« first day (4064 days earlier)      last day (33 days later) » 

1:06 AM
@BalarkaSen neat. constraints are definitely a thing for physicists, and i know that dealing with them can get tricky
though i mostly know they become tricky when it comes to quantization, e.g., en.wikipedia.org/wiki/First_class_constraint
 
1:29 AM
is there some analogue of reduced l.e.s. for cohomology groups?
 
1:42 AM
@robjohn @robjohn we can then take triangle POT and QOS and prove that the Line TS and PQ is parallel
And then say Angle RTS = RPQ and angle RST = RQP
Since it forms transversal
And prove the triangles are similar by AAA similarity criteria
@TedShifrin Your play with Angles gave a giant leap towards the solution for the problem :D
 
2:27 AM
@Ishwaran definitely false. Look at robjohn’s picture!
 
2:41 AM
@Ishwaran I had mentioned earlier that the line TS can't be parallel to PQ unless the base angles of original triangle are equal. It is best to draw a random triangle instead of a symmetric figure and then you can see why TS can't be parallel to PQ.
Also you should note that you are supposed to prove similarity of RPQ and RST. While proving similarity and congruence the order of letters is very important and that tell which vertex of one triangle corresponds to which vertex of another triangle. Here P corresponds to S and Q to T
So the angle at P matches angle at S. For parallel lines you should have angle at P equal to angle at T
 
here is a new picture, with $O$ included.
 
Hello @robjohn!
This time you have drawn straight lines instead of wiggly lines :)
 
Selling out!
 
That's right, I sold out.
 
2:58 AM
Also liked your proof which uses similarity of two right triangles with a common acute angle at R @robjohn. It appears to be rather very simple and direct.
 
Thanks. As Ted said earlier, the diagrams with isosceles triangles are just awful and lead to easy errors.
@ParamanandSingh are you sure?
 
Cool I like this one. The ones with straight lines are, what to say, too textbook-ish
 
 
1 hour later…
4:19 AM
Oh yeah
Error in the similarity
@robjohn looks legit
 
4:37 AM
In hatcher, for manifold $M$, there is a orientable covering space $M_R\to M$. Is this space have a name?
 
i think it's just called the double cover, or something like that. orientation cover?
this is not my field
oh, for a topologist
 
A geometer will do. The orientation double cover.
 
cool. i'm giving my answer partial credit.
but writing in red ink $\color{#C00}{\text{TRY HARDER}}$.
 
4:54 AM
Extra credit for principal $\Bbb Z_2$ bundle :)
 
Is it some kind of bundle?
 
5:06 AM
@ParamanandSingh Can you explain your proof little briefly ?
 
dinner at the kensington pub tonight.
 
i hope they boiled the food enough, to your liking
 
Potatoes and boiled cabbage?
 
i had a burger, suitable cooked for my taste :-)
outdoors, beside the circle :-)
 
that must've been great. i love that area.
 
5:14 AM
nice to have dinner w friends. especially since i am alone on the house for the week
hope the break is recovering peacefully
 
she's very fussy. she wants to do everything, but can't. i think she's healing. we haven't given her any pain relief medication in a week, and no complaints.
 
that is progress. it is tough all around, esp when they are small.
 
i talked to my dad on the phone today and she wanted to get on. she spent about 10 minutes telling him about her toys, and her leg, and the adventures of her imaginary friends.
 
I’m sure that was a fun conversation for grandpa.
 
he did enjoy it. they don't talk very often, so each time there is a lot of vocabulary development that has taken place since last time. as a narrative it was pretty hard to follow. she doesn't fully appreciate how people on the phone aren't in the room, so she changes subjects or emphasizes things by pointing at stuff.
 
5:22 AM
my granddad has little interest, my grandmother seemed to find every detail fascinating.
 
but she did respond to his questions. it wasn't a monologue. she doesn't ask questions yet.
 
it is good for both of them to interact.
the kensington air knocked me out :-), good night folks!
 
goodnight.
 
nothing to with the alcohol, of course
 
i was going to say.
 
5:25 AM
:-) gn
 
Night!
 
