Also you could've totally started from 4 :P

I thought about that, but I didn't want to further complicate it

also all those $<$ should be $\leq$

@Daminark I have the intuition of the eigen values of a transformation as the scaling that transformation performs upon something. Because of this I imagine a unit matrix multiplied by a scalar; which has only diagonal elements.

I also thought about that, but I also though meh it holds with $<$ so why type extra

wow

I cant imagine a case where the eigenvalues are not on the diagonal.

I mean, your matrix might have off-diagonal components which mess it up

Try this matrix, $\begin{pmatrix}0&1\\1&0\end{pmatrix}$

@TedShifrin your damn book is forcing me to wake up and turn my brain on! grumble

Characteristic polynomial is $t^2 - 1 = (t+1)(t-1)$

So eigenvalues are $\pm 1$

Neither of those numbers are on the diagonal

Veritasium's new video "The Science of Thinking" and the follow-up "How Should We Teach Science?" are good watches.

@MickLH Such is what happens when you're working with Ted's book

As well as the book he's basing this on, Thinking, Fast and Slow.

(Well, a good read rather than a good watch.)

"Book", "watches"

Oh lmao

And I suppose "Sweetwater" from the Westworld soundtrack is a good listen, though it's not very relevant

I have yet to hear of a food being described as a "good taste"

On that topic, I got this song stuck in my head randomly yesterday and I can't seem to burn it out, I listened to it again today and it's still there!

RIP my registration for smooth manifolds disappeared

And for text/performance

Tfw glitched

@AkivaWeinberger is mathematical induction in other words mean proving a single input for a function corresponds to its desired output, and afterwards proving that any (input +1) corresponds to a desired output based on the result we got proving the first input?

Apparently people seeing their classes already had a system-wide glitch

Whoa, Thinking, Fast and Slow is available for free online?!

(I am not a lawyer. Do not follow any of my legal advice.)

@CausingUnderflowsEverywhere I guess, yeah

@CausingUnderflowsEverywhere It doesn't have to just be +1, it could be any systematic transformation

@MickLH Explain?

Oh, like for showing something's true for all odd numbers, you add 2?

That's just adding 1 in disguise

You replace $2n+1$ with $2(n+1)+1$

Think it's time to introduce transfinite induction?

oops

those arrows are backwards

fixed

@Daminark No.

@user400188 That's probably just confusing for someone who doesn't already have a good grasp on how induction works (or how to read logical symbols)

I didnt have the best gasp beofore I read it

But, yeah, it is kinda neat.

I just knew an english definition, which was kind of vauge

@AkivaWeinberger Ok using the notation where $f^n(x)$ represents $f$ iterated $n$ times, you could "disguise" the +1 into that

whats a land and for all

Do you not have LaTeX on? (It's "logic and")

yes I do have it on, I just didnt want to call it A with no strike and upside down A haha

"If it's true for $0$, and if whenever it's true for $n$ it's true for $n+1$, then it's true for all $n$"

is what it'd be in English

Now lets say that function $f$ happens to be extremely horrible (lets say ... SHA-256) then I'd have trouble calling it just a "masked addition" because it seems that you could fit anything into that pattern, though I guess that's justified since induction is so powerful

Strong induction is also a thing (though it can be proven from the normal version)

@AkivaWeinberger wouldn't "whenever its true the other is true" translate to the biconditonal $\iff$ as opposed to $\implies$ ?

I guess what I'm really saying is, induction can still be applied even when there's no nice closed form expression :P

No @user400188, that's a one way implication

@user400188 "Whenever I'm hungry, I eat." I could still eat when not hungry.

Ah I see

Well I still like the symbolic definition. The whole thing tells you what induction is; and the brackets tell you what order the steps go in when you perform it.

Its a fall back if I ever forget the correct way to do things.

Another way to think about it: You have an infinite ladder. You know you can reach the bottom rung. Also, if you can reach rung $n$, you can reach rung $n+1$. Induction says you can reach any rung.

Oh that's a good metaphor

You can reach the millionth rung, for example (though it may take a while).

Nice

Indeed. That is the expanation I would give when I was talking about it for the first time to someone.

So now, generalizing, say you have an arbitrary well-ordered ladder... :P

I'm gonna keep that one in my pocket: "Prove that you can step on the ladder (base case), now prove that you can make one step up the ladder (induction case). You've now proven you can reach the top of the ladder."

You can't reach "the top", since an infinite ladder wouldn't have a top, but each individual rung is reachable.

Eh it's a metaphor, you can just disclaim it for infinite ladders

Anyway: i think its a good time to point out a simplification of my epression for induction.

hi @Akiva

@Daminark …And if you can reach every ladder before a limit ladder, you can reach a limit ladder?

do you go to any math circles?

@ZachHauk No, I don't think I know what those are

Did Causing leave the conversation?

oh :/

@AkivaWeinberger It's like a euclidean distance, except it equals $r$.

fails to keep a straight face

Success.

My work here is done. Have a good existence, earthlings!

