Good morning everyone, I know that any square matrix $A\in\mathbb{R}^{n\times n}$ can be written as the sum of symmetric and antisymmetric parts $$ A = \frac{A + A^T}{2} + \frac{A - A^T}{2} $$ I'm wondering whether we have similar results or similar notion for non-square matrices?
maybe? you'd need some generalization of what 'symmetric' should mean in that setting.
if S is any vector space and f: S to S is any map whatsoever, then for any x in S you have x = (x + f(x))/2 + (x - f(x))/2. so that alone isn't so interesting. if f is a linear map from S to itself that satisfies f(f(t)) = t for all t, then the decomposition just given has the form x = a + b where f(a) = a and f(b) = -b.
and this decomposition is kind of unique. if a+b = a'+b' where f(a)=a and f(a')=a' and f(b)=-b and f(b')=-b' for this linear f, then applying f to both sides of a+b = a'+b' you deduce a-b = a'-b', and adding these last two equations together you deduce 2a = 2a' and hence a = a' and b = b'.
so corresponding to any a linear "involution" [= map satisfying f(f(t)) = t] on a vector space, there is a decomposition of every element of that space into a sum of something fixed by the involution, and something that changes sign under the involution.
if you do this for the involution sending the function g(t) to the function g(-t) on the vector space of functions from R to R, you get the decomposition of such a function into a sum of an 'even' function and an 'odd' function in the usual sense
i can't think of any interesting involutions on the set of non square matrices, but maybe you can
Suppose that I have the following set R={(x,y): x,y are in R^n-{0}, y,x are linearly dependent}. I note that if we form a matrix [x y] (n+1 by 2 matrix) with columns x and y, then this matrix has rank at most 1.
So every 2 by 2 minor of the matrix must vanish.
Can I conclude from this information that R is finite?
if you're maybe talking about the set of ranks of elements of R, or reduced row echelon forms or something, you could maybe squeeze finiteness out of this, but not otherwise.
yes, it's a zero set of polynomials, so it is closed, but these are polynomials in multiple variables, so there's no reason for the zero sets to be empty
(with my f.lux on, without looking at names, sometimes I get confused between Leslie and Thorgott)
@Thorgott Let's fix a $t\ne 0$. Let's look at all (x,y) such that y=tx. We want to find such (x,y). So we solve the equations given by minors n(n-1)/2 of them.
I agree that this set is closed.
But then we have to union over all such t's.
and union(infinite) of closed sets need not be closed.
If you fix two rows i and j you have a function $f_{ij}(x,y)$ assigning to the given two vectors the corresponding minor. Every $f_{ij}$ is continuous and you are looking at the intersection $\bigcap f_{ij}^{-1}(0)$.
Does the construction of the universal cover stated in Hatcher have a name? I mean 'the construction is the usual construction of something more general stuff'.
Is it true that every positive integer $N$ has at most one representation $$N=F_m+F_n$$ with positive integers $m,n$ with $1<m<n$ , where $F_n$ denotes the $n$ th Fibonacci-number ? If yes, how can it be proven ?
Confusion in 'attaching of spaces via a map': Suppose that X,Y are given spaces. Let A be a closed subspace of X, f: A-->Y be a map. Then define ~ on $X\sqcup Y$ as follows: x~y iff one of the following happens: 1) x=y, 2) x,y both are in A and f(x)=f(y), 3) x is in A, y=f(x).
How do I show that ~ is an equivalence relation?
In particular, why is ~ symmetric?
Suppose that u~v. Suppose we have that u is in A, v=f(u). From here, how do we get u=v or f(u)=f(v) or u=f(v)?
there is another terminology for (1),(2),(3) combined, which is equivalence relation generated by the relations a~f(a) for every a in A.
But I don't understand this terminology yet so I'm working with the alternative (equivalent?) definition.
But the problem is in showing ~ to be an equivalence relation.
If I define ~ using partitions of the disjoint set then I get why ~ would be an equivalence relation. But using this equivalent characterisation does not give me the equivalence relation.
I’m assuming that $A$ is in fact a subset of $X$, not $Y$.
Let’s simplify matters a little by assuming that $X$ and $Y$ are disjoint, so that we don’t have to form $X\sqcup Y$; I’ll return to the general case later. We have disjoint spaces $X$ and $Y$, a subset $A$ of $X$, and a continuous $f:A...
I have posted one comment there requesting clarification.
@Thorgott: I'm trying to understand definition 13.13 in Bredon's AT book.
Suppose that we have u~v and it is known that u is in A and that v=f(u), $u\ne v$. If v~u were true, then this should give either f(u)=f(v) or 'v is in A and u=f(v)'.