6:09 AM
@Ishwaran: since $P, Q, S, T $ lie on a circle the angles subtended by $PT$ on both $Q, S$ are equal. Hence angle $PQT$ is equal to angle $PST$
Subtract these angles from 90 degree to get angle $TPQ$ equal to $RST$
 
Thank you @ParamanandSingh
 
 
1 hour later…
7:34 AM
I was making a math expression parser
it was not working for negative numbers so what I did before parsing was to replace each minus sign of the negative numbers in the string with ~ (advice from SO)
now I dont want to publish it to github
I think I should modify the algorithm, rather than doing this weird hack
 
 
2 hours later…
9:20 AM
@robjohn I think that would work if $du + i (pdx + dy) = gdz$ held globally in $\Omega$, but this only holds locally in $\Omega$, for not necessarily the same holomorphic $g$ at different points, unless I misunderstand your answer
 
whats the name of the rule we use here? I know it has got to do something with leibniz and that's all
 
9:33 AM
fix some $x_0 \in \Omega$, then $f(z) = \int_{\gamma_{z_0 \rightarrow z}} du + i(^{\ast} du)$ (mod $2 \pi i$), and in a small disk around $x_0$, we may write $f(z) = f(x_0) + \int_{\gamma_{x_0 \rightarrow z}} du + i(^{\ast} du)$ (mod $2 \pi i$), which is equal to $f(x_0) + g(z) - g(x_0)$ (mod $2 \pi i$), then exponentiating we get something of the form $C exp (g(z) - g(x_0))$ valid in a disk around $x_0$ @robjohn im guessing this is what you were saying?
 
9:46 AM
uh actually sorry, it should technically be equal to $f(x_0) + G(z) - G(x_0)$ (mod $2 \pi i$), where $G' = g$ in a disk around $x_0$
so after exponentiating we get $C exp(G(z) - G(x_0))$ valid in a disk around $x_0$
 
10:20 AM
0
Q: How to solve the problem on allegation by the assumption?

gateprepQuestion: " The average marks of the students in four sections A,B,C,D together is 60 %. The average marks of the students of A,B,C,D individually are 45 %,50 %,72% and 80 % respectively. If the average marks of the students of the sections A and B together is 48 % and that of the students of B n...

Somebody please explain the answer to the question in a more theoritical context.
The explanation seems to tough to comprehend.
My question is absolutely same.
 
 
1 hour later…
11:23 AM
@porridgemathematics you have that $\mathrm{d}f=g\,\mathrm{d}z$ locally. Then use monodromy to get what I said.
@porridgemathematics right
 
 
1 hour later…
12:29 PM
need help on 2nd sub division
 
12:52 PM
@Ishwaran are you in 11th or 10th?
 
10th
 
This is a really good question, that i never saw while I was in 10th
 
@AdilMohammed Actually this question is not from Examination point of view
 
Makes sense why i never saw it lol
 
Its something like a Brain teaser from the chapter trig
but it teases my brain a lot :P
 
12:56 PM
The trick here is to notice there are 2 different angles, and the 3 formulas you learnt in grade 10 (sin2x+cos2x=1 and the rest) can only be used if they have the same angles
 
@AdilMohammed yeah thats where Iam stuck
 
So that rules out multiplying numerator and denominator by some fancy functions and then using the 3 rules , so that leaves us only with one choice ahead- convert Tanx into sinx/cosx and proceeding
 
why not trying it by moving the whole denominator to RHS
 
Oh you cant do that, if you did you would be saying LHS is already equal to RHS.... we are supposed to prove they LHS=RHS but we cant make the assumption they are already equal (at least in these sort of questions)
 
then.. any other way ?
 
1:05 PM
if you continued along that path you would be getting 1=1, if i am right... which technically means the LHS=RHS (I think) math teacher don't like if you do like
 
of course... no maths teacher would
 
@Ishwaran try turning Tan theta into sin theta/cos theta and take the LCM while adding the fractions
 
so we must find an expression to multiply it
@AdilMohammed Let me try it
 
@Ishwaran Oh no, i forgot to mention but the idea behind finding an expression to multiply the fraction is so that you can use the 3 formulas
(basically if you have a linear thing, you need to make it into a square thing before you use any of the 3 formulas, this is where a^2-b^2 formulas and all come into play)
but as we have already concluded... we cant use the 3 formulas... so there's no point finding the expression to multiply it
 
@AdilMohammed then I will go on with this
 
1:17 PM
yup did you get it?
 
working on....
 
