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12:03 AM
Hold on, I think I get it
Well, wait. You say that we need to find a $\delta$ such that $g(y_0 - \delta) > g(y_0) - \epsilon$, but how did you get that in the first place?
 
12:27 AM
i.imgur.com/GtKT3uK.png Isn't this automatic because $f$ is the inverse of $g$?
 
12:39 AM
@user2103480 Sorry, here's what I really meant to say instead of a lucid picture: If $X$ is a top. space with an equivalence relation $\sim$ and $Y\subset X$ such that $Y$ is saturated wrt $\sim$ and $\pi\colon X\rightarrow X/\sim$ is open, then a) the composition $X\setminus Y\rightarrow X\rightarrow X/\sim$ factors through a continuous map $X\setminus Y\rightarrow(X/\sim)\setminus(Y/\sim)$ and b) this map is an identification map, so that $(X\setminus Y)/\sim=(X/\sim)\setminus(Y/\sim)$.
Now $X=S^2$, $Y=$strip about the equator
@politeproofs the first equality cause $f,g$ are inverse, the inequality because $f$ is monotonic
 
@Thorgott this makes more sense thanks
so you're saying what ted says but more complicated
smart
 
facts don't care about my pictures
what I just said is the formalized version of the picture I said above
it's effectively the same as Ted's, but I'm not choosing representatives at any point
 
@Thorgott i know i know
useful theorem for pedants
so indeed useful to me
 
the theorem is ugly, I just made it up
but there is importance in checking that it doesn't matter whether you take quotients before or after doing something else
it usually works, but in pathological cases it doesn't
gotta be able to justify my picture
 
@TedShifrin throwback to the time dana scott suggested spankings for justin bieber
quick get it thor before I delete it
 
12:51 AM
lol
 
the universe works in mysterious ways
 
1:36 AM
"It is now possible (with some effort) to choose..."
you love to see it
 
 
1 hour later…
2:53 AM
Did anyone here work through principles of mathematical analysis?
If you did, how long did it take you, roughly?
 
3:06 AM
I think the record here for any% (no corollaries, exerciseless) is 7:57 here
 
@BigSocks Well, of course you need to do exercises when going through a math textbook
Otherwise it's kind of pointless
Maybe not every single exercise, but at least half
 
@BigSocks I recently discovered a game-changing chapter 6 skip
 
lol
 
3:24 AM
Does anyone have a clever way of writing $(\prod_{1 \le i \le n} f_i)' = f'_1f_2 \dots f_n + f_1f'_2\dots f_n + \dots + f_1f_2\dots f'_n$ in summation notation?
 
$\sum_{i=1}^nf_i^{\prime}\prod_{\substack{j=1\\j\neq i}}^nf_j$
 
Oh cool, thanks
 
ok that typesets horribly, but the point is clear
 
Doesn't typeset all that bad IMO
 
or $\sum_i f_1\cdots f_i'\cdots f_n$
 
3:26 AM
substack doesn't belong in in-line math
I'm basically committing a felony
 
correct
but we knew you categorical people were heathens
To be honest, I would just put $j\ne i$ and forget the $1$ to $n$.
 
so would I usually, but I wanted to be precise this once
 
Heresy.
Thor, are you being like a @Balarka and staying up all nights?
 
well...yes
But I actually am about to go to sleep
 
3:43 AM
Have a nice sleep!
 
Schlaf gut!
 
$$(\prod_{1 \le i \le n} f_i)' = (\prod_{1 \le i \le n} f_i)\sum_{i=1}^n (f_i'/f_i)$$
hmm, that looks bad in that form
$(\prod_{1 \le i \le n} f_i)' = (\prod_{1 \le i \le n} f_i)\sum_{1\leq i \leq n}^n (f_i'/f_i)$
better
 
You can use \left( \right)
 
yeah, i'm just struck by it looking worse with double-dollars than single
 
123
3:58 AM
Hi All....
What is free vector, fixed vector and sliding vector???
Pls explain and example.
Hello @TedShifrin @copper.hat @Thorgott
 
Oh, ugh. Free = sliding, fixed means tail of the vector must be at the origin. For the others, you can put the base anywhere you want. Only physicists talk about this.
When you add $\vec a + \vec b$ you move the tail of $\vec b$ to the head of $\vec a$ and join the tail of $\vec a$ to the head of $\vec b$. We all do this, so we all allow ourselves to slide the vector over to do that.
 
it's not terminology i use either
 
Never ever in my 50 years of teaching have I used it.
When I write a vector in coordinates, of course its tail is at the origin. But I'm allowed to draw vectors $\overrightarrow{AB}$ whenever I want.
I taught out of a book that insisted on using parentheses for a point and brackets for the corresponding vector. Other than conforming to the book's notation when I taught that course, I never made the distinction. I identify a point and the vector from the origin to it.
I realize that I can add a vector to a point and not vice versa, but geez ...
 