Thanks. I read someone on MSE say that when self-studying, you should have an open problem in mind and try applying the techniques to it/learn techniques that would be beneficial for that problem. There is also the David Hilbert quote "He who seeks for methods without having a definite problem in mind seeks for the most part in vain" That is meant for people who are already research mathematicians though, right?
I mean, unless you are trying to be super f'ing pedantic. I guess, technically, $X \sqcup \{y\}$ is a space which consists of ordered pairs $(x,0)$ (with $x \in X$) and $(y,1)$ (with $y\in \{y\}$). Thus $(X\sqcup \{y\}) \setminus (A \sqcup \{y\})$ is the set $\{ (x,0) : x \in X \setminus A\}$.
My adviser, who was one of the pre-eminent mathematicians of the 20th century, always said that one should be working on a problem and then go learn what's needed for that.
But I don't think that anyone ever really cares about that kind of annoying detail. Since $\{ (x,0) : x \in X\setminus A\}$ is "the same as" $X\setminus A$ for all practical purposes.
@TedShifrin Does that apply for a current undergraduate studying ahead though? In that case, you should just try to cover the topics in your courses ahead, not try to learn techniques to solve open problems, right?
You are correct. By definition of the adjunction space, if $Y$ is a one-point space then $X \sqcup_f Y = X/A$ . In this case, the resulting space is a sphere, $D^2/S^1 = S^2$.
Indeed, let $p$ denote a point in $S^2$. By the stereographic projection, there is a homeomorphism $\operatorname{int}...
@TedShifrin Well, you will probably get more out of the lectures since you already know the material. Since you get familiar with the type of problems, you will know how to solve them faster or tackle harder ones.
@ILike What's the point if you "already know the material"? I don't get it. I think it makes more sense to go further in depth with things you've already studied than to mislearn and/or be bored in the next course.
As a teacher, I wanted to take my students on a voyage of discovery, and students who already "knew" it and showed off their knowledge were not much fun.
@Koro This is a philosophical issue, I guess. What does it mean to say $X=Y$? Literally identical.
The point would be grades. "I think it makes more sense to go further in depth" So you would recommend picking an open problem in that area, like your advisor said, and learn what's needed for it?
He was talking about graduate students and more advanced. But my opinion is that even the students who have gotten A's from me in undergraduate courses (including challenging courses) had plenty left to learn to master the course, and there was plenty of room left for exploration. Go back and work challenge problems from the course; really understand things you only got 75% during the course.
My view after 40 years of teaching is that students generally did less well in courses than I would have liked because they really didn't know the prerequisite courses. In differential geometry, they really didn't know multivariable calculus and linear algebra sufficiently well. In algebra, they really didn't know basic proof skills and linear algebra well enough.
@Koro Where $f$ maps $A$ to the one point, I presume?
If they had covered the course beforehand though, studying ahead, worked through the problems beforehand, they should've seen what they were missing though and could've learned that before taking the course. Doing this takes a lot of time though, yes
Honestly, most of the courses I taught students needed me (or someone) to teach them. A student who could learn them on his own belonged in graduate school.
@ILike Have you gotten past standard calculus-type courses to real mathematics?
I'm saying that most of us use $=$ to be interpreted in the "category" in which one is working. I write $\Bbb R^k\times\Bbb R^\ell = \Bbb R^{k+\ell}$ without hesitation. But it is not a literal equality.
I don't, @Xander, unless I'm teaching into to higher math, in which case I follow the textbook. I use $\subset$ for $\subseteq$ and write $\subsetneq$ if I want to mean that.
I'm currently in high school, we are doing basic calculus and will be doing some linear algebra (if you could call it that, it's only vectors, some operations with them, ...) and probability soon. For now, I'm mostly self-studying ahead for next year and undergraduate. So no.
though I admit that earlier this semester I annoyed my professor by asking whether $\lim\mathrm{GL}(n)$ is formed by filling in $1$s top left or bottom right
Yes, @ILike. And certainly in high school calculus and supposed "linear algebra" there is plenty to master beyond the level of the course. So many problems and sections not covered.
Well, there are some people who've watched my videos who complain that the course moves too slowly. That said, it moves at double or triple the speed of most US courses and with far greater depth of material. Even my best students, who have gone on on to do PhDs, were plenty challenged.
I would say that for students like @Thor in his youth, my course was probably too slow/easy.
let me explain the under $X$ part perhaps: there's a canonical map $X\rightarrow X/A$ and a canonical map $X\rightarrow Y\cup_fX$. the homeomorphism $X/A\cong Y\cup_fX$ commutes with these maps.
you often care about these maps as much as you care about the quotient/adjunction space, so this can matter
e.g. you might wanna know when $X\rightarrow X/A$ is a homotopy equivalence and this is the case iff $X\rightarrow Y\cup_fX$ is a homotopy equivalence, for this reason
When I was in college and graduate school, I never finished a course thinking I had mastered it ... even if I got an A. Only studying for qualifying exams and then teaching did I start to really master courses.