1:30 PM
@AdilMohammed Still ending up in 1 = 1
 
Yahhi did you take the denominator to the other side? Or anything to the other side, for that matter? the correct method is turning LHS into RHS and then telling "Tada, look I turned LHS into something that looks like the RHS, therefore the equation is true"
Ofcourse others also prefer "Hence proved" to conclude
 
Alternatively, $\tan x=\frac {\sin x}{\cos x}$ and $\cot x=\frac {\cos x}{\sin x}$ and then using $\cos (a+b)=\cos a \cos b-\sin a \sin b$ gives the result directly.
If U and V are finite dimensional vector spaces $T\in L(U,V), S\in L(V,W)$ then dim null ST$\le $ dim null S +dim null T. How do I prove this?
 
@Koro i thought about using that formula, but in his grade its not yet introduced so i decided against using it, but I think you can skip it if I am right (i just did an eyeballing of the solution, i am not sure)
 
@AdilMohammed I think by class 10 (in India), the formulae for sin (a+b) and cos (a+b) are introduced. Right?
 
Nope, not as far as i know
maybe in coaching foundation programs, but not in general
 
1:40 PM
@Koro nope
 
Well ,by 2012 I had those in my syllabus.
 
@Ishwaran wait ill write and show you the answer, but later... you shouldn't be getting 1=1 when doing proofs
@Koro whaaaaaaaaaaaaaat
 
@AdilMohammed I proved phi = theta
 
ok, the syllabus has probably changed now.
 
hence I guess I proved the sum
 
1:41 PM
anyways again that sum formula is not needed at all.
 
@Ishwaran god lord, how did you get that
 
Did you simplify cot and tan in terms of sin and cos and do the cancellation?
 
yes exactly, i may not have been great at communications but that's what you do
 
@AdilMohammed Wait a min I share my work
hope my hand writing is ok enough to understand
 
oh exactly what i suspected (i used to do this a lot too)
 
1:55 PM
:D
 
Do LHS separately, do not make it "interfered" with the RHS
Also dont outright write LHS=RHS like you have done in the beginning
 
yup I need to rewrite it neatly
 
@Ishwaran If x=x then it does not imply that x=1 so your last line in image needs work.
 
@Koro hmmm in trig functions too ?
coz for each value of theta the ratio is fixed ryt
 
Without shifting terms from LHS to RHS. Just write LHS in terms of sin and cos, and you'll see cancellation and eventually you'll get RHS.
 
1:59 PM
@Koro Yes.....
 
@Ishwaran No. Note that $\sin \frac \pi 4=\sin \frac \pi 4$, but each side (being the same) equals $\frac 1{\sqrt 2}\ne 1$
$\cot \theta +\tan \phi=\frac {1}{\sin \theta \cos \phi} (\cos \theta \cos \phi+\sin \theta \sin \phi)$ so $$\frac{\cot\theta +\tan\phi}{\cot \phi+\tan \theta}=\frac{\sin \phi \cos \theta}{\sin \theta \cos \phi}\Big(\frac{\cos \theta \cos \phi+\sin \theta \sin \phi}{\cos \theta \cos \phi+\sin \theta \sin \phi}\Big)=\frac{\sin \phi} {\cos \phi}\frac{\cos \theta}{ \sin\theta}$$
 
@robjohn thanks :)
 
@Koro I dont know why I am seeing evrything with syntax
 
@Ishwaran do you see mathematical expressions in chat?
If not, then you can install mathjax bookmark to see maths here in chat.
 
how to install it ?
 
2:04 PM
If you're on laptop, see the top right section and get the bookmark from LaTeX in chat link.
 
