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3:01 AM
I'm trying to prove two matrices $A$ and $J$, square, are similar. I managed to prove $A^2 = J^2$, is that enough to conclude $A \sim J$?
 
theyre also invertible
:( darn
 
Think of $(-I)$ and $I$
Both square to $I$
But they're def not similar
 
actually i have that $A^2 = J^2 = -I$ but I'm not sure that's relevant
 
@GFauxPas same argument applies
 
3:04 AM
darn, okay
 
What's the context? Maybe something else about the situation is helpful
 
Let $A$ be a square real matrix such that $A^2 + I = 0$
 
real
 
part 1, which i proved
 
$\Huge{real}$
 
3:05 AM
A is $n \times n$, prove $n$ is even
it's italicized in the assignment :P
 
$\Bbb{REAL}$
 
so I did tht
Lol
next is to prove that $A$ is similar to
 
$\mathfrak{reaL}$
 
$\mathfrak{REAL}$
 
$\mathscr{REAL}$
 
3:07 AM
Oh lawd
 
$\begin{bmatrix} J_0 \\ & J_0 \\ & & J_0 \\ & & & \ddots \\ & & & & J_0\end{bmatrix}$, where
$J_0 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$
 
Whoa whoa don't go through all that nonsense
$A^2 = -I$
 
so , thats what I have to prove
I just noticed that $J^2 = -I$
 
Don't bother with that
What can you compute about $-I$?
Just list some facts about it, they'll apply to $A^2$ as well
 
The minimal polynomial of $A$ is $(x^2+1)$, thus the associated $\Bbb R[x]$ module must be isomorphic to $(\Bbb R[x]/(x^2+1))^{\frac{n}{2}}$, the normal form follows if we take on each summand the basis $\overline{1}, \overline{x}$
 
3:10 AM
eigenvalue is $-1$ with eigenvector anything
 
@MatheinBoulomenos ...
 
Math that's using tools we haven't learned
 
@GFauxPas it's really easy, just think about it
 
@Mathein you got it!!!!
 
no eigenvector nonsense
 
3:11 AM
but I'm not trying to conclude stuff about $A^2$, i want to know if $A \sim J$
 
That's a bad way of going about the problem
 
I have forgotten how to do LA without module theory lol
 
Like we're saying to scrap that idea and do something easier
 
except we didnt learn about R-modules
 
@GFauxPas no similar nonsense
it's really easy
 
3:12 AM
hmm
 
What I have in mind is literally a number you pull out
 
pull out of what, J?
-1?
 
@GFauxPas seriously, give it a thought without those complicated machineries
 
No no no, forget $J$
 
3:14 AM
that's what I'm trying to prove, that $A \sim J$, thats the assignment
 
Like drop it from your memory forever
 
that's the problem
 
Wait you said you had to prove that $n$ is even
 
I said I proved that
GFauxPas
part 1, which i proved
 
oh, lol trolled
 
3:14 AM
XD
 
Rip in pieces
 
so back to $A \sim J$
I said that I noticed that $J^2 = A^2 = -I$ but that's not enough
you have that $Jv = [v_2, -v_1, v_4, -v_3, \cdots, v_n, -v_{n-1}]^\intercal$
which is weird but okay
 
@GFauxPas what criteria for diagonizability do you know?
 
$n$ distinct roots to the characteristic equation, uh
similar to a diagonal matrix
most of them boil down to the first one
in particular that if it has $n$ distinct eigenvalues then it's similar to a matrix with those eigenvalues on the diagonal
 
You have $0=A^2+I=(A+i\cdot I)(A-i\cdot I)$, you can use this to show that $A$ is diagonizable over $\Bbb C$
 
3:23 AM
why does it imply that?
 