I got less than an A several times. I mentioned in here earlier that one B was in the material of the multivariable math book; I think I'm relatively good at that stuff :P
When I see a transcript with nothing but As, my immediate assumption is that this person never stretched themselves. They never took a class outside of their interests, and never took a "hard" class.
@ペガサスSeiya When I am on hiring committees, I much prefer a transcript that has a C- in World History (but otherwise As and A-s) over a transcript with nothing but As.
@TedShifrin Oh, I get that. But in much of the rest of the world, folk are never given a classical liberal education. I want to hire people with classical liberal educations. :D
With respect to my current position, I am at a community college. A lot of the students we teach are severely disadvantaged, and I want to know that anyone we hire has something in their background which is going to help them empathize.
I will note that, contrary to many of the CCs in Arizona, we require two semesters of writing for any degree, including things like "welding" and "cosmetology".
And the broader community here loves us for it---our graduates are often complemented for being able to write and communicate well.
@TedShifrin Are you referring to your Multivariable Calculus and Linear Algebra course? I'm doing self-study on it right now. Thanks for putting it up! Just finished Day 1.
@GratefulDisciple Yeah, I was referring to that. Still, without the course/book one does not have the exercises, and to me that is the major part of the course (together with feedback thereon).
By the way, if you do get the book, you can find my homework assignments hidden away on the web (unless they're finally gone).
OK, cool. The only comment I'll make is that the assignments are mostly theoretical/proof stuff, because all the computational problems were done on WebWork (and graded automatically)!
@TedShifrin I see. Thankfully, there are answers to the selected exercises in your book. I'm taking this for math background for machine learning (I'm in Computer Science).
There are plenty in the book, @user726941. I mostly modeled the WebWork problems (which literally took me hundreds of hours to code and troubleshoot) on problems in the book. There are a few exceptions. I still have the WeBWork problem code, but you would need access to WeBWork to enable it.
"then why take the course?!" I'm not familiar with how the course-system is yet, but don't you need to take some courses to receive a degree? Or is there a way to take some test that proves that you know the material?
@Koro That is definitely not equal. I would say that $I^4 = I^2\times I^2$, but I would write $D^4\cong D^2\times D^2$ :D
@ILikeMathematics My point is that very few students know the material as well as they think. We had a course challenge system in place at UGA (not at all levels). Almost every challenge that happened while I was associate department head (8 years) was a failure. I think out of many tries, one student got credit for the course that he thought he knew so well.
I think people would usually write $D^4=D^2\times D^2$ in the context of smooth manifolds (where it is understood that products of manifolds with boundaries are automatically smoothed!), but not in other contexts
I studied Quotient topology today from various sources including Loring Tu's. I feel like I understood many things. I can now create/think about examples of quotient maps which are not open/closed :-). There was also an exercise on 'sphere with a hair' is not a manifold. I understood it :-).
I even stopped giving credit for linear algebra courses taken at community colleges, because they tended to be completely computational and without proofs.
I have a question. In computer science, we have an algorithm that works for a some subset (lets say A) and it has been proven that will take this subset as input and returns an output.
Now I created a new algorithm that takes a larger subset (lets say B such that A is a subset of B) and it works. But how to prove compatiblity with the previous algorithm?
I think I need to say the new algorithm is well-formed and extends the previous algorithm.
What is this process called in Math and is there an example for it?
@TedShifrin It's great you still have the WeBWorK files. If you like, I would be more than happy to try to host a WeBWorK system, since it appears that WeBWorK is an open source project. That way, the students of your YouTube lecture recordings can have their homework graded automatically.
All different issues. Plenty of ADHD. Some with dyslexia. Several with autism and Aspergers. They tended to be a bit disruptive.
@GratefulDisciple Interesting. I wonder if the various updates in WeBWork over the past decade will cause compilation issues with my code. The problem is that a course needs to be set up with an instructor, who then gets bombarded with questions from students working the homework. I'm not willing to be that generous :D
@TedShifrin And after you've worked on the challenge problems and feel like you understood everything from the course, would you then study ahead the courses you will take next?
I’m assuming that $A$ is in fact a subset of $X$, not $Y$.
Let’s simplify matters a little by assuming that $X$ and $Y$ are disjoint, so that we don’t have to form $X\sqcup Y$; I’ll return to the general case later. We have disjoint spaces $X$ and $Y$, a subset $A$ of $X$, and a continuous $f:A...
You are correct. By definition of the adjunction space, if $Y$ is a one-point space then $X \sqcup_f Y = X/A$ . In this case, the resulting space is a sphere, $D^2/S^1 = S^2$.
Indeed, let $p$ denote a point in $S^2$. By the stereographic projection, there is a homeomorphism $\operatorname{int}...