A "shortcut" to circumvent that is simply copying the equation and pasting it on the "Ask a question" section, or give answer section to get the latex "translated"
 
@TedShifrin Let $(\Sigma, \mathcal{A}, \mu)$ be a probability space. Let $(A_n)_n$ be a sequence in $\mathcal{A}$. Let $f: \Sigma \rightarrow \mathbb{B}$ be measurable such that $\int_{\Sigma} |\mathbb{1}_{A_n} - f| d\mu \rightarrow 0$ as $n \rightarrow \infty$. Prove there exists $A \in \mathcal{A}$ for which $f = \mathbb{1}_A$ a.s.

The claim is easy if $A_n \searrow A$, or $A_n \nearrow A$. But if, say, $A_{2n} = [-2, 1]$, and $A_{2n+1} = [-1, 2]$, then I don't even see a set $A$ that satisfies the requirement $\mathbb{P}(A_n \triangle A) < \epsilon$ for any $n$.
sorry, $f: \Sigma \rightarrow \mathbb{R}$.
in fact, the claim follows if $(A_n)_n$ converges, not even necessarily monotonically.
 
SMC
2:23 PM
I hope the question is not too much naive but there is any way to get this paper w/o paying? I need to consult it for my thesis but I'm not found on how pubblication works londmathsoc.onlinelibrary.wiley.com/doi/10.1112/plms/…
 
2:34 PM
isn't this it?
@JoeShmo it suffices to prove the set $f^{-1}(\mathbb{R}\setminus\{0,1\})$ has measure zero
 
SMC
It is but my uni is not listed as Institution
i guess no hope
 
oh, I'm stupid, I didn't realize I was logged in there...
have you tried scihub?
 
3:14 PM
@Thorgott I was planning on doing that, actually, but I still can't see an explicit set in the example I gave
 
if the set where $f$ takes values different from $0$ or $1$ has measure zero, there's an obvious candidate for an indicator function $1_A$ that is a.e. equal to $f$
 
suppose $f = \mathbb{1}_A$ for some $A$ in the example I gave. What is $A$?
$A_{2n} = [-2, 1], A_{2n+1} = [-1, 2]$
I'm not even trying to prove the claim at this point, just verify it.
 
3:47 PM
@JoeShmo $\int_{\Sigma} |\mathbf{1}_{A_n} - f| d \mu \rightarrow 0$ implies $\mathbf{1}_{A_n}$ converges to $f$ in measure, since you are working in a probability space (a finite measure space), this implies that there is a subsequence of the $\mathbf{1}_{A_{n_k}}$ converging to $f$ almost everywhere, clearly then, $f$ is equal to $\mathbf{1}_{\liminf_{k} A_{n_k}}$ almost everywhere
 
@porridgemathematics, so in the example, $\lim \inf A_{n_k} = [-1,1]$?
 
i dont know, because you haven't given me an $f$ in your example
 
if you take a subsequence involving both the evens and odds
 
the subsequence I took depends on $f$
 
ok
well the claim is that $f = \mathbb{1}_A$ a.e. for some $A$, no?
the choice is pretty much on the set $A$
 
3:53 PM
yes, and I claim that for some subsequence $A_{n_k}$ of your sets $A_n$, the candidate $A = \liminf_k A_{n_k}$ works.
 
ok, well we don't have much choice luckily
 
that subsequence itself is extracted depending on how the $\mathbf{1}_{A_n}$ converge in probability to $f$
 
either $\lim \inf = [-1,1]$, or $\lim \inf = [-2,1]$, or $\lim \inf = [-1,2]$
 
none of these work
 
3:56 PM
cause there's no $f$ in your example
 
cool, thanks to both of you!
 
your question started with 'there is an $f$ to which the $\mathbf{1}_{A_n}$ converge in $L^1$, but with the sets you have, there is no candidate for $f$
np
 
gotcha. meaning the condition would never hold on these sets to begin with
 
thats right
 
4:02 PM
How can I show : If $M\setminus\{p\}$ is orientable then $M$ is orientable ?
My definition of orientability is defined using homology (local orientations)
dimension of $M$ is greater than 1
 
4:19 PM
pick a ball around $p$, every point in the ball except $p$ has a local orientation. argue these all come from the same orientation on the ball. then, if you equip $p$ with the local orientation induced by the orientation of that ball, they all glue together to an orientation of $M$.
the dimension assumption comes into play when you use the deleted ball is connected in dimensions >1. the claim is still true in dim 1, but for different reasons
 