123
@TedShifrin But we use force as sliding vector along line of action. Whether in concurrent forces or coplanar. Why
 
In physics, we always draw the force acting at the point, velocity at the point, etc.
I have no idea what you just wrote.
I defer to Semiclassic.
 
4:13 AM
i don't use that terminology, so i have nothing to say about it
 
LOL, smart man.
We remove ourselves from this discussion, 123.
Go bug your teacher.
 
123
Pls see the link I have uploaded complete mechanics book here. See chapter-2
 
No.
 
123
@TedShifrin why you guys angry from my questions. What I am asking wrong here.
 
4:15 AM
Not angry. We both said we don't use that terminology and don't like it.
We are not obliged to answer everyone's questions.
 
123
I am just trying to understand the phenomenon by asking questions expert like you. I don't know what I am asking wrong here.
 
You're not wrong. You're asking people who are not interested. You need to talk to your physics teacher.
I wrote lots of stuff (which you probably did not read) explaining why I do not care about this terminology. I'm done.
 
Consider the annulus. Identify antipodal points on the outer circle. Also, identify antipodal points on the inner circle. Is it Klein bottle?
 
123
@TedShifrin problem is that we have this definition in our text books. But Google does not help me also neither teacher.
 
That is your teacher's job.
@User873110 No, I don't think so.
 
123
4:19 AM
Our teachers are just superficial to every topic, they just follow the formula for computation.
 
Look, @123. I've spent plenty of time helping you. I do not want to discuss this topic, and I'm not going to change my mind. Did you read everything I wrote above? I doubt it.
 
123
@TedShifrin anyway Thank you so much for answering me. And clear me that these are not topic mathematics.
 
I likewise don't discuss points of logic, set theory, etc., that do not interest me. I am not obliged to talk about everyone's questions.
Indeed, I'm not obliged to talk about anything.
@User873110 I have no idea what your background is. Do you know homology, etc.?
 
4:34 AM
@123 i am unable to read the links.
 
123
Hello @copper.hat my English is not very good. Is there any problem with the link. Or you are not interested to read the link.
 
well, i can look at a page or two, but i am not going to cycle through 17 pages looking for something. if you have a specific question i can try to answer. you should realise that notation used by physicists, mes, ees, etc, are not uniform and often not consistent with that used by mathematicians.
 
@TedShifrin Yeah I know this
 
123
@copper.hat thanks for your interest. This is maths mechaincs book. My specific question is following.
 
(also, unfortunately i am not a mathematician.)
 
123
4:44 AM
I am trying to understand vector rules for translation and rotation. Sometimes book used it as free vector in case of concurrent forces. Sometimes they used it as sliding vector move along the line of action.
 
@User873110 @TedShifrin It is the Klein bottle.
 
@123 where are the terms free vector, sliding vector defined?
 
Cut the object along the center circle of the annulus. Then you are left with two copies of Mobius strips, glued along their boundary.
 
Yeah I think so, double of Mobius band
 
Correct.
lol copper.hat
 
4:47 AM
:-) was afraid of flagging
 
123
I presume free vector means can be added as head to tail rule. Sliding vector means effect of force is same along the line of action does not change.
 
@123 that is too vague for me to be able to answer a question about.
 
123
@copper.hat thanks for your interest to help. I can not ask very good question about it. I am confused.
I got confused.
 
@123 sorry, i cannot help unless i understand what you are asking.
 
@BalarkaSen This is my idea: consider the annulus with an intermediate circle, if I use this math.stackexchange.com/questions/1284674/… then due to intermediate circle we have two Mobius band each having the intermediate circle as the boundary, i.e. we are considering the identification of two Mobius band along boundaries
Am I right?
 