Generally if two matrices $C$ and $D$ over $\Bbb C$ satisfy $CD=DC$ and $\operatorname{Ker}(C) \cap \operatorname{Ker}(D) = \{0\}$ then $\operatorname{Ker}(CD) = \operatorname{Ker}(C) \oplus \operatorname{Ker}(D)$
You can apply that to $C=A+iI$ and $D=A-iI$, then you get that $\Bbb C^n = \operatorname{Ker}(A+iI)\oplus\operatorname{Ker}(A-iI)$
 
god
 
:(
 
Morning everyone
 
okay Math how does that help
 
3:40 AM
Elements of $\operatorname{Ker}(A-\lambda I)$ are eigenvectors to the eigenvalue $\lambda$
 
sure
 
so if you have $\Bbb C^n = \operatorname{Ker}(A+iI)\oplus\operatorname{Ker}(A-iI)$, then there is a basis of $\Bbb C^n$ consisting of eigenvectors from $A$
 
solutions to $Av = -iv$ and $Av = iv$
 
exactly
 
so there are $n$ distinct eigenvalues and $A$ is diagonalizable, great
 
3:42 AM
no
you only have two eigenvalues
 
there are only $2$ distinct eigenvalues
sniped
 
err n l.i. eigenvectors
 
but for every eigenvalue there are "enough" eigenvectors
 
@GFauxPas we've already gone through this yesterday
10 hours ago, by Leaky Nun
a matrix can be diagonalized iff it has 10 distinct (linearly independent) eigenvectors, not eigenvalues
 
yes yes youre right Leaky don't rub it in
it has $n$ l.i. eigenvectors
:)
 
3:43 AM
right
 
which implies $A$ is diagonalizable, wonderful, so what?
 
lol
 
The next thing to realize is that $\operatorname{dim}(\operatorname{Ker}(A+i I)) = \frac{n}{2}$
 
agreed
 
I don't
how do you know?
 
3:44 AM
look at the characteristic polynomial of $A$
 
oh, well, we have $rk(A+iI) + rk(A-iI) = n$
 
that has coefficients in $\Bbb R$
 
@GFauxPas and then?
 
well, they look like they have the same rank, so they must
 
great
 
3:46 AM
such math
very amaze
 
gr8 m8 i r8 8/8
much rigor
 
More seriously, I would look at the characteristic polynomial of $A$
 
y so serious
 
this has to be of the form $(x-i)^a(x+i)^b$
where $a$ and $b$ are the dimensions of the eigenspaces to $i$ and $-i$ respectively
 
@MatheinBoulomenos your proof can't be better than GFauxPas' proof
his proof is the best
just give up already
 
3:49 AM
If $a\neq b$, then $(x-i)^a(x+i)^b$ can't be real, cause our "i"s aren't real
 
so maybe the characteristic polynomial is complex valued?
 
but the matrix is real
 
there's no way anyone in my class got this problem, its way harder than anything we've done so far
 
you literally derive the char.poly from the matrix by taking determinant
 
maybe I'm doing this too complicated and there's an easier way
Well, if you know module theory, there's an easy way
 
3:51 AM
I dont :(
 
but you do get the solution @GFauxPas
 
I don't see how any of this has to do with $J$
 
...
 
Okay so we have found that there are $\frac{n}{2}$ eigenvalues for $i$ and $\frac{n}{2}$ eigenvalues for $-i$
 
the i -i i -i thing is similar to J
 
3:53 AM
Question
 
Answer
 
Is measure theory not that popular?
 
Answer
I don't like it that much
 
?ralupop taht ton yroeht erusaem sI
 
I've seen a lot of discussions on differential geometry/topology in this chat, along with a lot of algebra, but not much of measure theory or functional analysis
2
 
3:54 AM
sisylana lanoitcnuf ro yroeht erusaem fo hcum ton tub ,arbegla fo tol a htiw gnola ,tahc siht ni ygolopot/yrtemoeg laitnereffid no snoissucsid fo tol a nees ev'I
 
@LeakyNun you're just being silly
 
I do it sometimes. Next quarter I'm taking functional analysis so...
 
@MatheinBoulomenos thanks, captain obvious
 
yllis gnieb si ykael
So, measure theory isn't that important/useful?
 
I wouldn't say that
it's both important and useful
 
3:56 AM
@Daminark Disgusting
 
I use measure theory all the time
 
Then, it's just not that fun?
 
that's a subjective thing
 
Apart from Sard's theorem, I've never seen much measure theory pop up in Diff Top or ATop
 
@orbit-stabilizer it's the foundation of integration
you know, a foudnation
 
3:58 AM
I don't know... I mean, I don't think I've met someone who thinks what we're learning in Baby Rudin right now is fun. It's plausible to me that, on average, more undergrads find group theory more interesting than real analysis.
 
