and $Q$ are invertible, then this is just that $A$ and $B$ have the same rank. That notion of "equivalence" is dubious, as are several archaic bits of terminology about matrices, dating back to times when coordinate-independent understanding of linear algebra did not exist.
There are reasons for considering relations $B=Q^\top A Q$ and $B=Q^{-1} A Q$. The case with inverse is change-of-coordinates/basis for a linear map $A$, where the $Q$ is the change-of-coordinates/basis map. The case with transpose is change-of-coordinates/basis for a _quadratic_form_ $A$, with $Q$ again the change-of-coordinates/basis map.
The cases $B=Q^\top AP$ and $B=Q^{-1} A P$ are change of coordinates of, respectively, a bilinear map $A:V\times W\rightarrow \hbox{scalars}$, and a linear map $T:V\rightarrow W$ given by $A$, where in both cases one changes coordinates in possibly-unrelated fashions on both vectorspaces.