## Transformations of Bilinear Forms

Sorry for the delay, but at least now everything is in Maryland. Now I just need a job and then I’ll have to move back *out* of my parents’ house again

Anyway, now we want to see how linear maps between vector spaces affect forms on those spaces. We’ve seen a hint when we talked about the category of inner product spaces: if we have a bilinear form on a space and a linear map , then we can “pull back” the form along the map. That is, we take two vectors in , apply to both of them, and then stick them into the form on .

When we write this out with our Dirac notation, we don’t really make much of a distinction between the vector spaces, relying on context to tell which vectors belong to which spaces, as well as which pairing we’re using. Luckily, when we write something sensible, it usually doesn’t matter how we parse things. Dirac notation is very robust! So, given vectors and in we have the form given by

as we’ve seen over and over. But now let’s take two vectors and in . First, we hit each of these vectors with . The ket vector clearly becomes , while the bra vector becomes . Now using the form from before we get

This looks like a form on that’s described by the linear transformation . Just as a sanity check, we can verify that sends to , then sends to itself, and the adjoint sends back to . So this is indeed a transformation from to itself that can be used to represent a form.

As a particular case, we might consider an automorphism and consider it as a change of basis. When we did this to a linear map we got a new linear map by conjugation , and we say that these two transformations are similar. In one sense, this is “the same” linear map, but described with different “coordinates” on the vector space. But when we apply it to a bilinear form we turn into , and we say that these two are “congruent”. In the same sense as before, they describe “the same” form on , but using different coordinates on the vector space.

Like similarity, congruence gives an action of on the space of bilinear forms on . Indeed, if and are both automorphisms, we can act on first by to get , then then by to get . But this is the same as , which is the action by the product .

“To say that two transformations are “the same” may be up to congruence. But in infinite dimensional space, don’t we have other options, related to homotopy? Two mathematical objects are said to be homotopic if one can be continuously deformed into the other.

As a 2nd generation professional science fiction author and editor, I have a love-hate relationship (self-adjoint?) with “fandom.” The best fans are wonderful friends. The worst, well, I classify them as thinking: “I hate the world because it won’t let me be God Emperor of the entire Galaxy, and meanwhile I have to live in my mother’s basement.”

Comment by Jonathan Vos Post | July 25, 2009 |

Not sure what the “fandom” thing has to do with anything.

Yes, there are all sorts of notions of equivalence, but the one I’m focusing on here is the idea of when the same transformation looks like two different transformations because of a choice of basis. That’s our clearest non-canonical choice around, and the one that is most important in linear algebra to pay attention to the side-effects of.

Comment by John Armstrong | July 25, 2009 |

“Not sure what the ‘fandom’ thing has to do with anything.” — context for digruntlement of your “I’ll have to move back out of my parents’ house again :/”

Sorry, I was trying to cheer you up by showing that geeks and nerds feel each others’ pain. Ance once in a while a Mathematican/Science Fiction Author reference shows up on (for instance) n-Category Cafe about Greg Egan, or Vernor Vinge, or Rudy Rucker.

Comment by Jonathan Vos Post | July 25, 2009 |

So not only have I failed to get an academic job and been turned away by the quasi-academic government job that was going smoothly up until a week ago, but now despite my disdain for that whole image I’m put into the same category as inert, egomaniacal, emotionally-stunted escapists who cannot deal with life on their own?

Gee,

thanksJon…Comment by John Armstrong | July 25, 2009 |

Quite a number of those goofy teenaged fans became fairly successful. Comic-book-crazed Ray Bradbury used to bicycle around Hollywood getting stars to give him autographs. He had to operate a coin-operated public typewriter in the basement of the UCLA library to write “Fahrenheit 451.” This weekend he spoke to a standing-room only crowd (a subset of of the 125,000+ fans) at San Diego Comic-Con. There are many more examples, of head-in-clouds fans becoming best-selling immortal authors. So, although you don’t like the lower bounds, the fact that there is NO known upper bound should give you some cheer. Your good times are yet to come. Your next book, expanded from this blog, may not be “Fahrenheit 451” but at least your PC doesn’t require you to feed in coins.

Comment by Jonathan Vos Post | July 26, 2009 |

[…] Given a form on a vector space represented by the transformation and a linear map , we’ve seen how to transform by the action of . That is, the space of all bilinear forms is a vector space […]

Pingback by Orthogonal transformations « The Unapologetic Mathematician | July 27, 2009 |

[…] the one basis to the other, and which thus takes the one form to the other. That is, the forms are congruent. So as long as we have some inner product, any inner product, to talk about lengths and angles, and […]

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