Mathematics - 2015-05-25 (page 4 of 4)

Soham Chowdhury

6:00 PM

Zero

Clarinetist

Oh really! Props! @SohamChowdhury

Soham Chowdhury

I know some theory, is all.

Clarinetist

Congrats to you on finding this stuff on your own :)

I remember one time...

I went to a gathering and there's this church that I used to be a part of - everyone was just used to pop-music-like structures, 4 chords and everything

Soham Chowdhury

I finished learning this today, after almost exactly one year of starting out.

Clarinetist

The piano we had was horribly out of tune. I remember hearing someone play To Zanarkand (Final Fantasy music) on the piano and just thinking, ugh, that sounded disgusting

Soham Chowdhury

6:02 PM

I have a six-string acoustic, so I did a leetle bit of 8ve transposition on the bass notes.

Clarinetist

(this was for a talent show)

and then I came after

and started going all Henry Cowell, if you know what I mean :P

My sister was not happy :P

Soham Chowdhury

Can you listen to what I linked you to, please?

Clarinetist

Yep, it's pretty @SohamChowdhury

Just finished it

Soham Chowdhury

Wait a sec.

Have you heard of Animals as Leaders?

Clarinetist

Nope, haven't

Soham Chowdhury

6:09 PM

Listen to this and tell me when you're done. I have an awesome surprise for you after that.

ok? @Clarinetist

Clarinetist

K

Soham Chowdhury

Tapping = piano on a guitar <3

Clarinetist

Ready

Soham Chowdhury

Piano cover!

Clarinetist

OMG

O_O

Soham Chowdhury

6:21 PM

2:36, man. How do you like it?

:)

Vrouvrou

@TedShifrin please can you tell me if this sequence

is bounded in $W^{1,p}_0$

Clarinetist

@SohamChowdhury Hah, crazy

Soham Chowdhury

You want some ultra-rhythmic metal?

Theses have been written about this band.

Clarinetist

@SohamChowdhury I'll have to stop after this one, sorry :P Gotta do some linear algebra

Soham Chowdhury

Then skip that and give this a shot. youtube.com/watch?v=jdYJf_ybyVo

Here's the metal if you'd prefer that: youtube.com/watch?v=jdYJf_ybyVo It's actually mindblowing (and angry!).

Have fun. What book are you using?

Clarinetist

6:25 PM

@SohamChowdhury I'm reviewing out of my stats book. Starting a M.S. Statistics in August and I'm pretty rusty plus I have to learn a lot of new things

Soham Chowdhury

Grab Axler if you'd like a "real" linalg course. It's sorta fun.

Clarinetist

@SohamChowdhury I intend on buying Axler when I have the time. I learned out of that 3 years ago but I don't have a copy of the text myself. You know, he did release a new edition in January

Soham Chowdhury

it's free online, legally i guess

Clarinetist

@SohamChowdhury I like having physical copies. I suck at learning via a computer screen

Not sure why

Soham Chowdhury

I like them too. I never did get very far with the book, though. I dropped it midway to start with a "real algebra" book and learn linalg along the way.

Do remember to look at the first link I gave you later, you'll love it.

Clarinetist

6:30 PM

@SohamChowdhury I can see how that would work. It does frustrate me a bit that a lot of textbooks I've read seem to treat abstract and linear algebra as two totally different subjects, but I really could see them being integrated together (in a course covering fields, I suppose)

Soham Chowdhury

No, scrap all that, listen to the first 45 seconds of this, please.

He actually retunes strings on the fly. Which is hard.

Clarinetist

WOW

Haha

You have to tune it just right too

Soham Chowdhury

Acoustic guitar = drum + strings, lol

Clarinetist

And he sings while doing that simultaneously...

-____-

:P

Soham Chowdhury

Can you make harmonics on a clarinet easily?

Clarinetist

6:36 PM

@SohamChowdhury You can, but strings have a much easier time

Soham Chowdhury

Of course.

Let me just say "chords" in a superior voice while I'm at it

Clarinetist

Hah

Ever heard of throat singing?

Soham Chowdhury

You have multiphonics, yes

But you can hardly do an Em9.

Clarinetist

I hope you've seen this :)

polyphonic overtone singing - Anna-Maria Hefele

►

True, overtone singing is limited

Soham Chowdhury

I hope you've seen this: youtube.com/watch?v=7gphiFVVtUI

Clarinetist

6:40 PM

I will favorite that and check it out later

Soham Chowdhury

This too, then. Paco, McLaughlin and di Meola. youtube.com/watch?v=ADwfyxpriAM

@Clarinetist Mahn. That is cool.

Clarinetist

Linear Algebra question: if $A \in M_{n \times n}$ is singular, is $A^{T}$ singular as well?

Soham Chowdhury

@Clarinetist Singular = noninvertible?

Use dets, I guess.

Clarinetist

OH DUH

Thanks

Because $\det(A) = \det(A^T)$

Soham Chowdhury

I can't believe what just happened lol

Clarinetist

6:45 PM

$A$ singular $\Longleftrightarrow$ $\det(A) = 0 = \det(A^T)$ $\Longleftrightarrow$ $A^T$ singular

This statistics book doesn't like working with determinants for some reason, so naturally, they're the last things I think of

Soham Chowdhury

Grab Axler ASAP. Nobody says singular anymore.

Oh, and he introduces determinants in the last chapter :)

Clarinetist

@SohamChowdhury Yeah, I remember that. Do people say noninvertible or is there a better term? I never did like singular/nonsingular

Soham Chowdhury

Dunno.

