@Huy I managed to reduce the problem to the following: show that for a $n \times n$ matrix $A$ and a $n \times n$ invertible matrix $C$, $r(C^{-1}AC) = r(A)$.
@Huy If I can figure out how to prove this, I'm done
@Clarinetist Thanks for notifying about the free Springer books, btw. There are over 10,000 free books available at Springer website. Amazing. I wonder whether this is temporary or not...
To show this, I start by showing their column spaces are equal
This is obvious by definition of $P_X$, since $P_X = X(X^{T}X)^{-}X^{T}$, $C(P_X) \subset C(X)$, where $C$ is the column space
Furthermore, $X = P_X X$, so $C(X) \subset C(P_X)$, hence equality of column spaces and therefore ranks
By a theorem I proved months ago that I haven't touched, since $P_X$ is symmetric, it has a singular value decomposition of $p \times p$ matrices (assuming $X$ is $n \times p$), given by $P_X = P \cdot D(\lambda_i) \cdot P^{T}$, where $D(\lambda_i)$ happens to be the $p \times p$ diagonal matrix of the eigenvalues of $P_X$.
Anyway, due to $P_X$ being idempotent, for a eigenvector $\mathbf{x}$ of $P_{X}$, $$P_X \mathbf{x} = P_X^2 \mathbf{x} = P_X(P_X\mathbf{x}) = P_X(\lambda\mathbf{x}) = \lambda(P_X\mathbf{x}) = \lambda^2 \mathbf{x}$$
Obviously, $P_X\mathbf{x} = \lambda\mathbf{x}$, so $\lambda = \lambda^2$, which means that $\lambda$ is either $0$ or $1$.
@Clarinetist: I get what you're doing, but is there any reason you're doing it this way instead of following the answer you received in your question? just out of curiosity.
So we are justified in saying that $$r(D(\lambda_i)) = r(P^{-1}P_XP) = \sum_{1}^{p}\lambda_i$$ since the number of linearly independent columns is the number of columns that do not have all $0$s (since $D(\lambda_i)$ is diagonal) and $\lambda_i$ can only be either $1$ or $0$ @Huy
@Huy I quite frankly don't understand the approach in the answer very well, but I think I'm starting to realize the approach I'm doing and the answer I have are very much related. I just didn't know how to connect the pieces given what I know/remember
your current approach needs much deeper results (spectral theorem), the other one was basically just applying definitions, but if it doesn't matter to you, use whichever you like :P
actually, it's almost exactly the same argument, except that you assume you know the spectral theorem already, whereas the answer doesn't use it but just immediately diagonalizes (because it's easy in this particular case)
@Huy So now, what we need to do is prove that $r(P_X)$ is equal to the number of nonzero diagonal elements of $D(\lambda_i)$, equivalent to the sum of the $\lambda_i$. However, to get here, I have to show that $r(P_X) = r(P^{-1}P_XP)$, which is then equal to the sum of the $\lambda_i$ (as shown previously), and then everything is all fine and dandy
@Huy The book I have here says that this rank equality is a consequence of the following theorem: if $A$ is a $n \times n$ nonsingular matrix, then for any $n \times p$ matrix $B$, $R(AB) = R(B)$ and $r(AB) = r(B)$ ($R$ being the row space). Similarly, if $B$ is a $n \times n$ nonsingular matrix, then for any $m \times n$ matrix $A$, $C(AB) = C(A)$ and $r(AB) = r(A)$. I don't see how to use this
I don't really work that way, unfortunately. I can already tell you that one of my professors made an error in his lecture notes that no one has probably caught yet (which I'll bring up when I take his class if he hasn't changed the notes). I'm teaching myself the next class' material before taking it
@Huy I've definitely heard of the Spectral Theorem, but I didn't know that was the Spectral Theorem
@Huy Given that what I'm doing here is going to be pretty much what I'm going to be doing in my stats program, do you have a text recommendation? Closest thing I've found is a statistically-oriented linear algebra text
It's been very difficult for me to find a text which substantially covers SVD, projection matrices, and generalized inverses other than maybe a page or so each of explanations
@Clarinetist: there are several versions of the spectral theorem, most of them are very similar though. the most basic one is that for any symmetric matrix, there exists an orthonormal basis for the whole vector space consisting of eigenvectors of the matrix.
@Clarinetist: unfortunately, in my first year all lectures were still in German, so I only know good German books / lecture notes
ask Ted when he's around, I think he even wrote an intro level book on linear algebra
"a geometric approach of linear algebra" or so maybe :P
Linear Algebra: A Geometric Approach - by Theodore Shifrin bingo
@Clarinetist: the current prof teaching linear algebra is writing his lecture notes in English, so maybe those will be good after he's finished. but that'll probably be too late for you :P
@Clarinetist: https://people.math.ethz.ch/~kowalski/script-la.pdf this is its current process, will probably get updated again around February (when the next term starts)
@Huy As far as I know, a generalized inverse of $A$ is any matrix $G$ satisfying $$AGA = A$$ I didn't know the term "pseudo-inverse" existed until I came on MSE asking questions. I've taken them to mean the same thing, but that's (quite likely) wrong
I found now that it is related to the Weingarten map. A principal vector $t$ satisfies the relation $W(t)=\kappa t$, where $\kappa$ is the principal curvature. @robjohn
@anon Here's a possible rigorous defn of ambiguity group. $X$ be an ambiguous figure. Let $Q_i$ be subsets of $X$ such that $\bigcup Q_i = X$ and each $Q_i$ are projection of possible and unambiguous objects $A_i$ in $\Bbb R^3$ to a certain hyperplane. Then you can align the $A_i$'s appropriately and project to realize $X$ as projection of a possibly disconnected but unambiguous/possible set in $\Bbb R^3$ with connected components being $A_i$.