Haar measures are useful in number theory
 
@MatheinBoulomenos haar?
 
huh?
 
haar measures are useful all over the place
 
3:59 AM
har?
 
haar is a name
 
In mathematical analysis, the Haar measure assigns an "invariant volume" to subsets of locally compact topological groups, consequently defining an integral for functions on those groups. This measure was introduced by Alfréd Haar in 1933, though its special case for Lie groups had been introduced by Adolf Hurwitz in 1897 under the name "invariant integral". Haar measures are used in many parts of analysis, number theory, group theory, representation theory, statistics, probability theory, and ergodic theory. == Preliminaries == Let ( G , â‹… ...
 
4 mins ago, by MatheinBoulomenos
@LeakyNun you're just being silly
 
@orbit-stabilizer I use lots of hard measure theory all the time and tbh Im not excited by it, to me it's mostly just language, a tool to express other things that are actually interesting
 
@EricSilva yeah, that sounds right.
I'm interested in learning it for probability
 
4:02 AM
yeah it's important there, again as your foundational language for expressing interesting things
I guess I've studied some geometric measure theory for it's own sake but that's kind of a different beast than what I think someone just seeing basic measure theory might think of when they say the words measure theory
 
But your milage may vary, what's just a tool & language to one can be an object of its own interest to others
 
@MatheinBoulomenos true, some people like number theory for goodness sake ;)
 
I'm mildly offended
 
What do you do as a number theorist? Add really big numbers? Don't we have calculators for that?
 
Basically
There's multiplicative and additive number theory
 
4:06 AM
I've never heard of anyone doing research in just measure theory in my life
 
Well, there's ergodic theory. I have a professor who does research in that.
 
Some people specialize in multiplying big numbers and other specialize in adding big numbers
 
that's a completely different thing than just "measure theory" though
 
@EricSilva right
 
4:08 AM
i think 90% of the hits you might get if you look up measure theory in the arxiv or something will be functional analysis or geometric measure theory stuff
 
Only the best mathematicians can count really high. Tao can count to infinity iirc.
 
I once leafed through a five volume set on measure theory by a guy called Fremlin - I guess he must have specialied in that area; martingale theory in financial math heavily relies on it too.
 
Tao can count to inaccessible cardinals
5
 
@MatheinBoulomenos @Narcissusjewel hahaha
 
4:09 AM
Tao's power levels are clearly over $\aleph_0$
 
@MoziburUllah i forgot about that guy, I guess he's as close as one could get to an abstract measure theorist
 
god why would anyone ever read this
 
I don't know why, but measure theory just seems so dry! gah!
Differential geometry sounds sexy
2
 
lots of differential geometry uses hard measure theory lol
 
4:12 AM
I actually like dry stuff
just not measure theory
 
@orbit-stabilizer drink some water
 
"Yes, I study measure theory - no I don't use the imperial system, I'm from Canada."
 
"Yes, I study in Imperial - no, I don't use the imperial system"
 
i think maybe the amount of measure theory present in your work is inversely proportional to your skill as an expositor
 
4:14 AM
@LeakyNun will do
What sexy thing does measure theory have
Manifolds sound cool to me from differential geo
 
varifolds are cool
 
Manifolds are dank
 
And I can visualize them
 
I find manifolds confusing
 
Oh wow, that's a thing
 
4:18 AM
@MatheinBoulomenos In what way?
 
In mathematics, a varifold is, loosely speaking, a measure-theoretic generalization of the concept of a differentiable manifold, by replacing differentiability requirements with those provided by rectifiable sets, while maintaining the general algebraic structure usually seen in differential geometry. More closely, the varifold generalize the idea of a rectifiable current. Varifolds are one of the topics of study in geometric measure theory. == Historical note == Varifolds were first introduced by L.C. Young in (Young 1951), under the name "generalized surfaces". Frederick Almgren slightly modified...
 
they have better compactness properties
 
Anyone have any idea how to approach this $$\int_0^T dt_1 \int_{t_1}^T dt_2 \cdots \int_{t_{n-1}}^T dt_n \, ? $$
 
Gah, compactness
 
when i say compactness properties i mean that spaces of varifolds behave better under limit procedures and min max constructions than spaces of manifolds do
 
