Whoa. I can retire on that, @Zach. Oh, wait a minute.

hah

And all ends up in the sum...

No, I'm not asking for formulas in terms of entries of the matrix, @Maks.

The crucial property of matrix multiplication is that it is associative, even though it is not commutative.

Pizza for dinner!

@Vishal: I taught for more than 40 years. So I have experience with questions and annoying students :P

and not frozen pizza ;)

@kfrz note that in that scenario you'd have both [[1,2]] and [[2,1]] as elements of your array

Good, @Zach. And I hope no pineapple on it. Yuck.

@TedShifrin Oh yes

I can write it as A(BX)

why does everyone hate on pineapple pizza?

i haven't even had it

Now think, @Maks :)

And BX has at least a non-trivial solution

because it's wrong.

Acuatlly, it has inifnite solutions

i'm sure it's good

right ?

$BX=0$ has ... yes.

@TedShifrin this made me go on shady parts of net and download munkres. I guess I owe you 20 xD

@Alessandro and Ted get to be arbiters of taste re Italian food :P

Then A(BX) = 0 has to have

infinite solutions too ?

So 1+2 would be a different input than 2+1

Well, @Maks, what is $A(0)$?

one of the kids in my school; "Alessandro"; has a family who owns a pizza restaurant

@VishalGupta where shady parts of the net is the first link on google :P

@TedShifrin This is what future has in store for me then xD

and actually, before I was born, my whole family went there, and got food poisoning

So, tidbit of the day

@TedShifrin A times the null matrix?

I dont get that notation

@Maks, I meant $A$ times the $0$ vector (you put in an $X$ with $BX=0$, right?)

@AlessandroCodenotti its that easy? i didn't google, went straight for the kill :p

@TedShifrin Then its 0

In my books I use different notation for the 0 vector and the 0 matrix. I suppose I should here, too, but it hasn't come up.

Right, @Maks. So $(AB)X = A(BX) = 0$ has amongst its solutions all the vectors $X$ with $BX=0$.

@VishalGupta I was wrong, it's the second link apparently. You don't even need to write the title of the book, just "Munkres" in the search bar

@TedShifrin We use the arrow for the vector and $0_{m \times x}$ for the matrix

@Ted By the way, I found the Riemann Sphere pretty interesting. anywhere I can read up on that?

OK, @Maks :)

There's some discussion of it in the last section of that chapter of mine you're reading, @Zach.

@TedShifrin Then again, it has infinite solutions, or am I wrong ?

If you agree with the sentence I wrote, then why do you ask that?

Because there may be something I'm not seeing that doesnt allow it to have infinite solutions hahaha

Then you do not agree with what we said?

Just to make sure

@TedShifrin I do

And I understand it

The justification is that $A\vec 0 = \vec 0$.

But I want to make sure I'm not missing anything

You know, i'm still confused about what actually happened in the shining

that was a weird movie.

@AlessandroCodenotti when google knows what we are thinking. scary times :p

@Vishal: And, worse yet, the government and Putin know too.

Facebook already puts up ads based on spying.

adblock for life :)

I did decide to encrypt the hard drive on my desktop computer, finally.

my HDD has always been encrypted

especially my server's HDD

oh wow, I do not even know how to do that

I wasn't sufficiently paranoid until Nov 20, @Zach.

Macs let you opt to do it automatically, @Vishal.

@VishalGupta You can use Windows' Bitlocker tool

if you're using windows

@kfrz the usual tactic for counting stuff like this is 'stars-and-bars'. Wikipedia almost certainly has a page.

and if you're using Linux, it should be pretty easy

yeah, i am on good old windows. will search for it

@Semiclassic: Unless I call it stripes-and-bars :P

when I was setting up my server, they actually asked "Would you like to use encryption" and I was like "why not"

Pffft

that's weird... so you know how the end of a colored pencil is usually like flat

no eraser, just flat

Counting in that fashion isn't too hard once you see the trick

i was looking for the right shade of blue, and so i found it

and both sides were flat

isn't that odd...

Maybe these pencils need a pencil sharpener.

some pencils are sold like that

I bought some when I was your age that needed sharpening.

Saves points breaking in transportation.

@Ted when did you first work with computers?

During the age of punch-card programming? ;)

By contrast, when order doesn't matter it's a lot harder to count ways to put integers together

I bought my first Mac in 1988. I think I had one in my office just before that.

Oh, I actually programed in FOCAL in high school (1969 or so) and a bit of Fortran, too. But I avoided computers until I could put one on my desk.

@ZachHauk its not available on win home sadly. will have to look for something else

So yes to tape and to punch cards. And that drove me away from it for 15 years.

hey @TedShifrin you can probably answer me this

heh, nowadays I only program in assembly

Barely says hello ... just throws me a question. Typical.

unless it's something that i need to script

@TedShifrin Hi :D

sorry haha

Uh huh. Hi.