The ambiguity group of $X$ is the collection of tuples $(f_1, \cdots, f_n)$ of isometries $f_i : \Bbb R^3 \to \Bbb R^3$ such that $\bigcup f_i(A_i)$ projected to that hyperplane gives the figure $X$ as before. Group operation is simply slotwise composition of isometries.
Pretty sure this works.
(e.g., take the tribar and $Q_1, Q_2, Q_3$ as in the paper)
I would just multiply it all out by hand. There's no reason to use a calculator for that expression, and the calculator is likely to have trouble with the 10^{-18} anyway.
I have a curve representing the error margin between an approximation and a function. How do I find the average error? I think I need to take the integral of the domain of the function and then divide by the range of the domain?
I used absolute value on the error margin so it would always be above 0
We have that $$\sigma (u,v)=\gamma (u)+v\delta (u)$$ and $$K=\frac{-(\dot\delta \cdot \textbf{N})^2}{EG-F^2}$$ I want to show that if $\gamma$ is a curve on a
surface $S$ and $\delta$ is the unit normal of $S$, then $K = 0$ if and only if $\gamma$ is a line of curvature of $S$.
$$$$
Doesn't t...
@robjohn I finalized an amazing generalization of a problem that I sent to a magazine and it was published. Trying to explore the generalization I got although I cannot say at the moment I have some brilliant ideas to get something more amazing.
@AndrewThompson: I'd say a lot (but very hard to guess how many, probably 50%) of MSc students have a TA position which pays a little bit, but not even remotely enough to live from. Apart from that, 90-95% of the students don't work and just focus on uni. Some probably get a scholarship and the others get money from their parents to live. As you probably already know, tuition costs are really low, so just the living costs are high.
Personally, I decided to move out of my parents' house last year and I didn't want any help financially, so I work as a TA and also as a teacher at high school.…
I have seen most everything he's done - but not twin peaks yet. Me and my buddy are slowly watching it. By slowly I mean we've watched 3 episodes in a quarter.
Everything's fantastic but I love Blue Velvet and eraserhead and Mulholland Dr (probably in that order).
Never got into Twin Peaks, but I thought it was pretty great... in a weird, surreal way.
Not sure I remember that one, but I remember the highway scene with this song [Chris Isaak - Wicked Game (Official Instrumental) ](youtube.com/watch?v=KCPJCCLengk)
Oh, right, it's definitely interesting. I feel the same way, just not sure I find it that interesting and fun, to imagine I would ever pursue a masters degree in it hah.
especially if I wasn't decided on pursuing a field like this professionaly, for many years.
@Jake1234: Well I'm already teaching at high school and I like doing that. And over here, to get a permanent job as a high school teacher, one must have a MSc in the subject they want to teach.
I think it really pays off. I'm always surprised when I talk to other maths teachers about maths how much they still know about it even though many of them didn't continue studying or restudying.
at least 3 of the maths teachers at my current high school even have a PhD. unfortunately, they did their thesis about stuff I have no idea of. :P
I think having a MSc in the subject one wants to teach guarantees that the teacher at least really understands the stuff they need to teach and also a bit more. on average, they obviously (hopefully) understand a lot more.
where are you at right now? still in high school? @Jake1234
@Jake1234: you see, over here the school system works a bit differently than in many countries: every kid goes to elementary school for roughly 6 years (age 7 to 12). after that, they have to choose between something called "secondary school" and something called "high school". the former has no entry requirements and takes 3 years. the latter has an entry test (which more than 50% fail) and takes 6 years.
everyone who has finished high school can go to any university in Switzerland. people who have finished secondary school usually do some apprenticeship and then work.
Oh right, that's pretty similar to how it is here. Thought elementary is 7 to 14, then high school with some focus/gymnasium (4 years) and "secondary school" (4 or 3 years).
MSc in teaching is still enough to teach at high schools here though.
I used to focus on theoretical physics, I took classes in functional analysis and diffgeo to prepare for quantum mechanics and general relativity which I also took. I was going to take QFT this year but it really didn't fit in my timetable so I decided to just learn some more maths. so now I'm just learning a bit about Lie groups/algebras and geometric topology.
The prof I have just covered so much in 1 semester... that I'm not sure I can handle this sort of stuff. He also keeps mentioning things that aren't really explained any simple books.
I suppose the approach of this prof is really.. well, set theoretic. Lots of things like ultrafilters, uniform spaces.. they're not mentioned in a lot of the beginner books.
Yeah, same here, taking 3rd semester of real analysis at the moment.
There will be probability + stats, some differential geometry, there were 2 semesters of programming, I think the rest is usually free to choose, with respect to what you will focus on.
We have that $$\sigma (u,v)=\gamma (u)+v\delta (u)$$ and $$K=\frac{-(\dot\delta \cdot \textbf{N})^2}{EG-F^2}$$ I want to show that if $\gamma$ is a curve on a
surface $S$ and $\delta$ is the unit normal of $S$, then $K = 0$ if and only if $\gamma$ is a line of curvature of $S$.
$$$$
Doesn't t...
