I think my conclusion was not totally correct, there's not just one trivializing open set, but all trivializing open sets must have the same cardinality.
does anyone here has a work-sheet or something else on "Stolz–Cesàro theorem", my syllabus doesn't include this but still i want to practice?(i am unable to find on internet)
those are properties of an operation, not something that can be closed (but they are inherited by subsets which are closed under the operation, as can be directly checked)
@MikeMiller Well, sort of, it's kinda complicated. We have 5th grade to 12th grade in here, most of which classifies as high school, but it's partitioned into 5th-10th and 11th-12th. The first is called "secondary education", the second is called "higher secondary education". To get from the first to second one needs to go through an exam, and one has to do like really well there because everyone does.
I've been sick and not doing much for the past couple days. I'm starting to feel better, so I'll probably do some light reference-chasing related to my project and some light unrelated reading.
@tobias I guess a better question that would solve my confusion would be: does there exist non-modular operations that allow for a finite set to be closed under the operation?
right now I'm sitting at a campus pc with my laptop. both the pc and my laptop have mathematica open running the same kind of code for different parameters
Fine; I'll call them sickening. I do not have much respect for history as taught in my school, although I am not rebuking the subject itself if that was the impression you got - that was not my intention.
well, there's two ways for a theorist to get 'experimental data'
one is to rely on actual experimentalists. that takes too long :P
the other is to program stuff and call that your data
i say that a tad sarcastically, but to the extent that one uses numerics both to get simulations and to test analytical approximations that's pretty much true
It's got a chapter on probability & statistics at the very end. That I do not know, and would like to know. But other than that, yeah. Also no ambiguous word problems anymore - which is nice.
i find it a little funny that probability and statistics are always stated together. mostly because i like probability much more than i like statistics
I got a seminar talk tomorrow that is supposed to last for 45 mins, problem is I've got 6 pages of text for the board and at least 2 if not 3 more are gonna come to it D:
i had a talk today that was 1 hour for 6 pages, talk before that i had prepared 16 pages for 2 hours and only did 8, so i guess the speed is variable lol
What do math majors do? Do they all teach? Is there research and if so what is the research for? Is it science related? (sorry just starting out university but computer science, not math)
for some reason I'm not bothered though, I'm going to do the stuff thats cool in detail and if there isn't time for the rest then there isn't time. This is probably exactly the wrong attitude and the listeners might be annoyed...
I like manifolds, which are higher-dimensional versions of curves and surfaces. I would like to know which 3-manifolds embed in 4-dimensional space. I try to invent "obstructions", usually numbers associated to the 3-manifold so that if the number isn't zero, it doesn't embed. Defining and calculating those obstructions is one part of my research.
Hi there! Does anyone know the difference between the kernel of a linear transformation and the basis of the kernel (again, of a linear transformation)?
Thanks guys! I have reduced the matrix to a rref. Next I have identified the variables of the R3=>R3 room. x1 = 2*x2, x2 = independent/arbitrary, x3=0. Thus I can conclude that the basis of the kernel is [2, 1, 0], right?
Even being pretty broad it's a hard sell. The closest thing I can think of is "one can be defined using instanton homology groups", where instanton homology groups are groups that count instantons on Y x R somehow.
i suppose the most obvious stumbling block to a physics example is that 4D there tends to mean 3+1 spacetime, i.e. having metric tensor $dt^2-dx^2-dy^2-dz^2$.
which is definitely 4D but is pretty simple on the face of it
i imagine a string theorist could think of some nice brane example, but um
dont know what rref matrix is, but if you have that $A(x_1,x_2,x_3)=0$ if and only if $x_1=2 x_2$ and $x_3=0$ then $\{ (2, 1, 0) \}$ would a be a basis of the kernel, but ´ $\{ (1, 1/2, 0) \}$ would also be a basis
@Fredefl the kernel is not $(2, 1, 0)$ but the SPAN of $(2,1,0)$, this means all elements of the form $(2x,1x,0)$ where $x$ is a number from your field (probably the real numbers in your case!)
@Semiclassical This is also only about smooth embeddings, not isometric embeddings, metric tensors appear in what I do but artificially so (the resulting object doesn't depend on the input metric)
Yes, real numbers. I do not need to determine the kernel per-se. I need to find the basis of the kernel, which I have done, however, I need to find a vector that is not included in the range of T, which I have found to be [0, 3, -1] and [1, 3, 0]. That would be the null space, which would be the kernel, right? Or am I just completely wrong @s.harp?
