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5:00 PM
so odd order matrices it is clear that one of the eigenvalues of $I+A$ will be 1
and hence we can write $Ax = 1.x$
but am thinking what for even order matrices
then simply $det(A) = 1.det(A)$
 
so odd order matrices it is clear that one of the eigenvalues of $I+A$ will be 1
How it is coming?
 
why not
 
isn't -1
 
choose that eigenvector $x$ for which $A$ has eigenvalue $1$
so $(I+A)x = (1+\lambda) x$
and $\lambda = 0$ one of the eigenvalues
so $(I+A)x = x$
@ManeeshNarayanan
 
and $\lambda = 0$ one of the eigenvalues
A+I
how it coming?
 
5:05 PM
Perhaps you know $\lambda$ is an eigenvalue of $A$ if $Ax = \lambda x$
 
if $A$ has eigenvalue $\lambda$
then what is the eigenvalue of $I+A$?
its $1+\lambda$ right?
is that ok?
 
so $I+A$ has an eignevalue $1+\lambda$
right?
 
5:07 PM
where $\lambda$ is one of the eigenvalue of $A$
?
 
now you have $\lambda = 0$ as an eigenvalue of $A$
so what will be an eigenvalue of $I+A$ now?
just substitute $\lambda = 0$ in $1+\lambda$
its 1
is that ok?
 
how did you substitute $\lambda=0$?
 
becasue I know that $A$ has eigenvalue 0
dnot we?
 
yes for, for odd case only
i the case where n is odd
Am I right?
 
5:12 PM
What is happening ?
 
@Astyx we r here first part and discussing the first answer
 
non-singular means invertible ?
ie of maximal rank ?
 
Yup
 
I miss this stuff :(
 
@ManeeshNarayanan you can see the even case in 3rd bit
there it gives a counterexample!
 
5:29 PM
0
Q: Automorphisms of $C(\{1,...,n\})$

user193319If I am not mistake, $C(\{1,...,n\})$ denotes all of the continuous functions from $\{1,...,n\}$ to $\Bbb{C}$. I read somewhere that every automorphism on $C(\{1,...,n\})$ is a composition with a homeomorphism on $\{1,...,n\}$. My question is, a composition with a homeomorphism and what else? Als...

 
@BAYMAX He had given for any matrix
How did he write like that?
 
Salut @Astyx
 
hi teed
 
Salut
 
and friend
 
5:32 PM
Hi @Ted
 
s
 
Heya @Meow
hi @Balarka
 
i did some spivak proofs
 
I'm glad to see that Demonark totally missed my point yesterday :)
 
last niht
 
5:33 PM
Have you done a few of those rational function limit ones?
 
Hi @TedShifrin
 
in your pdf? or
 
heya @Tobias — how go the lectures?
Yes, Meow.
 
one sec let me check the notebook
 
@TedShifrin Good. We will be starting on rings on Monday (this week was the fall break)
 
5:34 PM
This is a one-year course or only one-semester, Tobias?
 
Just one semester
 
Yikes. Crazy fast.
 
I will start out the rings by discussing the number of ways to write a number as a sum of two squares, motivating the Gaussian integers
 
@TobiasKildetoft Nice
 
(or more geometrically, the number of integer points on certain circles)
 
5:35 PM
Nice ... I always did that stuff at the end of the first semester at UGA, but very few instructors ever had time to cover it.
(You may not recall that, thanks to me and much to the ire of most of your algebra colleagues, my book did rings first and group theory second semester.)
 
The book covers the proof that those primes that can be written as a sum of two squares are precisely those congruent to $1$ mod $4$, and from there, it is a very small step to counting solutions in general
 
It's a nice application of the (third?) isomorphism theorem for rings.
 
@TedShifrin I can never decide whether I would prefer to do rings or groups first.
 
ok i tried it but my work was a clusterfuck
 
I had sound rationale for doing rings first. (The main motivation was the math education majors taking only the first semester ... and what they needed to understand to be better high school teachers. ... But then I realized $R/I$ is far easier for commutative rings than wrestling with normal subgroups, so why not do the stuff with more structure first. It's against Bourbaki, but who cares?)
 
5:38 PM
number 13
 
@BAYMAX Please help me
 
i think my work wont work but
here's what i was trying to do regardless
 
Of course something like #9 would be easier to start, Meow.
Or #11.
 