So we know by application of zorn lemma and some logic that $Nil(R) = \cap P$ where P is the intersection of all prime ideals. We can prove that

I was wondering is there a geometric way to see this ?

Karim: Do not ask me commutative algebra.

A more conceptual way ? Or is it just algebraic

I don't remember enough of it.

Eisenbud's book should discuss the geometry behind everything.

@Ted oh, me and Balarka were talking about geometry stuff a little bit ago

like, about paraboloids

Have you done Hilbert's Nullstellensatz? That should link this to geometry.

it was an interesting conversation

oh okay @TedShifrin I will check EisenBud more. No not yet

Yes, definitely look at Eisenbud.

@ZachHauk you mean "I and Balarka" :P

No, @Vishal, "Balarka and I" !!

If you're going to correct, correct correctly.

It's society's fault for not teaching what subject and object pronouns to use!!!

@TedShifrin I did not know the difference there. How is it so?

You've been taught, I'm quite sure, @Zach.

I know some people who use "Whom" in the incorrect circumstances, probably to sound smarter

@TedShifrin what is Hilbert Nullstellensatz ?

i.e. "Whom did this?" instead of "Who did this?"

Because in every language I can think of (English, French, German), it's "X and I," not "I and X."

@TedShifrin how are you doing btw ?

Whom is a direct or indirect object. Do NOT use an object as a subject! Geez.

Go look it up, Karim.

okay

@TedShifrin is it a rule? Ich und Tobias seems ok to me..

@Ted ahhhhh I didn't say it was me!!!!!

Yes, it's a standard custom, @Vishal ... sicher.

Oh dear.

I also forgot to say hi to everyone

Hi @ZachHauk @VishalGupta @Semiclassical

@Zach: Well, saying "It is I" has gone completely out of use because it sounds so stilted.

I should lure a set theorist in this chat with some trick. Random thought of the night.

Grammar is a lot more subtle than semantics

I usually say "Speaking" on the phone rather than "It is he" ... :P

@AlessandroCodenotti what do you want to discuss in set theory ?

LOL @Alessandro

I know a teacher at the high school who speaks only in third person

"We shouldn't eat our pencils, should we, @Zach?"

e.g. "Mrs. Teacher wants everyone to sit down and take out their homework"

Oh, wait, that's 1st person.

that's 1st person plural

Oh, referring to herself in third person? That's obnoxious.

@Adeek a proof that's giving me a few headaches and is very set theoretical

However, I do have a teacher right now who speaks as you just did.

But I suppose my schtick in class (and all over the videos) of saying "And so we ask ourselves, 'Self' ..." gets obnoxious, too.

@AlessandroCodenotti what is it ?

So maybe these people are trying to inject humor. Uh huh.

that is, like "Maybe we should sit down and do our homework, okay?"

There's only one time I can think of when I'd use third person

@TedShifrin I might get a dog this summer.

@TedShifrin never thought of it. a quick google search doesn't show a lot.

@Adeek get a cat

Karim: Do you not already have enough responsibilities and expenses?

So many people are making fun of my picture that I'll just replace it with my cats >:)

yeah @TedShifrin yeah I guess a dog would add a lot of expenses per month

Especially if he/she eats.

This is addictive, but I gotta sleep. Good night to us here in Europe and a good day to our American friends

Hello !

@Ted and to that I ask you... what Dog doesn't eat?

Gute Nacht, @Vishal.

Guten nacht @VishalGupta

@Zach: That was humor lite :)

Lol there was this problem that everyone was fretting over

Which is a complete troll

@VishalGupta Bis morgen :P

No name-calling, @Daminark.

@Adeek Gute Nacht, aber danke :)

@Adeek this theorem, which is needed in the proof of another theorem (the Sierpinski-Erdös duality theorem)

@Adeek I meant Gute not Guten and Nacht should be with capital N

@VishalGupta oh I see been so long since I written german thanks

Finding a good color in this colored pencil box is like finding a needle in a haystack... that has been put through @Ted's projective duality machine

But... I like name-calling! Especially wrt problems!

@Zach: Put it back through again :P

(Ctd from above) And that's when my students get a quiz back. I'll always frame it as "the grader chose to focus on this" even if I'm the one who graded that problem for the class

will i need to rent it from you?

Perhaps.

It's a bit of a pious fraud, of course: they know my handwriting

But I still stick to referring to "the grader" as distinct from myself

@Vishal: Here is one discussion.