4:21 AM
I like to study maths objects, but more focused on their instrinsic properties rather than applications
 
@DanielSank idk, I just find it easier to work with schemes. They're confusing me as a category, as they are a full subcategory of the category of locally ringed spaces, but somehow that doesn't help a lot, they're not even closed under fiber products
 
I hope they have a direct computation on why the irrationals have full measure, despite it does not contain any intervals
 
trying to do extrinsic differential geometry without invoking some sort of measure theoretic analogue of manifold is a tall order
 
Almost all proofs on that one rely on the fact that Q has zero measure, hence an indirect way to determine it
Currently one of the greatest mystery for me about measure theory is why do measures on uncountable sets so counterintuitive
 
why did someone even write this
 
@MatheinBoulomenos I'm gonna be a revolutionary and invent subtractive number theory
 
Wait the fuck thats a real tehsis?
 
4:28 AM
Wtf is Liberal vs apodictic?
 
... So what?
Why can't it be a thesis?
 
It totally can be
 
Maybe they're doing their masters in math education?
 
I revise my statement to
The fuck, somebody made that the title of their thesis?
 
I mean... it's accurate
 
4:29 AM
Its also really vague
 
I'm just stunned that someone analyzed the pedagogy of a bunch of really boring measure theory books
 
Somebody should write a thesis " ON SOME PAPERS ON SOME MEASURE THEORY TEXTBOOKS AND THEIR USE
BY SOME PROFESSORS IN GRADUATE-LEVEL COURSES"
 
A PRESENTATION OF SOME RESULTS, USEFUL TO SOME APPLICATIONS OF SOMETHING
 
$\Bbb{ON~SOME~PAPERS~ON~SOME~MEASURE~THEORY~TEXTBOOKS \\ AND~THEIR~USE~BY~SOME~PROFESSORS~IN~GRADUATE-LEVEL~COURSES}$
 
4
Q: Measure of the irrational numbers?

KvothealarI have read that the measure of the irrational numbers on an interval $[a,b] = b-a$. This both makes sense and doesn't make sense to me. If you consider that the union of the irrationals with the rationals are the reals, then if the rationals have measure 0, then the irrationals must have the sam...

 
4:32 AM
I guess it has to be done by someone though, these books all suck
 
Uncountability can only be justified with measure theory if powersets axiom don't exist
[More random]
 
I think my university uses Folland
 
that's one of the few ive never touched
 
There is ultrafinitism, finitism, infinitism. But perhaps we can split infinitism into two: Countabilism and uncountabilism.
All predicativiatists are countablists
Measure theory has this additivity, but it only works for sets with countable support. You cannot have an uncountable Lebesgue measure without something gone really broken
Jordan measure, being a finite union of rectangles, cannot handle certain sets like the cantor set because the inner jordan measure vanishes thus it does not have a measure
In general, for any given measure on any set, the set is measurable under that measure if its outer and inner measure agree
 
@Secret that's not true for any measure. Measures which satisfy this are called "regular"
 
4:50 AM
jordan measure isnt even a measure
 
5:01 AM
grrr, I need maybe 3 more years of practice in measure theory to fully make sense of it
IMO, The difficulty level of measure theory on uncountable sets is inaccessible cardinally higher than any other maths topics
 
idk, measure theory isn't easy, but at least the basics are not that hard
 
@MatheinBoulomenos right
good luck proving $\mu([0,1])=1$
where $\mu$ is the Lebesgue outer measure
 
How does one usually compute the ordinary cohomology of spaces? (Or CW complexes if there's a slicker way to do that)
 
Well, the axioms are easy to understand, and you can see how it is a generalisation of size, but the consequences are very counterintuitive, especially when uncountable sets are involved
For me, the most counterintuitive thing about lebesgue measure is you can have sets of nonzero lesbegue measure that contains no intervals
 
@LeakyNun ? I don't see how that's difficult
 
5:06 AM
@MatheinBoulomenos then how?
 