@TedShifrin Also, students have been working with multiplication and addition their whole lives so it seems intuitive to start with all the usual "stuff" and then strip away structure to get groups
 
Yeah, I started with integers, then modular arithmetic, then polynomials ... then general (mostly commutative) rings.
Heya @EricSilva.
 
5:41 PM
That sounds good
 
@ManeeshNarayanan I am trying,anyway i have also posted that linear algebra chat room
 
Hi @Baymax.
 
Hello @TedShifrin how are you?
 
Pretty good, thanks. What are you up to?
 
actually @ManeeshNarayanan asked this first part of question and I was trying to help,we did for odd order matrices but what about even order we are thinking
@TedShifrin
 
5:46 PM
So you're thinking about part a) only?
 
yes uptill now,i think part b is taking odd order and 3rd part is taking even order matrices
 
for #9 i have that $\delta = \min(1, 6\epsilon)$. we have that $|\frac{2x+1}{x+3} - 1| = |\frac{2x+1}{x+3} - \frac{x+3}{x+3}| = |\frac{x-2}{x+3}| < |\frac{6\epsilon}{6}| = \epsilon$
 
So odd ones have to be singular, @Baymax, and for $2\times 2$ a nonzero skew-symmetric matrix is easy enough to write down.
 
yes
 
i just have one question
 
5:51 PM
The proof in the answer for part a) is a bit glib at the last step. Do you see how to get the last step?
Oh, there you are, @Meow.
 
does there exist a non zero non singular skew symetric matrix of even order
 
so in that example we happen to get an $x-2$
lucky for us
but what if we cant take out an $x-2$ anywhere?
 
Slow down, @Meow. You skipped the scratchwork.
So $|x-2|<1$ tells me that $\frac 1{|x+3|} < \frac 16$?
Are you sure?
 
um my mathjax isnt rendern that
 
Yeah, I done messed up.
 
5:53 PM
2)why $<x,Ax> = <A^{t}x,x>$?

*) how $x =0 $ implies that $A-I$ is invertible?
 
wait
was i right the first time???
 
@Baymax: But the question was about $n=2$ only? You're doing the general case? Yes, there certainly can be a singular one.
 
damnit i think i was
 
I'm not a mind reader, Meow :)
 
i wrote $\epsilon/6$ at first in the minmum
then changed it
 
5:54 PM
so for first case it works for odd order matrices
@TedShifrin
 
@Baymax: The proof says — Suppose $(A-I)x = 0$. Then $Ax=x$. So then $\|x\|^2 = \langle Ax,x\rangle = -\langle x,Ax\rangle = -\|x\|^2$.
Slow down, @Meow. I'm questioning your 1/6.
 
Morning
 
hi Faust.
 
Hows Ted?
 
Other than continuing back pain, fine, thanks. How're you doing?
 
5:56 PM
Pretty stressed out gramas been having some troub;e
 
@Baymax: The question is this. How do we get from $A-I$ invertible to $A+I$ invertible?
 
i got a midterm on monday for AA so im trying to focus but its hard
 
Sorry to hear that, Faust. We old people suck.
 
alright, well no, it doesnt. you gotta flip the inequality sign
 
Uh huh @Meow.
That's important.
 
5:57 PM
@TedShifrin perhaps from $I-A$to $I+A$ invertible i got it
 
im confused where i flurped up in my scratch work
 
You like to flurp incessantly, Meow.
Yes? @Baymax
Hi Semiclassic
 
can i send you a picture of my work
 
Sure.
 
5:59 PM
no
@TedShifrin
$A = -A^{t}$
 
LOL, why not, @Baymax? :)
NEGATIVE
 
so $I+A = -(I+A)^{t}$?
 
Huh?
 
@TedShifrin i found out that i have dylexia and some kind of comprehension disorder in addition to ASD and the moevment disorder
been a ough week
rough*
 
i was visualizning it as addition of 1 to the main diagona;l
 
6:01 PM
Yes, so, @Baymax?
 
apperntly my word comprehension and sentence comprehension fall in the 8th and 9th percentile for the population but my ability to understand essays is in the 92 percentile which i find utterly confusing :S
 
At UGA students would get extra time and other accommodations for that if they university disability folks verified the problems, @Faust. That is tough going for math.
Wow.
 
Morning everyone!
https://imgur.com/a/lCKSI
My professor's answer (in the picture) is wrong, I believe. The Standard Deviation is supposed to be $\sqrt{10}$. But as you can see, he just used the number $10$ as the SD. Is he correct or is it a mistake?
 
