# Talk:Bilinear map

WikiProject Mathematics (Rated Start-class, Mid-importance)
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Field: Algebra

## Untitled

Whoever put this in:

"In other words, if we hold the first entry the bilinear operator fixed, while letting the second entry vary, the result is a linear operator, and similarly if we hold the second entry fixed."

thankyou!

How about another example for a matrix ${\displaystyle B\in M_{n}(F)}$ as a bi-linear form: ${\displaystyle B:F^{n}xF^{n}->F}$ defined by ${\displaystyle v^{T}Bw}$, ${\displaystyle v,w\in F^{n}}$

## map vs operator

I think the term "operator" should be reserved for maps of a space into itself (or by extension ExE→E), while for the general case (in particular bilin.forms) the term "map" should be preferred. MFH: Talk 17:46, 27 May 2005 (UTC)

Agree. At any rate, we should keep our naming consistent; witness multilinear map vs. bilinear operator. -- Fropuff 18:03, 2005 May 27 (UTC)
Seems reasonable. I was wondering about the same topic some time ago. This should be put at Wikipedia:WikiProject Mathematics/Conventions.MathMartin 11:03, 29 May 2005 (UTC)
I've partially fixed this, but haven't gone so far as to move the page: I would, however, support such a move. Geometry guy 16:35, 13 May 2007 (UTC)
Yes, I agree. Someone with operator privs should perform the move. --MarSch 17:47, 13 May 2007 (UTC)

## Pairing

How is this article related to pairing? It seems there is an overlap. Nageh (talk) 16:22, 26 May 2010 (UTC)

I think the question should be: Shouldn't this article be extended to cover [[module (mathematics}|module]]s? Meaning, we get R-bilinear maps, which should probably be covered in this article, and then pairing (which is evidently just an R-bilinear map) should be a redirect here. K-bilinear maps (K being a field, so this is what is currently covered by this article) are merely special cases of R-bilinear maps, which is the subject of Pairing. — Quondum 16:31, 9 September 2012 (UTC)

## Inverse of bilinear map

For the case, VxV->F, I guess the inverse of the bilinear map should be well defined because the map is commutative, if ${\displaystyle f(a,b)=c}$, then ${\displaystyle f^{-1}(c,a)=b}$ and ${\displaystyle f^{-1}(c,b)=a}$. Is this right? and is it meaningful? is it well defined? Jackzhp (talk) 05:54, 8 May 2015 (UTC)

The appropriate place to ask this would be at WP:RD/MA. —Quondum 13:02, 8 May 2015 (UTC)