I've got a question: Given a big (uncountable) algebraically independent set $A\subseteq\mathbb{R}$, can you say anything about the outer measure of $A$? My instinct tells me that the two are independent topics, and the measure could be trivial or infinite or anywhere in between, but I don't legitimately know
 
4:45 PM
it can be 0, no clue if it can be positive
 
5:09 PM
@Thorgott You mean if $p\in B$ is an open ball then there is a unique generator $\mu_B\in H_n(M,M-B)$ such that $i^x_*(\mu_B) = \mu_x$ for each $x\in B\setminus\{p\}$? Here, $\mu_x$ is a local orientation and $i^x:(M,M-B)\to (M,M-x)$ is an inclusion map.
 
yes
sufficiently small open ball, we should say
 
5:25 PM
@Thorgott Is that can be proved easily? I saw some similar statement for section in Hatcher (Lemma 3.27) which is not very easy to prove at first sight.
 
totally different statements, this one is easy
the point is that we know a priori that $H_n(M,M-B)$ is cyclic and $i^x$ induces an iso on $H_n$ for each $x\in B$, so the only potential issue is consistency, but local consistency holds by definition and passing to global consistency is a connectedness argument
 
5:53 PM
@Thorgott the global consistency you mean if $B'$ is another open ball that contains $x$ then $i^x_*(\mu_{B'}) = i^x_*(\mu_B)$ ?
 
no
I mean that you can find $\mu_B$ such that $i_{\ast}^x(\mu_B)=\mu_x$ for all $x$
 
Oh I knew that as a local consistency
 
 
3 hours later…
8:28 PM
@Thorgott seems there is a useful fact in Hatcher prop 3.25
 
8:46 PM
W is finite dimensional vector space. $T_1,T_2\in L(V, W)$. It is to be proven that null $T_1\subset $null $T_2$ iff there exists $S\in L(W,W)$ such that $T_2=ST_1$.
 
Hi, @robjohn. You doing better?
 
Hi professor Ted and professor Robjohn
 
So, "it is to be proven." Do so.
 
I proceeded like this:
For converse: Suppose there exists $S$ such that $T_2=ST_1$. Let $x\in $null $T_1$ then $T_1 x=0$ hence $ST_1(x)=S(0)=0$ (because every linear map maps 0 to 0) whence $T_2x=0$ so $x\in null T_2$. This shows null $T_1\subset $null $T_2$
 
Yes, that wasn't the interesting direction.
 
8:54 PM
Now for the other direction: Since W is FDVS, let $w_1,...,w_k$ be a basis for W. I define $S:W\to W$ by $Sw_i=w_i$ (identity) and then I claim that $T_2=ST_1$. For any $x\in V$ we have $ST_1(x)= T_1(x)$, I am stuck here.
 
yeah, you can argue using the double cover as well
the point in both arguments is roughly the same
namely that you can extend the orientation (a.k.a section of the oriented double cover) through the puncture if it has enough codimension
 
You're stuck because you're wrong.
 
taking S = I is unlikely to work unless T_2 = T_1.
 
^
 
yeah, $T_1$ does not have to be $T_2$. I figured.
 
8:57 PM
For example, $T_2$ could be the zero map.
 
hmm, I was wrong in taking S=I
 
You're too experienced to do something like that.
 
i dunno how to think about this without giving it away. S is gonna have to take stuff that T_1 acted on and move it around so it lines up with what T_2 does. maybe a starting point is to pick a basis for the range of T_1 (if it is nonzero; if it's zero that is an easy special case). and fiddle around with first defining S on that. bear in mind you will need to use the hypothesis about nullspaces somewhere.
 
Shhhh, @leslie. Hush.
 
Finite-dimensional nature of W has to do something with it and the fact that a linear map is determined completely by its value on basis of domain.
 
8:59 PM
i'll hold my tongue.
 
Go play with the munchkin, @leslie.
 
these two facts will show me the light.
 
Finite-dimensionality isn't actually relevant.
 
sadly, she's napping. tired out after a visit to the duck pond.
 
Oh, y'all went to the duck pond, cast and all? Cool.
 
9:00 PM
we put her in a stroller. hadn't done that in a while.
 