Local versus global?
 
i am confusing a bit
 
6:03 PM
@Dragneel: You appear to be correct, yup.
 
Alright great :)
Also, how do you insert a picture in the chat?
 
@TedShifrin apparently its really common for people with Autism that they can't understand short sets of instructions cause we see to many possible contexts for the sentence or word. but given enough sentences together we are able to narrow down what the intended instructions actually are
 
It never seems to work with me.
 
Upload button next to send button.
@Faust: That's sorta what I meant with local versus global. Quite fascinating, actually.
 
Odd. It's missing for me.
 
6:04 PM
Sadly math tests are words and sentences not paragraphs
 
Are you on a desktop or on mobile, Dragneel? It may not show up on mobile.
 
Desktop.
I'm not too worried about it.
 
theres a buttone beside "send"
 
Hmm, @Dragneel, it might take a certain amount of rep.
I think it requires 100 or 1000 rep.
 
It's possible, yeah.
 
6:06 PM
@TedShifrin the rest of my tutoring for calc I went really well! i was able t explain derivatives and limits and continuity and differntiability and lots of things related to to tangent lines really well +)
 
Good :)
 
Kids pretty bright but he didnt even know the the dam triangles
 
What triangles? You're talking basic trig?
Or slope of lines or ... ?
 
hopefully he passed his exam this time but i dont think he did super well...
the two basic triangles
like he courldnt find $arcsin (\sqrt3 /2 )$
 
Prof @Ted! \o
 
6:08 PM
Apperntatly his highschool allowed him to give answers in decimal form
 
Flurp: A Geometric Approach
 
He was so happy when i explained the $ \frac{f(x+h) - f(x)}{h} =f'(x) $
 
limiiiit
 
yeah
 
h could be the amount of flurps to flurp a flongy
 
6:10 PM
but im too lazy and u r smart semi
 
Heya @Studentmath
 
eh, i'm not exactly awake right now
 
i dunno how you go through two calculus classes without someone getting that across to you
 
@Faust: I've been repeatedly reminding my precalculus students to draw the $30^\circ$-$60^\circ$-$90^\circ$ triangle and double it to get an equilateral triangle. Perhaps that'll help your student. Yeah, everyone cops out and allows calculators too much.
 
hes a smart kid too
 
6:11 PM
I hope you put limit in there, @faust.
 
yea when we were doing it ofc
 
@Meow — are you flurping or are you emailing me?
 
@TedShifrin special angles: a geometric approach
 
i just told him when your evaluating it a good idea when your done is to check the derivative and sub in the values of x to see if its right and he was mind blown they were the same thing =)
 
flurp flurp flurp
 
6:12 PM
Yes, Leaky, of course.
 
im trying to get this flurping flone to work
 
I'd say "special functions: a geometric approach"
 
thinks the room has lost what few of its marbles it once had
2
 
except that that's basically how I do think about elliptic integrals/functions
 
@Jasper
 
6:12 PM
(in the sense of being elliptic integrals being particular integrals on an elliptic curve, and elliptic functions being their inverse functions)
 
@TedShifrin I’m reading a book that basically amounts to “galois theory: a geometric approach”, in all seriousness
 
@Faust Hi!
 
$(A+B)^t = A^t + B^t$ might be the key @TedShifrin ?
 
How you doing!
 
Are you a freshman dreaming
 
6:14 PM
@Semiclassical it is clearly transpose
 
are you in semi state @Semiclassical
 
@LeakyNun evidently it wasn't clear, since I didn't read it as such
but yeah, that makes sense
 
This chat is very confusing today to read line by line
 
Anyone familiar with Maximum Likelihood Estimator?
 
I am blabberring today,shld get som slp ..
 
6:16 PM
i'm tempted to take a nap myself
 
@BAYMAX Yes, of course.
 
Actually the brings up a good question WTH is the proper diffrence between maximal and maximum?
 
so $(I+A)^t = I+A^t$
and we know $A = -A^t$
 
@Faust: "Maximal" is an adjective; "maximum" is a noun.
Right, @Baymax.
 
I meant in the context of math ll
 
6:16 PM
@TedShifrin Maximum speed...
 
so $(I+A)^t = I-A$
 
I think that's a misuse of the language, @Semiclassic, but so be it.
 