Just use subadditivity
 
start from the definition :P
 
what is your definition of outer lebesgue measure?
 
the infimum of the sum of the length of the open cover
 
that's definitely easy to do
 
5:07 AM
yeah, I don't see how that's difficult
 
how?
 
it suffices to do show the inf over finite collections of intervals that cover $[0, 1]$ is 1
and once youve passed to finite intervals it's really easy
 
but the same for $[0,1]\cap \Bbb Q$ is also $1$
 
what
 
Nope
 
5:11 AM
that's nonsense
 
Enumerate the rationals $q_1, q_2, \ldots $
 
finite collections of intervals
give me 2 intervals that cover $[0,1]\cap\Bbb Q$
 
sure but it doesn't suffice to take inf over finite collections if you're only working with rational points
 
then why does it suffice for $[0,1]$?
 
because it's compact
 
5:12 AM
hmm
how does that help?
 
because you can always throw away all but a finite number in any open cover
 
aha
 
and it only decreases the sum
 
nice
 
I need to think how to directly prove the irrational has full measure and the rationals have zero measure...
 
5:16 AM
when you measure it
it says "full"
so it has full measure
 
and more generally, to understand why the measure of a set has nothing to do with whether it is totally disconnected
(This is how I screw up at that vitali set question)
Hmm... what is a cover for the irrationals...
The obvious one is the reals, but can we break this down further...
Actually let's restrict to $[0,1] \cap \Bbb{I}$
 
you only need to do the rationals, and then use the properties of measure to get irrationals
 
That's exactly what I found unintuitive, is there a direct way to do it on the irrationals?
 
why would you do that irrationals suck
 
My biggest confusion on measure theory is I don't understand why measure of a set has nothing to do with whether it is totally disconnected
e.g. cantor set has measure zero, and fat cantor set has measure tends to one, but both are totally disconnected like the rationals
 
5:27 AM
one, not tends to one
a measure is a number
 
fat cantor set is a term for a type of cantor set, it's measure can be arbitrary
 
This confusion is why the vitali set proof makes no sense to me since I still saw it as a set containing no intervals
So clarify that is need to understand why vitali sets are non measurable for me
 
nonmeasurability isn't really a visualizable thing
 
$\Bbb{NONMEASURABILITY~ISN'T~REALLY~A~VISUALIZABLE~THING}$
2
 
5:46 AM
@Balarka turns out, Peter was like "Eh, let's have a deadline, December 18th"
So I'm gonna be working on the Dold-Thom business right now, since I've got a good block of this week
 
what is he making you do
 
We're writing a paper on something related to AT
 
cool
 
Mine's gonna be on the Dold-Thom theorem, basically it boils down to this
If you have a space $X$, you have a natural action of $S_n$ on $X^n$ by permuting coordinates, so if you mod out, you get $SP^n(X)$
Let $x$ be the basepoint of $X$, then you can see that there's a natural inclusion $SP^n(X)\to SP^{n+1}(X)$ by $[x_1,\ldots,x_n]\mapsto [x,x_1,\ldots,x_n]$
So you can take the colimit over that, you get $SP^{\infty}(X)$
Sometimes called just $SP(X)$
Now, it turns out, $\pi_n(SP(X)) \cong H_n(X;\mathbb{Z})$
Which you can do cool things with. I'm only just starting but one thing you can compute is that $SP^n(\mathbb{S}^2) = \mathbb{CP}^n$
So $SP(\mathbb{S}^2) = \mathbb{CP}^{\infty}$
 
you gotta do only tuples of distinct points in X^n before modding right?
 
5:54 AM
But then that gives you by Dold-Thom that $\mathbb{CP}^{\infty} \cong K(\mathbb{Z},2)$
Oh yeah distinct points
 
right i knew this fact about $\mathbb{C} P^{\infty}$
 
Wait no hold on is it distinct?
 
i dont understand why it would be distinct points
 
Yeah I don't see why it has to be, and Aguilar doesn't mention anything about that
 
0
Q: Rules of transposes for inner products

ALannisterRight now, I am reading a text on Nonlinear Optimization for a class, and there is a minimization example where for $A$, a matrix of dimension $m \times n$, $c \in \mathbb{R}^{n}$ and $b \in \mathbb{R}^{m}$, we have that the Lagrangian is given by $$L(x, \lambda) = \langle c, x \rangle + \langle ...

 
5:58 AM
this fact is p interesting
it sounds harder than the application you gave though lmao
 
Which fact?
 

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