4:51 AM
This is completely correct.
 
123
@copper.hat I have read complete chapter solved all the problems. But trying to understand the behavior of forces in translation, rotation and both
@copper.hat book used the sentence "to avoid the reaction at point we take moment" what is meant by that
 
@123 i need more context to be able to answer that.
 
Actually, I made the same comment here math.stackexchange.com/questions/3996896/… and then deleted it.
 
123
@copper.hat I add picture in the link with highlight.
 
@123 what link???
 
4:56 AM
It is not immediately clear how to compute the fundamental group of the Klein bottle, so that would not answer OP's question. What they're doing is one way to compute $\pi_1(K)$
 
123
@copper.hat don't open any chapter, see below page I underline the context
 
Can someone consider this math.stackexchange.com/questions/3996949/… it is interesting to me.
 
123
@copper.hat direct link to the page
 
@123 there is not really enough context for me to answer, but i think they are just saying that the net torque is zero.
 
123
If you want to see complete example. Pls see cahpter-3 example-5, page 58-59
 
5:01 AM
@BalarkaSen I think so, OP wants to use van-Kampen without knowing it is Klein bottle.
 
One can give an axiomatic definition of cup product from knowing cup product on the "universal object" $K(G, n)$. I'm not sure how computationally useful that will be.
 
123
@copper.hat I am afraid you guys are very kind. But you guys got annoy from my questions. That's why I confused to ask.
@copper.hat whenever you have time you can answer. Thank you for your participation.
 
@123 like i wrote, i think it is just a torque balance equation. since this is a statics problem.
 
123
@copper.hat Oookay.. Thanks to clear me that point.
 
@123 The wording in the text is not terribly clear. Unfortunately that is a characteristic of many books of that nature.
 
123
5:09 AM
@copper.hat that's also was not very clear to me. Then asked question to you.
 
@TedShifrin 50 years of teaching... That's 50 times longer than I've spent learning math
 
surely you have one than one year of learning mathematics?
 
Can someone give me an example of a proper homotopy equivalence of plane? A proper homotopy equivalence can be defined analog way as homotopy equivalence, but here one has to replace all maps, including all homotopies, with proper maps. Of course, a homeomorphism is a proper homotopy equivalence, but other than that.
 
@copper.hat Nope! Learnt how to add fractions a little over a year ago, haha
 
5:29 AM
@User873110 Collapse a finite radius disk on R^2 to a point, and map the exterior to R^2 \ 0 in the obvious way
 
I am worried about the continuity of this map on the boundary of the disk
Probably, I don't understand the term "collapse" here
 
Constant map on the whole disk. There's no issue with continuity, if $D$ is the finite radius closed disk this is the quotient map $\Bbb R^2 \to \Bbb R^2/D \cong \Bbb R^2$.
 
6:04 AM
Something like this $\varphi:\Bbb R^2\to \Bbb R^2$ is defined as follows $$\varphi(x,y)=\begin{cases}0& \text{ if }x^2+y^2\leq 1;\\ \big(\sqrt{x^2+y^2}-1\big)(x,y)& \text{ if }x^2+y^2\geq 1.\end{cases}$$
Am I right?
 
sure why not
 
Now, convex combination with identity gives the proper homotopy, i.e. identity is the proper homotopy inverse.
 
thats right
 
Nice. Thanks
 
6:26 AM
@Thorgott @user2103480 @Astyx youtube.com/channel/UCo4au6lRX4-_gIczBneEZWA
 
6:40 AM
wow, that is borderline disturbing :-)
 
 
1 hour later…
7:42 AM
good night folks.
 
123
8:23 AM
Pls see the link. I don't understand the situation of question. Where is axis, cylinder, wall and flat beam location.
If anyone can help.
 
9:02 AM
I've finished proving that my function is an isomorphism, how do you show it's unique?
 