@TedShifrin maximal is more like with respect to a property
 
oh hey
60
Q: "Maximum" vs. "maximal"

nickfWhat is the difference in usage between maximum and maximal? When would you use one or the other? Maximum can be a noun or an adjective: This is the maximum it can be set to. This is the maximum value. whereas "maximal" is always an adjective: This is the maximal value. Is this ...

 
so $det(I+A) = det(I+A)^t = det(I-A)$
 
6:17 PM
for what it's worth
 
@TedShifrin
 
Maximum is using the $<$ property, Leaky, so that's worthless.
Right, @Baymax.
 
yeah thats not exactly what i was asking but its my fault for asking in an english context
 
BTW, @Baymax, I added a comment on the question you linked. The original answer to c) was extremely unclear.
 
so nexrt?
 
6:19 PM
Heya @Alessandro!
 
why are we focussing on 2x2 non singular skew symmetric , trending is can we find a 2x2 singular skew symmetric matrixx?
@TedShifrin
 
Like say we have a cut vertex set theres a diffrence between a maxium cut vertex set and a maximal cut vertex set
 
@Baymax: Look at my comment.
I have no idea what you're talking about, @Faust.
 
I am dreamy
I got that
 
6:21 PM
@TedShifrin Neithier do i =)
 
@TedShifrin $2 \times 2$ blocks of these and a block of zero down the main diagonal?
how can i write this
will the system be a 4 by 4 matrix?
 
@TedShifrin do you happen to have any homomorphism between groups questions or categorizing abelian groups in terms of cyclic ones lieing around in your pockets i can have?
 
Or larger, @Baymax. Think of a block matrix $\begin{bmatrix} A & B\\ C& D\end{bmatrix}$ with $2\times 2$ blocks.
 
flurp this ill just write it out
 
@Faust: Consider $G = \left\{\begin{bmatrix}a&b\\0&c\end{bmatrix} : ac\ne 0, b\in\Bbb R\right\}\subset GL(2,\Bbb R)$. Let $H$ be the subgroup of such matrices with $a=c=1$ and let $K$ the the subgroup of such matrices with $b=0$. Prove that $H$ is a normal subgroup of $G$ and that $G/H \cong K$.
 
6:26 PM
$|\frac{2x+1}{x+3}-1| = |\frac{x-2}{x+3}| < \epsilon$
$|x-2| < \epsilon|x+3|$
suppose $|x-2|<1$ (we set $\delta = 1$) then $|x+3| < 6$
So $|x-2| < 6\epsilon$
 
This doesn't quite fit what you said, @Faust, but it's important. Suppose $G$ is a group, $a\in G$, and the order of $a$ is $n$. Prove that $\{k\in\Bbb Z: a^k = e\} = \langle n\rangle\subset\Bbb Z$. Let $d=\gcd(\ell,n)$. Prove that the order of $a^\ell$ is $n/d$.
 
therefore $\delta = \min(1,6\epsilon)$
 
Yes it has zero determinant @TedShifrin
 
@Meow. You're not following my rules. Do NOT do anything with $\epsilon$ until the very end. I won't even look at that.
@Baymax: But if you make $B=C=0$ and both $A$ and $D$ nonsingular, then what?
(Don't do determinant. Think row rank.)
 
okay
ignore everything except the first step
and the third
 
6:29 PM
We had that before.
 
sure @TedShifrin but will sleep a bit
 
Night, @Baymax!
 
night!
Any connection b/w row rank and singularity?
 
@Meow: The point of mathematics is to write clear, correct sentences, not flurp around. That's one of the reasons I wrote that handout — to give my students a paradigm to follow, not to ignore.
 
then $|\frac{x-2}{6}| < |\frac{x-2}{x+3}|$
 
6:30 PM
@Baymax: To me the definition of singular is in terms of row rank.
 
I'm so excited to fail my Stats midterm (again) this Tuesday :D
 
@Dragneel: You could be excited to ace it instead.
 
Hmm. That never occurred to me.
 
any reference @TedShifrin to think the connection b/w row rank and singularity
I think if row rank < n then it is singular?
 
so yeah we wanna put a bound on $|x+3|$ so we just get $|x-2|$ and so bounding $|x-2|$ is sufficient
so we let $|x-2| < 1$ so that $|x+3| < 6$
 
6:32 PM
perhapsbecause of identical or identical upto multiplicity rows
 
@Baymax: Right. As I said, that's the definition I give and most books give. Nonsingular means row rank n (for square matrix, of course, in both cases).
 
then we get what i said above
 
@Faust By categorizing abelian groups in terms of cyclic groups, do you mean applying the classification of finite abelian groups?
 
And it's irrelephant, Meow.
 
what is?
 