Ah.
So, follow up on what you said, @Koro. What sort of basis for the domain should you take?
 
I'm trying to complete the solution professor Ted. I'll share that shortly.
 
Oh, well.
Did the ducks remember her, @leslie, or did the cast cast a pall on the matter?
 
they kept their distance, i think because she couldn't just go and hang out with them without us pushing her there in a stroller. they did not seem hostile.
 
Right, tolerating stroller + parents is too much for the ducks to contemplate.
I'd love to see a picture of her holding court with them :P
 
9:09 PM
i might be able to screencap a video.
koro's exercise does go through in the linear algebra sense for infinite dimensional spaces but can fail in normed linear spaces if you want "S" to be bounded.
 
Well, I wasn't thinking in such detail. But if $T_i$ are bounded operators, won't $S$ then be bounded?
If one of them isn't, of course ...
@leslie Re video, for munchkin's privacy, best not to make this public.
 
hmm, I believe this should go through in the bounded context, but you need to quote Hahn-Banach
 
@TedShifrin It seems complicated. My experience in linear algebra so far has been in matrices/determinants mostly. It's probably the first time I'm reading linear maps (not necessarily from R^m to R^n, wherein I could write linear map in form of matrix times a column vector).
 
my favorite approach to this question would be through quotient spaces, but that's probably a bit too fancy
 
Well, even if you use matrices, you still need to think about bases, @Koro. Go back and answer the question I asked. What's a smart way to choose a basis for the domain?
Too fancy by far, @Thor, but yes.
Why are you doing problems at random if they're not consistent with what you've learned?
 
9:18 PM
I read linear algebra for the first time few years back when I was at college. Now, I'm not at college. I started reading linear algebra and after going through vector spaces, now I'm at linear maps (chapter 3 in linear algebra done right).
 
Oh, my least favorite book. Well, then he should have taught the relevant tools in the chapter, eh?
Focus on the question I've asked twice.
 
basis for range $T_1$ can be considered first that is $T_1v_1,...,T_1v_k$ for some k and then this can be extended to a basis of W. Is that correct?
 
I asked for a basis for the domain, did I not?
An intelligent choice thereof.
 
domain of S, right?
 
No.
Domain of $T_i$.
 
9:31 PM
I am confused there because $T_i$ can be infinite dimensional also :(
 
i do think there can be a problem in the bounded case. T_1 can be injective but also crunch norms of nonzero inputs arbitrarily close to zero (e.g. "multiplication by x" on C[0,1] with inputs supported in tiny neighborhoods of zero). taking T_2 = I, it seems that your S would potentially have to blow the norms of some vectors up by arbitrarily large factors.
this might go away if you require T_1 to have closed range.
 
Ah, right @leslie.
Did you ever comment on that Laplacian question? I think your first objection was right, since the OP wrote down $\Delta u$ to start.
 
the real question is what the injective objects in the category of banach spaces are
 
@Koro I believe Axler's book is all finite-dimensional, so don't worry about this.
And I assume he's talked about bases, dimensions, probably nullity-rank (although you do not need that here).
If not, the question is inappropriate.
 
on review of that comment, i think he was just being sloppy with notation. but i'm not sure that his operator is closed, because he didn't impose anything resembling boundary conditions on the domain and you usually need to do that. i couldn't figure out if he was in R^1, where it might not make a difference, or R^n.
 
9:37 PM
I thought $\Bbb R^n$.
 
ok, I think I see what goes wrong
 
Now I'm gonna go back and look.
 
we're fine if there is a bounded projection onto the range
 
generally if you say "let the domain be the set of things for which both sides of this formula make sense," where the formula involves a differential operator, you don't get even a closable operator without boundary conditions or regularity conditions. even for a one dimensional candidate for d/dx.
but that's at the level of vibes and not real math.
 
Oh, right, closed on its domain.
He didn't specify where he's working, either.
It seems like standard Sobolev fiddling, but I can't remember enough to read the sloppy post.
Plus his "affirmation" and "actualization" must come from some other language. I have no idea what he's talking about.
 