@LHC2012 prove that it is unique at each step
 
Oh ok
@leak
is it generally true that for any isomorphism of ordered fields that has property if x<y, then f(x)<f(y), then the isomorphism is unique?
also I don't see how you show that it's surjective...
nvm that's easy
 
10:10 AM
@copper.hat our generation just calls such things "art"
Not that art wasn't like that before
Spending large parts of ones life in a neverending cabinet of curiosities, aka the internet, leaves its marks though
 
10:28 AM
i wonder, if not knowing anything about those curiosities leaves any marks, i.e. being overly naïve
 
@LHC2012 what you asked makes no sense
an isomorphism can't be "unique" per se
you need to qualify it, i.e. it is the unique isomorphism that satisfies property X
and if you're asking whether any order-preserving automorphism (i.e. isomorphism with itself) of ordered fields must be the identity map (i.e. sending every element to itself), then this paper provides a negative answer
 
11:05 AM
Could someone clarify the inequality of step of this, please?

$\displaystyle \left|\sin z\right|=\left|\frac{e^{zi}-e^{-zi}}{2i}\right|=\frac12\left|\frac{e^{2zi}-1}{e^{zi}}\right|=\frac12\left|\frac{1-e^{2xi}e^{-2y}}{e^{xi}e^{-y}}\right|\ge\frac12\frac{1-e^{-2y}}{e^{-y}}=\frac12\frac{e^{2y}-1}{e^y}$
$\;z=x+iy\;,\;x,y,\in\Bbb R\;$
 
 
1 hour later…
12:13 PM
@BalarkaSen gold mine
 
12:32 PM
@BalarkaSen lmao, I discovered that channel a couple weeks ago, it's so good
 
 
1 hour later…
1:45 PM
So, given a comm. ring $A$ and multiplicative set $S = A \ P$, $P$ a prime ideal, they denote $A_P := S^{-1} A$ the localization of $A$ at $P$, and then say that $S^{-1}P := PA_P$. Should I believe that $PS^{-1}A = S^{-1}P$?
 
you mean $S=A\setminus P$?
 
I guess things on the right are in (and only in) the set $\{ p/s \vert p \in P $ and $ s \in S \}$ and things on the left are sums of products of things in $P$ and fractions in $A_p$.
and yes, mb
 
Maybe someone here is interested in POMDPs. There's a bounty on this question.
1
Q: How do I learn the value function for a POMDP with a single-step horizon (bandit)?

jdizzleConsider a single-step POMDP, where an agent receives a random belief state $b(s)$ over some fixed set of contexts $s$, selects a single action, and receives a single reward, before the episode ends. I understand that the belief state value function, $V(b)$, is piecewise linear and convex, with a...

 
Hello.
 
both sides are just $S^{-1}A\otimes_AP$, no?
 
2:03 PM
gotta freshen up that tensor product knowledge
 
I mean, what you said above is completely correct and answers the question affirmatively as well
tensor product is just the abstract nonsense viewpoint
 
yeah but your description is cool too
I happen to like abstract nonsense
let's say I can see how the LHS is probably that. how do i see the RHS is also that
 
$S^{-1}M\cong S^{-1}A\otimes_AM$ for any $A$-module $M$
naturally
 
is that because of the universal property of the tensor product (of $A$-modules)?
 
you can write down inverse maps both ways by the universal properties like a normal human being, or you can note that we have $\operatorname{Hom}(S^{-1}A\otimes_AM,-)\cong\operatorname{Hom}(M,\operatorname{Hom}(S^{-1}A,-))\cong\operatorname{Hom}(M,-)\cong\operatorname{Hom}(S^{-1}M,-)$ on the category of $A$-modules on which $S$ operates bijectively and then apply the Yoneda lemma
 
2:17 PM
you mean the tensor-hom adjunction?
 
ye that's the first iso
 
yeah what is the second one
 
universal property of localization
 
oof, gunna have to look that one up. and the last one?
 
also universal property of localization
the point is that if $S$ operates bijectively on $N$, homomorphisms $S^{-1}A\rightarrow N$ are uniquely determined by where they send $A$ and each such possibility is realized
 
2:23 PM
Gunna have to stare at the sun while I think of this one, but thanks again man
 
think about the explicitly isos first
abstract nonsense is only to be enjoyed after one has done it explicitly
 
ok ok fair. so for $S^{-1} M \cong S^{-1} A \bigotimes_A M$... the LHS has fractions, but the right has equivalence classes of pairs of (fraction with numerator in $A$, element in $M$)
 
2:40 PM
@Alessandro Are you there
 
Why are amenable groups sofic?
My favorite definition of soficity is there are approximate homomorphisms to $S_n$ separating points
Oh maybe you get a quasi-action on the set of points in a Folner set
 
Hm not sure
what do you mean with approximate homs?
 
