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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
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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
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up vote 1 down vote accepted

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.

Edit: that is, (letting the other shoe drop), these are change-of-coordinates/basis expressions for four specific things: linear maps of a space to itself, quadratic forms on a space, bilinear maps on two spaces, and maps from one space to another. That is, doing such things to matrices is not mere conformity to some inexplicable set of rules, but, in fact, is doing an obvious thing. Understanding which is appropriate is much aided by understanding what the point is.

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Yes, thanks for writing up these clarifications. But what these "conjugations" actually mean, was more or less clear to me. I just wanted to know if the case under question has a standard name. I'm trying to write a proof where I have to change a lot of times coordinates in V and W in suitable ways, and I wanted to say: at every step I get a matrix which is "..." to the previous, using a standard term. Maybe i should just write "equivalent" (after defining what I mean). –  Hans Aug 18 '12 at 0:58
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@Filippo Oh, sorry, I didn't understand your question properly ... I think it is hopeless to rely on a standard terminology, outside of some context. That was the back-side of my discussion. I fear that one must specify in any given context, due to lack of truly standard ... short... names. Some people/sources will say that there are standard names, but this is not true. "Equivalent" and "similar" are worthless... –  paul garrett Aug 18 '12 at 1:08
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