9:44 PM
@TedShifrin Not worrying about V. range T1 is FDVS so let $Tv_1, Tv_2,..., Tv_k$ be a basis for range $T_1$. Given any $c_i$'s (scalars), if $\sum c_i v_i=0$ then $\sum c_i T(v_i)=0$ whence $c_i$'s are $0$ that is $v_i$'s are LI in V. Now, for any $v\in V$ there exist $d_i$'s such that $v=\sum d_i v_i$ which gives $T(v- \sum d_i v_i)=0$ that is $v-\sum d_i v_i \in null T_1$ so the smart way to write a v in V is to write it as sum of linear combination of $v_i$'s and an element from null T1.
no, there is still one mistake.
 
Not worrying about $V$?
Then you're doing the problem wrong.
Where is the nullspace of $T_1$?
 
@TedShifrin Not worrying about V. range T1 is FDVS so let $Tv_1, Tv_2,..., Tv_k$ be a basis for range $T_1$. Given any $c_i$'s (scalars), if $\sum c_i v_i=0$ then $\sum c_i T(v_i)=0$ whence $c_i$'s are $0$ that is $v_i$'s are LI in V. Now, for any $v\in V$ there exist $d_i$'s such that $Tv=\sum d_i Tv_i$ which gives $T(v- \sum d_i v_i)=0$ that is $v-\sum d_i v_i \in null T_1$ so the smart way to write a v in V is to write it as sum of linear combination of $v_i$'s and an element from null T1.
 
I'm giving up on giving hints. Start with a basis for the nullspace of $T_1$. Do NOT start with random bases.
 
@TedShifrin I mean its dimension can be infinite also.
 
OK, since you keep ignoring me, I'm just going to quit. Someone else can work with you.
 
10:11 PM
@TedShifrin Still feeling funny in the tummy, but food is not so offensive.
 
10:39 PM
koro, any thoughts on how to use your view of elements of V to define S?
 
doing homework in my head: let $E_{ij}$ denote the matrix whose $i, j$-th entry is $1$ and everything else is $0$. then $E_{ij}$ has all rows zero except the $i$-th row which is $e_j^t$, so $E_{ij} e_j = e_i$, and $E_{ij} e_k = 0$ otherwise.
$E_{ij} e_k = \delta_{kj} e_i$
 
is it really only in your head if you type it out? :)
 
i guess not
 
i absolutely have to write stuff like that out, even when i know it and the composition law is simple and 'easy to do' in one's head.
 
i can never matrix multiply in my head
 
10:45 PM
i do my part for the pen and pencil manufacturing industries.
 
Joe
Sorry to interrupt, but does anyone know how to pronounce this symbol? It appears after the word "collection".
 
$e_i^t \otimes e_j = e_i^t \otimes E_{kj} e_j = e_i^t E_{kj} \otimes e_j = (E_{jk} e_i)^t \otimes e_j = 0^t \otimes e_j = 0 \otimes 0$
Success!
 
joe, i'd just pronounce it C. it looks to be a script C.
 
What happened to you @robjohn? I hope you're doing better
 
Joe
@leslietownes: Thanks. What was Paul Halmos thinking using that symbol??
 
10:50 PM
in many books script letters are often used to denote things that are larger than sets and cannot be sets. that might be what is going on here. what is it a collection of?
if it's in his set theory book, that's probably what he's doing. otherwise, who knows. every author seems to like at least one form of alternate typesetting, whether script or fraktur or who knows what.
 
Joe
Yeah it's from his set theory book. I think he uses lower-case letters to denote sets, and then upper-case letters to denote sets which contains sets, and the weird letters to denote sets which contain sets which contain sets
 
sometimes called 'proper classes' although it would surprise me if he got into that in his book. en.wikipedia.org/wiki/Class_(set_theory)
 
Halmos was just a wild guy.
(Halmos ended up at Santa Clara University, I used to travel with a prof. from the math. department.)
(Since I am being serious for once, I should clarify that Halmos was not a wild guy in my sense of the word.)
 
11:14 PM
i never met him. he was about but in declining health during my time.
there's a nice photo of him with a cat out there. no mathematician who likes cats can be all bad.
 