$\varphi_n : G \to S_n$, Hamming distance of $\varphi_n(xy)$ and $\varphi_n(x)\varphi_n(y)$ goes to zero as $n \to \infty$
Separation of points is also asymptotic, $\varphi_n(g)$ is bounded away from the identity in Hamming distance
 
soficity, more like.. sophistry
gottem
 
2:49 PM
I bet if $\{K_n\}$ is a Folner sequence, $\varphi_n : G \to S_{K_n}$ given by $g \mapsto (x \mapsto gx)$ whenever $x, gx \in K_n$ and extending in whatever way outside are the approximate homs
 
idk @Thorgott like could you send $(m,s) \to ((1,s),m)$ and then going backwards could you do $((a,s),n) \to (m,s)$ where $an =m$ which you could find since $M$ is an $A$-module?
idk I think I have soggy brain, might need to think about something else briefly
 
yeah, that should be it
really like the only sensible choice, isn't it
 
completely floored at how I wrote something that wasn't actually just crap. but yeah, there can only be one reasonable answer since this is a natural thing was my guiding thought
 
3:19 PM
Ooh I never thought anyone would discuss soficity in this chat! What's next? Hyperlinear groups?
Note that Connes' Embedding Conjecture was resolved by quantum complexity theorists earlier this year. This is why operator algebras is the most amazing branch of maths---because it connects with so many other branches of math.
 
@user193319 What are those? $(U(n), d_n)$-approximable, where $d_n$ is one of Schatten/Frobenius/etc norms?
 
3:39 PM
So a curve is nonsingular when its partial derivatives are not zero at all points in $Z_f(\overline{k})$. How would someone reasonable check this? $Z_f(\overline{k})$ could be pretty big. I am guessing you do not try to show $\{(a,b) \vert f(a,b) = 0, \partial f / \partial x (a,b) = 0 , \partial f / \partial y (a,b) = 0 \} $ is empty
 
4:00 PM
I think it usually happens to be the case that the partials of $f$ don't vanish at most points and it only remains to check that the critical ones aren't zeroes
(which is more convenient than trying to first calculate all the zeroes and then check at each of them)
but I dunno
 
it's alright. sounds like a "easy most of the time but hard in general" situation
 
I guess
but I'm not an algebraic geometer, so what do I know
 
also lets say we got $f(x,y) = x^2 y - y^2 -2x + y$ and so we define a rational function $\alpha := \frac{x}{y-1} = \frac{y - x^2}{x-2}$. Book says you got a pole at $(2,1)$ "since $(y-1)$ vanishes and $x$ is inverible at $(2,1)$"
So yeah, I guess it makes sense that you get a pole since neither of the 2 versions of the rational function $\alpha$ are defined at $(2,1)$, but their rationale seems so odd. I guess my question is "how do we know $x$ is invertible at $(2,1)$ and how do we know there's not some other rational function equivalent to this one that is defined at $(2,1)$?"
@Thorgott coulda fooled me
I see that he refers to some points $(0,1)$ and $(2,4)$ as points where $\alpha$ is defined, and I noticed they are different from $(2,1)$ in that one or the other version of $\alpha$ is defined, meanwhile the other or the one version of $\alpha$ shows up as "0/0". When you put in $(2,1)$ you just get $2/0$, which is different.
maybe that's the crucial difference
I guess "invertible" is just "you get a constant that isn't zero on the numerator"...
 
4:45 PM
Ok so I know this is not closely related but this is the only remotely related place I could find. Well I want to ask a question, it is about fuzzy logic, anyone here familiar with fuzzy theory, I wrote a basic fuzzy program in Java but I am getting unexpected results. It only has 2 input variables and 1 output variable.
If it's suitable to ask it, just tell me, I don't want to ask an off-topic question so I am asking to ask :D
 
Hey all, I asked a question really late in the morning without thinking about how it’ll definitely be like 3 pages down before the morning...I don’t want to repost the question, but can anybody help with this diff eq problem?
0
Q: Classifying a Bifurcation?