6:33 PM
@Meow: What you just wrote.
That's what I assumed, @Tobias.
 
so if $B $ and $C$ are non singular then $A$ is non singular?
 
what do you mean
 
Huh? @Baymax
 
aah
 
You've forgotten that you already understood your mistake an hour ago, @Meow.
 
6:34 PM
broke there a bit
 
i did?
 
nods
 
now@TedShifrin
 
@Faust In that case, to start: Let $G$ be a cyclic group and consider the set $H$ of all homomorphisms from $G$ to $\mathbb{C}^*$ as a group with pointwise multiplication. Show that $G\cong H$.
 
i need more practice with this
its all so confusing to me
 
6:35 PM
Remember you said inequalities flip?
Yeah, you definitely need practice.
There's a reason I wrote so many exercises :P
 
is this a normal thing to have trouble wiht?
 
Yup.
 
or am i just bad at it for whatever reason
 
After 10 times teaching the course, I wrote those notes for a reason, even though I did the stuff in class on the board.
You need to trust me and try to follow the outline every single time. And don't screw up inequalities.
 
@TedShifrin $\begin{bmatrix}
a & b \\
0 & c \\
\end{bmatrix} \begin{bmatrix}
1 & b' \\
0 & 1 \\
\end{bmatrix} \begin{bmatrix}
\frac{1}{a} & -\frac{b}{ac} \\
0 & \frac{1}{c} \\
\end{bmatrix}=\begin{bmatrix}
a & ab'+b \\
0 & c \\ \end{bmatrix} \begin{bmatrix}
\frac{1}{a} & -\frac{b}{ac} \\
0 & \frac{1}{c} \\
\end{bmatrix}=\begin{bmatrix}
1 & (ab'+b)(-\frac{b}{ac}) \\
0 & 1 \\ \end{bmatrix} $ since $(ab'+b)(-\frac{b}{ac}) \in \mathbb{R} $
i think i amy of done a lot fo work for nothing cause i ddint define a homomorphism
 
6:38 PM
was that incorrect @TedShifrin
 
OK, so you verified it's a normal subgroup, @Faust.
I still don't understand, @Baymax. Are you talking about the block matrices I wrote down way up there? ^^
 
yes
 
now i need to find a homomorphism s.t $ker \phi = H$
?
 
The four blocks are totally independent of one another.
Homomorphism from whom to whom, @Faust?
 
$ \phi : G \to G/H $
?
 
6:40 PM
Um, no.
 
but isnt that how you use the FIT?
 
I don't know what those letters mean, but reread the question.
oh isomorphism
 
First isomorphism theorem
 
No, it's not :)
 
can i use that fact that its a normal subgroup somehow?
the only theorem i know won't work cause K isn't normal
 
6:44 PM
Forget for a moment that $K$ is given to you as a subgroup.
 
i think you want me to $ \phi: G \to K $ s.t $\ker \phi = H $
 
Precisely.
 
so finding H was normal was a waste of time lol
 
Indeed, you could have finessed that by doing the second thing, right?
There are lots of math exercises like that.
 
well once i figure out how to constuct $\ker \phi =H $ i show containment both ways it and it gives H is normla in a line
 
6:47 PM
Most likely you're supposed to just state that kernels are normal subgroups ... unless you're specifically asked to prove that statement. :)
What else do you need about $\phi$ to apply FIT?
 
$\phi :\begin{bmatrix}
a & b \\
0 & c \\
\end{bmatrix} \to \begin{bmatrix}
a & 0 \\
0 & c \\
\end{bmatrix} $ ?
but then im not sure i get the right kernel
 
Well, work on it. But answer my question. What do you need to check on $\phi$?
 
$\phi :(\begin{bmatrix}
a & b \\
0 & c \\
\end{bmatrix} \begin{bmatrix}
a' & b' \\
0 & c' \\
\end{bmatrix}) $ looks like a homo morphism
surjective
 
Right, you need to check it's a homomorphism and ?
There you go.
I'm done. You're on your own ;)
 
surjective is easy
and i think showing its homo morphism will be easy to
but the kernel im not sure
 
6:52 PM
Really?
 
it looks like the $\ker \phi = \begin{bmatrix}
a & b \\
0 & c \\
\end{bmatrix} $ for all non zero b
 
You need to review what a kernel is.
 
but i want a and c
dammit
nvm im an idiot
thanks =)
 
You're welcome. I gave you another one to practice on, and Tobias gave you one too :) Happy studying.
 

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