@MikeMiller I don't know anything, but I've heard that many geometrical things have some applications in (maybe higher dimensional) fluid dynamics, where solutions appear that encode this geometry (even though it may be unstable). Ie here youtube.com/watch?v=YCA0VIExVhg they do knots as vortices in real fluids
But you should know that if you have a map from R^3 to R^3 so that there are two linearly independent vectors that do not lie in the image (which you are saying I think), then the kernel is at least two dimensional
remember formulas like if A: V to W is linear, dim(ker A)+dim(im A) = dim(V)
ok actually sorry I have said some non-sense, its late, the statment "if you have a map from R^3 to R^3 so that there are two linearly independent vectors that do not lie in the image (which you are saying I think), then the kernel is at least two dimensional" is completely wrong
Right - so let's make i a little more hypothetical. So, I have to find a vector that is not included in the range of a linear transformation. The only vector that is not in the range, would be the null-space, right @s.harp?
Alright - I might not really have a good grasp of this. If I have the basis of the range (image) of T, how do I then go about finding a vector, that is not included in said range? @TobiasKildetoft
Taking the cross product is a trick that works in R^3. Another consideration is that you have a set of three linearly independent vectors, namely the standard basis, and you know that at least one is not an element of the range. So, if {v1,v2} is a basis for your range, and {e1,e2,e3} is your standard basis, one of the equations av1+bv2=ei is inconsistent.
@KarlKronenfeld The cross product of k-1 vectors in R^k still has mostly the same properties as the original and is defiend in more or less the same way
If I have a finite field of cardinal $2048=2^11$, I take an element $x\in K*$ and $x\ne 1$, the question is what is the degree of $x$ in $F_2$? What does it means ? Because $x$ is not necessarily algebraic in $F_2$.
@Fredefl Just look up cross product and find the formula. (This may be over your head, if you are unfamiliar with dot products) It is a way to produce a vector orthogonal to the plane that is the range of your linear map. Saying that v1 is orthogonal to w is the same as saying that the row vector v1 times the column vector w is 0. So, w is orthogonal to v1 and v2 if and only if it is a solution to the equation Mw=0 where M is the 2x3 matrix with rows v1 and v2.
Ah, two vectors that span the range, of course. Imagine the range as a plane, and the two vectors on that plane. The cross product of those two vectors will be orthogonal to the plane.
The assignment with finding the vector that is not in the range of T comes right after finding the base of the image of T - thus I am wondering, can I use that same base of the image to find the vector? @KarlKronenfeld
- even though crossproduct of the span seems to do the trick
A vector orthogonal to the vectors in that basis is orthogonal to the image. So it will be orthogonal to any pair of vectors that make up a basis for the image.
Seems like that might not be the "preferred solution". The context of the assignment is that we have two (ordered) bases for the same subspace, B={u1, u2} and C={v1, v2}. Furthermore, the change of base matrixes of B<-C and C<-B have been calcluated. I am unsure how that fits in, though.
Unfortunately, the only usable description I have found of said matrix type has been in the documents provided by my institution in danish - I'll see what I can find in english.
I guess I will simply use the matrices M and N from the start. Recall that the columns of M are the elements of the ordered basis B, and similarly for N and C. Then, $MP_{C\leftarrow B}=N$.
Now, you have equations representing the vector $\mathbf x$ using the matrices $M$ and $N$.
The ability to swap $M$ and $N$ makes it possible to solve for $a$ and $b$.
Seems like something is wrong here @KarlKronenfeld, wolframalpha.com/input/…*%7B%7B-3,1%7D,%7B7,-2%7D%7D%3D%7B%7B2,1%7D,%7B-1,0%7D,%7B7,3%7D,%7B16,7%7D%7D.
actually when you expand (a + b)^n in a commutative ring, the coefficients aren't really in Z. They are sums in the ring, using that ring's addition, of finitely many copies of 1 in that ring. @KarlKronenfeld
I don't seem to, now. Quite frankly, I do find this assignment to be way above my current understanding of linear algebra - really struggling understanding how to deduce a and b.
Think of a and b as being the coordinates for your vector $\mathbb x$, which may as well be a point in the plane spanned by v1 and v2.
You already have a pair of coordinates for $\mathbb x$, namely 1 and lambda. Unfortunately, it is relative to a different system of coordinates, based on u1 and u2.
@EricStucky The best I can think of is using a permutation on the index set $F$ of the local trivialization. But I know that this shouldn't work for $n\geq 3$ (because it's actually the second part of the exercise, to find a counter-example).
I would believe that for $n\geq 3$ because there's no canonical choice of how to swap the sheets, and so it's hard to 'be consistent'. But for $n=2$ this problem doesn't exist.
@EricStucky So, why would it work on $n=2$? It's the only notion I think of for "swapping sheets". I've been thinking about working with the uniqueness of liftings, but that requires extra assumptions over the total space.
I have fiddled around a lot with it, and I still cannot seem to wrap my head around it @KarlKronenfeld - I am afraid I'll have to give up on it in a moment.
No worries Karl, I much appreciate your help, it was great! If you happen to have some strange, long, all-inclusive equation for it that works - I'll happily take that and try to wrap my head around it tomorrow.