@AkivaWeinberger we just started trig and everyone asks me how to compute sine and cosine w/o a calculator

however, they get an answer that is meaningless to them :]

Make sure to remind them to eat their pies two at a time

@MickLH we don't even learn radians here..

lol

The ancients used the half-angle formulae to get values for integer multiples of $3^\circ$

that's cool

and then linear interpolation to get a good approximation of other degrees (though I suppose you could divide $3$ in half further for better accuracy)

unfortunately we don't really cover any formulae besides law of sines and cosines

So they teach you that trig is a thing and leave it at that?

Basically @Akiva

Well no

And then everything changed when series expansions were discovered. (Especially for logarithms, which were also a hassle to compute)

we learn how to use it to find angle measures and stuff

Also, the short answer is, "You don't compute it, you bought a table from someone who computed it for you!"

I guess I'm being silly, that's pretty much the main point of trig. I guess the rest of its use is more like a happy accident

'Cause the real way most people did trig was just to look it up in a table.

The thing about tables is that the hard work of actually computing things only needs to be done once.

I got a new notebook

lol sorry if that's not exciting

@AkivaWeinberger I'm doing some linear algebra

here's a question Ted proposed to me during one of his exercises

Find a linear map satisfying $A^2 = I_2$ such that $A$ does not preserve distance

@ZachHauk sorry, I was trying to learn about Jordan forms and my adderall kicked in lol

Nah it's fine :P

Oh, I have an idea...

I mean, I have an advantage because I was working on a bigger problem before

But I guess it kind of balances out cuz you're smarter

Hm. Does $\begin{pmatrix}1&2\\0&1\end{pmatrix}$ work?

Wait, that doesn't make sense

Never mind.

Just think geometrically

Whoops. $\begin{bmatrix}1&2\\0&-1\end{bmatrix}$

what?

oh

Missed a sign.

let me check

What geometric operation is that?

In fact, $\begin{pmatrix}1&a\\0&-1\end{pmatrix}$ should work...

So on a real finite dimensional Euclidean space, a linear transformation is angle preserving if and only if it is a scalar multiple of an orthogonal matrix

One of the problems on our first pset this quarter was to characterize this

@SimplyBeautifulArt I only skimmed it, but the first comment ("It's absolutely countable. You can describe all of these ordinals using a finite string of symbols in a finite language (specifically, English). There are only countably many such strings") seems accurate

Just order them by length of description, and alphabetical order when the descriptions have the same length

:-/ okay

@AkivaWeinberger ah, it's a shear reflected over $x$-axis

That's lame

yep, you're right

Have you seen orthogonal matrices @Zach? How long have you been doing this?

doing what

the problem?

Bye everything. Thanks for the help @Daminark

I already knew the answer, and for a bout an hour I've been playing my games and eating food

@user400188 Bye!

instead of actually continuing with the problem

@ZachHauk I was thinking, $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ (a flip along the $y$-axis) works. But it preserves distance.

You got that quicker than I did, with absolutely no context

Well, if it works, $E\begin{bmatrix}1&0\\0&-1\end{bmatrix}E^{-1}$ (some transformation, a flip, undo the transformation) should work, right?

Yes it does

Fun fact: Typesetting matrices is annoying.

That's what I was thinking

So, anyway, I made $E$ a shear.

I said shear also

In fact, wouldn't it be any reflection?

In the middle? Yeah, I suppose

What's a shear?

@ZachHauk In fact, this all was somewhat related to the chapter of Ted's linear algebra book I was reading. (Projection maps.)

@Daminark Turns a rectangle into a parallelogram, intuitively.

That's the text I'm using

More formally, $\begin{bmatrix}1&a\\0&1\end{bmatrix}$. @Daminark

The determinant of a shear is always $1$

^

Oh, we saw smth like that when we did Jordan form

So any linear transformation with determinant 1 is a composition of shears, right @Akiva

(For $2\times 2$ matrices)

But we didn't talk about it geometrically at all

And HK doesn't geometry at all

The codes for bmatrix and pmatrix are backwards. B is a soft sound, P is a hard sound!

Stupid bouba/kiki effect.

(Yeah, I know it's for parentheses and brackets. That only means there's something wrong with those words, too.)

@AkivaWeinberger So a countably finite amount of operations that always map $\alpha_n$ to $\alpha_{n+1}$ for $\alpha_n<\alpha_{n+1}<\omega_1$ will result in some $\alpha_\beta<\omega_1$?

I think parentheses comes from "parent" but I'm probably pulling it out of my ass

Yeah probably not because it sounds Greek

anyways, @AkivaWeinberger can I borrow your brain for the rest of my life?

Sorry, edited.

@ZachHauk Good life goals man

@ZachHauk Jokey answer: "I kind of need it..." More serious answer: It's not being smarter. It's just more experience in this area. (Having thought about this sort of thing literally earlier today helped.)

Can we get that brain/computer plug from the matrix going?

$\begin{bmatrix}\text{Like}&\\&\text{this?}\end{bmatrix}$

I've brought this upon myself :'(

I think my brain basically has two modes when it's idle. Thinking about math, or playing a song in my head / thinking about music.

√(♫) = Akiva