@leslietownes yes, taking inspiration from what I want to prove ($S_2=ST_1$). Defining S:W$\to W$ as $S(T_1v_i)=T_2v_i$ (for all $1\le i \le k$) determines S completely and uniquely (from a theorem that says that a linear map F:X->Y, where X is FDVS is completely determined by its values on basis of X.).
Now for any v in V, there exist $c_i$’s (scalars) and v’ in null T_1 such that $v=c_1v_1+…+c_kv_k+v’$, which gives $T_1v= c_1 T_1v_1+…+ c_k T_1 v_k$ so $ST_1v= T_2(c_1v_1+…+c_kv_k)$. Since v’ is in null T1 which is a subset of null T2, it follows that v’ is in null T2. So we get $ST_1v = T_2 (c_1v_1+…+c_kv_k)+T_2 (v’)= T_2( c_1v_1+…+c_kv_k+ v’)= T_2 v$
1 hour ago, by Koro
@TedShifrin Not worrying about V. range T1 is FDVS so let $Tv_1, Tv_2,..., Tv_k$ be a basis for range $T_1$. Given any $c_i$'s (scalars), if $\sum c_i v_i=0$ then $\sum c_i T(v_i)=0$ whence $c_i$'s are $0$ that is $v_i$'s are LI in V. Now, for any $v\in V$ there exist $d_i$'s such that $Tv=\sum d_i Tv_i$ which gives $T(v- \sum d_i v_i)=0$ that is $v-\sum d_i v_i \in null T_1$ so the smart way to write a v in V is to write it as sum of linear combination of $v_i$'s and an element from null T1.
 
koro, note that your recipe seems to define S only on the range of T (note that it is only well defined at all because the T v_i are linearly independent). but you are definitely on the right track.
 
@leslietownes this is the central idea.
 
yes, indeed.
 
@leslietownes ah yes, you’re absolutely right. The working above can be fixed if I just remove the word S is uniquely determined. We know S exists (by the existence theorem for linear maps) and that’s it.
Also the representation of v in V as the sum $v=(c_1v_1+…+c_kv_k)+v’$ , where v’ is in null T1, is unique as sub space spanned by v_i’s and the subspace null T_1 have only 0 in common.
 
11:25 PM
is that clear from your construction, or is it something you proved?
 
I proved it. Let x be in span ( $v_1, v_2,…, v_k)\cap $ null T_1. Then $x=\sum c_iv_i$ which gives Tx=0 (as x is in null T_1 also) that is $\sum c_i Tv_i=0$ whence all ci are 0 as Tv_i ‘s are LI. So x=0.
 
Algebra people: Divisible abelian groups are projective + injective. Is the other direction true? Just for my own personal edification, not homework.
I don't care enough to think either just T/F would do lol
 
Leslie, I hope my answer to my question is complete now?
 
11:47 PM
i think so. if i were writing this up for future personal reference i might re-order the argument and make the various steps more distinguished from one another. things tend to spill out in fits and starts in a chat window.
news flash, my daughter has announced that after she turns into a cat she is going to turn into an elephant.
 
@Koro Here is another, more direct way (I think): One direction is immediate. For the other, let $v_1,..,v_p$ be a basis for $\operatorname{im} T_1$ and add $v_k$ as necessary to make a basis for $W$. Suppose $x_1,...x_p$ are such that $v_k = T_1 x_k$ (note that the $x_k$ are linearly independent), and define the linear $S v_k = T_2 x_k$ (and zero on the rest).
Now pick some $x$, we need to show that $T_2x = ST_1 x$. Note that for some $\alpha_k$ we have $T_1 x = \sum_k \alpha_k v_k = T_1(\sum_k \alpha_k x_k)$ and hence $x-\sum_k \alpha_k x_k \in \ker T_1 \subset \ker T_2$. Hence $T_2 x = T_2(\sum_k \alpha_k x_k) = \sum_k \alpha_k S v_k = \sum_k \alpha_k ST_1 x_k = ST_1 x$.
(I'm procrastinating from work :-).
 
copper, interesting piece in the chronicle this morning. an editor was cleaning out some old papers and found letters from the unabomber asking about places to go in south america (the editor had written a book on traveling there). before he began unabombing.
 

« first day (4064 days earlier)      last day (33 days later) »