COsborneFor the system of differential equations $$\dot{x}=y-x, \dot{y} = \mu x - y$$ with Jacobian matrix $$J = \begin{bmatrix}-1 & 1 \\ \mu & -1\end{bmatrix},$$ is a stable spiral for $\mu < 0$, a stable node for $0 < \mu < 1$, and a saddle for $1 < \mu$. However, how would the bifurcations between the...

 
What exactly is "the area element" in a surface integral? Until now I thaught the dx after an integral is just a notation that indicates that we are integrating "with respect to x". Now we had surface integrals in calc III and I seem conpletely lost. We have to do some calculations with those dσ named surface elements.
.. and I have no Idea what that means. I can write what exactly is the "definition" we had and what are the tasks when I'm at home, but maybe someone can give a general answer, I hoped
 
5:10 PM
@T_01 Forget surfaces for a moment. You covered line integrals already, right?
 
 
1 hour later…
6:20 PM
yes, now that you say that, we had them in calc 2
 
6:33 PM
@BalarkaSen The norm is the Hilbert-Schmidt norm on $M_n(\Bbb{C})$, which is $||x|| = \tau(x^*x)^{1/2}$, and $\tau$ is the normalized trace on $M_n(\Bbb{C})$. Also, a group $G$ hyperlinear if and only if it can be embedded into the unitary group of the hyperfinite $II_1$ factor if and only if its group von Neumann algebra $L(G)$ can be embedded into the hyperfinite $II_1$ factor.
 
7:09 PM
Hello!! Let $V,W$ be $\mathbb{R}$-vector spaces and let $\phi:V\rightarrow W$ be a linear map. Let $1\leq k\in \mathbb{N}$ and let $v_1, \ldots , v_k\in V$.

If $V=\text{Lin}(v_1, \ldots , v_k)$ then does it follow that $W=\text{Lin}(\phi (v_1), \ldots , \phi (v_k))$ ?

We have that $\phi (V)\subseteq W$ and since $\phi$ is linear we get that $\text{Lin}(\phi (v_1), \ldots , \phi (v_k))\subseteq W$, right?

So we get $\subseteq$ instead of $=$, or not?
 
I am assuming Lin is the linear subspace spanned by those vectors?
 
@anakhro Yes!
 
You are right with the subset, but try a few examples. Like try R --> R^2 by sending x to (x,0).
 
So we have that $v=1$ and $\mathbb{R}=\text{Lin}(1)$. Then $\phi (1)=(1,0)$ but $\mathbb{R}^2=\text{Lin}((1,0),(0,1))$,or not? @anakhro
 
@MaryStar Right idea. Ae you familiar with the dimension of a vector space?
 
7:21 PM
@anakhro In the general case : The dimension of $V=\text{Lin}(v_1, \ldots , v_k)$ is $k$ (since all vectors are linear independent), then the dimension of $\text{Lin}(\phi (v_1), \ldots , \phi (v_k))$ is also$k$. But the dimesion of $W$ can be greater, as in the example you gave. Is that correct?
 
Yes. :D
But one minor error.
What if phi is not injective, is the dimension of the Lin still k after applying phi?
Say we go from R^2 --> R by projecting to the first coordinate.
 
@anakhro But isn't every linear map also injective?
 
Well what about the example I just gave?
Is it linear, and is it injective?
 
@anakhro You mean $\phi:\mathbb{R}^2\rightarrow \mathbb{R}$ with $\phi ( (u,v) )=u$ ?
 
Hello, can someone help me with a problem related to linear transformation?
 
7:30 PM
@MaryStar Yes.
@Physicsfreak Just go ahead and ask and someone will help if they can
 
Yeah ok
 
7:46 PM
$\phi ( (u_1,v_1)+(u_2,v_2) )=\phi ( (u_1+u_2,v_1+v_2))=u_1+u_2=\phi ( (u_1,v_1) )+\phi ( (u_2,v_2) )$ and $\phi (a(u,v))=\phi ((au,av))=au=a\phi (u,v)$ so $\phi$ is linear.

But it is not injective, is it?
 
So basically we have 2 position vectors R1 = x1î+y1j and R2= x2î+y2j. If matrixA=( x1 y1)
( x2 y2)
Then we need to prove that det(A) is invariant under rotation of an angle(theta) about x-axis.

My teacher says simply state that A'=RA and then apply determinant on both the sides, but that would have been possible only if the components of R1 and R2 were along the column of matrix A i.e if A were.
(x1 x1)
(y1 y2)
Is my teacher right in saying so?
 
