This is a linear algebra problem, @anakhro. Given a subspace $W \subset V$, the normal space of $W$ to $V$ is $V/W$. There is no canonical embedding $V/W \to V$ if you don't have an inner product on $V$. That's all.
ya the problem is I don't know anything about Riemannian geometry. Is there like an isotopy between any two embeddings then that arises from a connection between the metrics, orrrr....
For the linear algebra question, they are related by an invertible linear transformation of $V$, by the change of basis matrix the inner products are related by
@manooooh Any function is a relation. Any relation has an inverse relation.
The inverse relation of a function is a function by definition when a function is bijective.
The conditions of "surjective" and "injective" are basically the conditions "total" and "single valued" for the inverse relation.
Keep in mind the finely corrected statement that "any function (seen as a relation) has an inverse relation".
Saying "any function has an inverse" is ambiguous.
And 9 times out of 10, people will tell you that you are wrong as it colloquially would sound like you are saying any function has an inverse function.
@anakhro but I am not saying that, the people need to read carefully, I said "any function has inverse, this inverse is always a relation" which is the same as "any function has an inverse relation"
Hello! Numerics question from non-mathematician : let Ax = y be over-determined real finite system of linear equations. The number of linear independent equations is of the same order as variables, each about 10^5. Now, what happens to me is I can find many "equally-good" approximate solutions. I don't know is this typical behavior or specific to my problem. For an example, for different initial guesses solver gives me different solutions,
its like the least-square solutions are kinda degenerated. This gives me the opportunity to setup additional constrains to pick up solutions that have properties that I like and still be almost "the best" least-square approximate solution. I LIKE this but it bothers me that I get free stuff by imposing additional constrains on already over-determined system.
I understand that question is about something that in nowhere near of mathematical rigor, but shows up consistantly. Anyone care to comment?
So, this is what I thought that over-determined. incosistant system is - such M lines that don't intersect and my solver is looking for a point that is as close as possible in a least square sense
Hello! Recently I've posted a question in Physics community, but I'm not sure whether the theme is proper for it or for Math community. I won't duplicate the question in Math, guess, it's against the rules. Can I leave the link on my question here?
I can't understand some steps in obtaining the collision term in the Boltzmann equation for plasma. For the first time it was made by L.D. Landau in his article "The kinetic equation in the case of Coulomb interaction" (Zh. Eksper. i Teoret. Fiz. , 7 : 2 (1937) pp. 203–209)*.
I also can't find p...
An extreme situation is the matrix A with 100 rows, 1 variable, each row being the equation 0 = 1.
I don't know how you really are approximating such a system, but like I mean, making it overdetermined doesn't change any information than leaving it as not over-determined.
Whether it was $\begin{bmatrix}0&1\\\vdots&\vdots\\0&1\end{bmatrix}$ or $\begin{bmatrix}0&1\end{bmatrix}$, your information is the same in both cases, but over-determined only in the first.
So if your approximation depends on some other information, and you throw more information in there, then that's why you are getting something different.
In the non-approximation setting, inconsistency sort of makes the problem pointless/trivial (since you will always have 0 solutions for an inconsistent system, no matter how much information you give).
Yeah definitely :) I used a few chapers later on in the book for my disseration because it describes lots of what is going on in Kummer's $\Bbb Z[\zeta_p]$ work
rather than just using post-Dedekind ring theoretic language I guess