calculation of bilinearform

[bhelmes]

A peaceful evening for all,

is there a possibility to calculate a quotient of a bilenearform in another clever way ?

Example: let be A = Einheitsmatrix
(1 0) = A
(0 1)

f(n)=(n, 1) A (n, 1) where n is element of N


f(n)
----- = ?
f(m)

example:
f(7) 50
---- = ---- = 10
f(2) 5

Is there a possiblity to make a reduction in the bilinearform ?

Greetings from the primes
Bernhard

[ewmayer]

What does 'A (n, 1)' mean? You started by defining A as the 2x2 identity matrix, so you don't get to parametrize it in terms of some integer variable n afterward.

And 'f(n)=(n, 1) A (n, 1)' - is that supposed to represent a length-2 row-vector left-multiplying a 2x2 matrix? If so, the result is another row vector, not a scalar function as you have on the LHS.

[Dubslow]

He almost certainly a row vector multiplied by A (taken to be any of a variety of forms, such as inner products or metrics or norms or positive definite non-degenerate symmetric bilinear forms) multiplied by the column vector corresponding to the row vector (the dual).

(See e.g. https://en.wikipedia.org/wiki/Inner_product_space or https://en.wikipedia.org/wiki/Metric_tensor#Definition or https://en.wikipedia.org/wiki/Norm_(mathematics))

Such is a very common construction in many areas of math and physics, as you surely know. Obviously in this case A=I is the standard Euclidean metric, and (n, 1) a vector in R^2. f(n) is the norm (or norm squared depending on the given terminology) of said vector. Given the lack of easily produced ways to express row vectors vs column vectors, it's a reasonable shorthand to write it as such. (I would probably have written it differently, but I believe it's easily understood as is.)

On the face of it, I don't see division of bilinear form outputs to have a particular shortcut. The actual numbers in the matrices and vectors of course depends heavily on the chosen coordinate basis, but the result of the operation should be coordinate independent.

[science_man_88]

Originally Posted by Dubslow

He almost certainly a row vector multiplied by A (taken to be any of a variety of forms, such as inner products or metrics or norms or positive definite non-degenerate symmetric bilinear forms) multiplied by the column vector corresponding to the row vector (the dual).

(See e.g. https://en.wikipedia.org/wiki/Inner_product_space or https://en.wikipedia.org/wiki/Metric_tensor#Definition or https://en.wikipedia.org/wiki/Norm_(mathematics))

Such is a very common construction in many areas of math and physics, as you surely know. Obviously in this case A=I is the standard Euclidean metric, and (n, 1) a vector in R^2. f(n) is the norm (or norm squared depending on the given terminology) of said vector. Given the lack of easily produced ways to express row vectors vs column vectors, it's a reasonable shorthand to write it as such. (I would probably have written it differently, but I believe it's easily understood as is.)

On the face of it, I don't see division of bilinear form outputs to have a particular shortcut. The actual numbers in the matrices and vectors of course depends heavily on the chosen coordinate basis, but the result of the operation should be coordinate independent.

in PARI/GP that would be the shorthand for the matrix transform ~ so [1,2] is a row vector and [1,2]~ is the column vector equivalent.

[ewmayer]

Originally Posted by Dubslow

He almost certainly a row vector multiplied by A (taken to be any of a variety of forms, such as inner products or metrics or norms or positive definite non-degenerate symmetric bilinear forms) multiplied by the column vector corresponding to the row vector (the dual).

Ah, I did not consider that the rightmost (n 1) might be a column vector - some simple notation to that effect like (n 1)^T would've helped. Since A is the identity, why even include it? Just write f(n) := (n 1).(n 1) = (n^2+1), a scalar. I.e. there is no hifalutin 'quotient of bilinear forms' involved at all, just the quotient of two scalars.

[Dubslow]

Originally Posted by ewmayer

Since A is the identity, why even include it?

The metric tensor article is a good place to see examples. Like I said here, A=I is the standard Euclidean metric, but if that was the only possible metric we wouldn't have bothered naming a whole class of things as "metrics" or such as the long winded name above. In many cases, especially in differential geometry (think general relatively and the bending of spacetime) or certain linear algebra applications (Hilbert spaces, found in e.g. quantum mechanics) non-Euclidean/non-identity metrics are quite necessary to do anything useful.

And as according to the OP's title, this is about "bilinear forms". That means something which takes two inputs and is linear in both; in the specific instantiation of bilinear forms on vector spaces, all such bilinear forms can be written as a matrix multiplied on either side by the input vectors, as suitably if crudely depicted in the OP, and there's a lot of interesting forms [be they actual "metrics" or not] that are more than the identity matrix. OP merely used I to ask his general question. (Of course metric tensors and the other articles linked often use other notation [angle brackets, bra-ket notation, all sorts of stuff], even for vector spaces, but the crude depiction in the OP is accurate if one allows for the presumptions I made.)

The question as posed, at least about reflexive inputs (each input is the same), gives the ratio of the magnitude of two different vectors (one vector as both inputs to the form, the other vector as both inputs to the form, the results being their respective norm). As I said, the ratio of the norms of two vectors is independent of the coordinate basis chosen (meaning the exact numbers in the vectors/matrices can differ but "mean the same thing"), so I don't think there's any computational shortcut to be had here.

To support my presumptions, this is my context basis for it: as I've tried to emphasize, this concept of a bilinear form taking two inputs, where the inputs are "dual" to each other in any of a variety of senses (conjugates, transposed, dual-vector-functionals) is a very common underlying theme pervading a lot of linear algebra/differential geometry/functional analysis, hence my immediate presumption that the two vectors in the OP are meant to be dual (that is, in the euclidean-vectors-over-reals case, transposed).

In fact a lot of "convolutions" or "transformations" can be viewed as instantiations of a certain bilinear operator which integrates its two inputs multiplied over R, with a specific choice of the second input giving rise to a specific convolution. If we call this bilinear form "A", then for instance the Fourier transform of a function f is merely <F|A|f> (in bra-ket notation), where F = e^(-2 pi i x t) is the standard multiplicand (Fourier kernel?); and the (bilateral) Laplace transformation is <L|A|f> where L = e ^ (-s t) is the "Laplace kernel" (or multiply by the heaviside step to get the more common unilateral Laplace transform). And in a slightly different sense, these transformations can be viewed as maps from one vector space (of suitable functions) onto a sort of dual space, where the result of the transformation is really nothing more than a dual-vector of the original space (analogous to row-column transposition in the Euclidean-vector case).

[ewmayer]

Well, I choose to remain grumpy based on the ill-posedness of the original question, coupled with the crappy notation of the example. Don't try to cheer me up with your happy DiffGeom chatter, Bill Winslow! :)

Irrespective of the metric, a ratio of two norms is still a ratio of scalars. Why make it harder than that?

[Were I a certain long-departed poster I would gratuitously add 'gibberish!'. Luckily for the OP I'm merely a grump, rather than a misanthrope.]

[Dubslow]

Originally Posted by ewmayer

Irrespective of the metric, a ratio of two norms is still a ratio of scalars. Why make it harder than that?

Indeed, I am of the same mind here.

Originally Posted by ewmayer

[Were I a certain long-departed poster I would gratuitously add 'gibberish!'. Luckily for the OP I'm merely a grump, rather than a misanthrope.]

Although I wouldn't go so far as to call it gibberish, he would certainly be very grumpy (and worse) about this. I don't see much in the way of substance.