@TedShifrin I found this confusing footnote in Spivak, after he wrote down the definition of the distance between two points in the plane : "The fastidious reader might object to this definition on the grounds that nonnegative numbers are not yet known to have squre roots. This objection is really unanswerable atm - the definition will just have to be accepted with reservations, until the little point is settled.'' What on Earth did he mean ?

@Mahmoud: He means that it'll take stuff in chapter 8 to prove that every positive number actually has a square root.

@TedShifrin Question is : How is that in need of a proof ? :o

Well, it's far from obvious!

In high school math (and typical college calculus) one just assumes it, but it's far from obvious.

If our world were just the rational numbers, it would be most false.

Everything that isn't true by definition should be provable

Even falsehoods?

Bah

Just drop the principle of non-contradiction

And then @Semi's statement becomes true

I'd amend that to "Any true statement..." except that them I'd run afoul of Godel

Uh huh ... And you'd get thrown out trying to score.

ahhh

i have to draw my hand now

how do i do that???

And : Why is $f(x+1)$ a left shift and not a right one ? I mean my humble intuition tells me that adding $1$ makes your number go to the right.

Trace it, holding your pencil in your other hand?

not allowed to trace for this

i suck at art

I do too, Zach, except for drawing math pictures.

To get f(0), which x should you plug into that? @Mahmoud

But yeah the problem was to show that if $x_1,\ldots,x_n$ are unit vectors in some normed linear space, assume that for some $\epsilon \in (0,\epsilon)$ such that $\|\sum_{j=1}^n \lambda_jx_j\| \le (1+\epsilon)\max_{1\le j\le n}|\lambda_j|$

No, that's not right, @Semiclassic. Let $g(x)=f(x+1)$. To get $g(0)$, what $x$ do I plug into $f$?

Then prove that $\|\sum_{j=1}^n \lambda_jx_j\| \ge (1-\epsilon)\max_{1\le j\le n} |\lambda_j|$, followed by showing linear independence of the $x_j$

Is there going to be a point in mathematics where the preparation needed to start research takes longer than the human lifespan?

Seriously, @Daminark, $\epsilon\in (0,\epsilon)$?

Oh wait

$\epsilon \in (0,\frac{1}{2})$

@Ted LOL

I mean "plug in" to be literal replacement of x in f(x+1)

My typos are just...

It's too confusing without naming both functions, @Semiclassic @Mahmoud.

It's $x=-1$, so $g(-1)=f(0)$

So the graph of $g$ is obtained by shifting the graph of $f$ one unit to the left.

hmm

i need to draw a word that makes me happy

$\mathbb{MANIFOLD}$

Lol

$\mathbb{NO}$

@Daminark: Did the problem say distinct unit vectors? Else, linear independence ain't so true.

;)

i've used \mathbb for most things but i saw people using \Bbb and it was shorter

so i just used that

anyways, pizza eating time. adios

It didn't explicitly say "distinct", but I think the requirement of the sum would break for the right values of $\lambda_j$

LaTeX gets angry with the "old" AMSTeX Bbb, but MathJax and ChatJax don't protest.

Another way to put it: To get the same output from g(x) as f(x) you need to 'undo' the effect of x->x+1.

I dunno, @Daminark. Honestly, I'm not thinking about it at all.

Fair

@TedShifrin nothing concrete again. sounds either a politeness issue or an issue of expecting verb to follow first person pronoun; not a rule unfortunately. and yeah, I could not resist coming back for more :p

I mean it is possible given the second sum expression to directly prove linear independence

It's basically a rule of standard usage, @Vishal, at least in English and French. I will continue to research it. Go to sleep!

One of the problems was about proving that only the $\ell^2$ norm gave a Hilbert space

Part a of the problem actually used manifolds in the finite dimensional case to show it was impossible for $\ell^1$ and $\ell^{\infty}$

It was a nice argument

Hmm, all you need is a violation of the parallelogram law.

In the general case, yeah, and you just use $(1,0,0,\ldots)$ and $(0,1,0,\ldots)$

But the idea in the special case was that given any inner product $[\cdot,\cdot]$, you could express $\|x\|$ in that as $\langle Px, x\rangle$ for some symmetric, positive definite matrix $P$

Where $\langle \cdot,\cdot \rangle$ is the standard Euclidean inner product

You mean $\|x\|^2$, of course.

Right, yeah

Sorry, I make a lot of typos

But yeah, then you consider the smooth function $Q(x) = \langle Px,x\rangle - 1$

Anyhow, I'm outta here for now ...

The unit ball is the 0 set of this function, and thus a smooth submanifold of $\mathbb{R}^n$

But for $\ell^1$ and $\ell^{\infty}$, that's impossible due to the sharp corners

Thought you might appreciate it :P

Well, see you @Ted!

hows it going @Adeek