So you're writing rows to mean columns?
You are correct that the matrix should multiply the column vectors to get the new column vectors. And your teacher is correct, too, because the determinant of the transpose of a matrix is the same as the determinant of the matrix. But it's more natural to use columns as you want to.
The matrix equation the teacher stated only works if it's with the transposes.
I'm guessing the teacher was writing row vectors and you're writing column vectors.
 
Yes so instead she should have given column vectors right?
For the result A'=RA to hold true?
 
Yes. But perhaps her notation $(x1 y1)$ is really the first row of the matrix. That's what I would assume. You seem to be making rows into columns.
For example, when you input a matrix in a computer algebra system, typically {{1,2},{3,4}} gives the matrix with those rows.
 
@TedShifrin yes that's how it was given in the question. But the results A'=RA would hold only when the components of R vectors were along the column right?. I understand that in this case since we take determinant it hardly matters but saying A'=RA won't be correct right ?
 
7:55 PM
If you interpret what's written as rows, it is correct. "How it was given in the question" is too vague for me.
Where is the notation defined?
 
In the question matrix A was given as the following
(x1 y1)
(x2 y2)
Basically components of vectors along the row of the matrix
 
I'm telling you that to the rest of the world, that is indicating rows.
In that case, notice what the column vectors are. They're what you want.
So unless you type that your teacher/textbook defined that notation to mean columns, she is right and you are wrong.
 
Column matrix is
x1
y1
 
@anakhro But what does it mean that we have a counterexample? I got stuck right now.
 
Oh, wait. Now I wasn't paying attention because you switched from what you said originally.
Compare what you just typed and what you typed originally. Anyhow, you know the correct thing to do, so I'm going to quit now.
 
8:03 PM
i no longer quit when i'm ahead. now i quit when i am a butt.
 
My only problem is that for the result A'=AR to hold, she should have given A as
(x1 x2)
(y1 y2)
Rather than
(x1 y1)
(x2 y2)
Am I right @TedShifrin
 
If this is the convention that I am saying with row vectors typed, you're right. If she's typing the column vectors as rows (which is what I thought you wanted to do), she's right. I've asked you three times where the notation is actually defined.
 
@MaryStar Well what I was saying is that the projection map here doesn't preserve the linear independence of the vectors.
So the dimension before and after are different---it goes down.
 
@TedShifrin Oh ok so in the matrix A the elements are Aij where i is row j is colum and what I am saying is that the elements should have been A11= x1, A12= x2, A21=y1 and A22=y2
 
Yes, I agreed with that ages ago. What we have not established is what the parentheses your teacher/text are using mean. If they mean what I usually use, then you are correct. I am not going to discuss this any further.
 
8:12 PM
@TedShifrin yes the parentheses mean what is generally used as a convention. Thanks for your time :)
 
@anakhro What if $W=\text{Lin}(\phi (v_1), \ldots , \phi (v_k))$ tehn does it follow that $V=\text{Lin}(v_1, \ldots , v_k)$ ?
 
@MaryStar Have you tried some examples to test your idea?
 
@anakhro If $\phi (v_1), \ldots , \phi (v_k)$ are linearly independent then $v_1, \ldots , v_k$ are alsolinearly independent, right?
 
Just try a few examples. When you are testing something, also try to break it. Try to figure out why it works, or why it doesn't.
 
I have shown that $v_1, \ldots , v_k$ are linearly independent iff $\phi (v_1), \ldots , \phi (v_k)$ are linearly independent @anakhro
 
8:32 PM
Hello, everyone. This is a question I asked on the main site, but it never got any answers. We know that the real ordered field can be characterized up to isomorphism as a complete ordered field. However, that is a second order characterization. A natural question is, is the first order theory of the real ordered field axiomatized by the axioms for ordered fields and an axiom schema stating that every nonempty parameter-free definable set which is bounded above has a least upper bound?
Here, this is the question I asked on the main site: math.stackexchange.com/questions/3827456/…
 
8:53 PM
oh hey, you're the transfinite derivative person
did you ever catch my final remark in that conversation?
 
Yes, I saw.
The current question is yet another one that I asked on the main site, but did not get an answer to.
 
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