Clarinetist

The only reason I've started using singular is because statistics likes using the singular value decomposition of a matrix

Soham Chowdhury

Yeah, that's like the standard name for that.

You just get used to saying SVD. :)

@BalarkaSen are you online?

Clarinetist

6:58 PM

0

Q: $A$ is positive definite if and only if $Q$ is invertible for every choice of $Q$

I proved the following theorem already: $A$ is nonnegative definite if and only if there exists a square matrix $Q$ such that $A = QQ^{\prime}$. Now I have to prove the following corollary: $A$ is positive definite if and only if $Q$ is invertible for every choice of $Q$, The boo...

fubal

is calculus == analysis?? I often read ppl talking about (on reddit especially) taking their "calculus x" lecture

anon

calculus is a subset of analysis with lower standards of rigor

fubal

do they teach that to math students or is that something for other students?

Balarka Sen

@SohamChowdhury I am now.

anon

@fubal math students take calculus, but are liable to go beyond that into analysis or other areas of math. other general students are more liable to stop at calculus, or even before that at college algebra.

Vrouvrou

7:09 PM

@anon have an idea about my problem please

Balarka Sen

analysis is calculus with steroids

anon

@Vrouvrou sorry but I'm not in the mood to relearn the things in that post on the chance I could help

LeGrandDODOM

@DanielFischer do you know how to prove that determinant is a log concave function on positive definite matrices ?

fubal

@anon so you start with calculus and then do analysis in some universities? because as I know it the first two lectures you are hearing as a math student always are analysis 1 and analysis 2

Daniel Fischer

@LeGrandDODOM Not off the top of my head.

Paul Plummer

7:13 PM

@anon Want to take a stab at most recent question? :P (sort of a joke, although you might have fun attempting it)

Mike Miller

@DanielFischer: Did you see the harmonic function question I pinged you over last night?

Daniel Fischer

@MikeMiller No, and I didn't see the ping. Link ready?

Mike Miller

Nah, I'd have to hunt through the logs, it was posted here.

Let me check if it showed up on main.

Oh, I just noticed the author is here

Did you solve that question @fubal?

fubal

@MikeMiller no I didn't yet, I continued repeating the pde lecture and then did exercises

The question was:

nTuply

Hi all.

Clarinetist

7:22 PM

Is my comment correct?

So basically, it's just: Assume $Q$ is noninvertible. Then there exists $v \neq 0$ such that $Q^{\prime}v = 0$, or $v^{\prime}QQ^{\prime}v = 0$. Take $A = QQ^{\prime}$. Then $A$ is not positive definite. Then use the contrapositive? — Clarinetist 4 mins ago

nTuply

Can anyone please answer this question for me? I'm really struggling with L.A

1

Q: Understanding the difference between Span and Basis

I've been reading a bit around MSE and I've stumbled upon some similar questions as mine. However, most of them do not have a concrete explanation to what I'm looking for. I understand that the Span of a Vector Space $V$ is the linear combination of all the vectors in $V$. I also understand tha...

linear-algebra vector-spaces span

fubal

ok, to anyone with some understanding of pde. Let C = {x in R^n, |x| < 1, x_n < |x|*a} for some 1>a >0 (like a ball without a cone I would say). Now u is a solution to the dirichlet problem, u = |x| on the boundary of C, and u harmonic inside C. Why is u(x) >= |x| for x in C?

Mike Miller

@DanielFischer: There.

fubal

@MikeMiller thanks for still chasing the answer!

Mike Miller

Sure, I'm interested

Ramanewbie

7:26 PM

@Sawarnik

Sawarnik

@Ramanewbie :DD

You're hippa, sure? @raman

Ramanewbie

@Sawarnik Hi, you wanted to talk to me yesterday ?

@Sawarnik No, I'm Ramanewbie.

Sawarnik

@Ramanewbie dang, then can you get him online please?

Daniel Fischer

@fubal $\lvert x\rvert$ is the Euclidean norm?

Ramanewbie

@Sawarnik Ok.

Mike Miller

7:30 PM

Yes, @DanielFischer

Ramanewbie

@Sawarnik But why the hell did you think I was Hippa ?

fubal

@DanielFischer yes it is

Sawarnik

@Ramanewbie Do you live in Paris?

Ramanewbie

@Sawarnik Yes.

Daniel Fischer

@fubal The Euclidean norm norm is subharmonic.

Sawarnik

7:33 PM

@Ramanewbie What is your first name in real life? I know that raman stands for ramanujan.

Ramanewbie

@Sawarnik Ramanewbie -_-

@Sawarnik Why would you care ? You won't know me...

fubal

@DanielFischer isn't it superharmonic?

oh damn -laplace >= 0

every time the minus damnnn. I tried that, but forgot the minus. Thanks @DanielFischer

thought it would be superharmonic

Daniel Fischer

@fubal Not sure what that minus sign does there. $\Delta f \geqslant 0$ means subharmonic.

fubal

yes, we defined that as $-\Delta f \leq 0$ means subharmonic. I checked for that and forgot the minus, thus thought it would be superharmonic

Daniel Fischer

Ah.

fubal

7:38 PM

But I have a follow up

Ramanewbie

Is there a chat for that room electronics.stackexchange.com

fubal

Later in the same proof we define $D = C \cap B_{1/2} = C* 1/2$. Let v(x) = u(2x). Then u = (1/2) * v on the boundary of D.

That is obviously true on the boundary of the cone, but I can't see why it is true on the boundy of $B_{1/2}$

@DanielFischer If you still have patience it would help me a lot if you could look over that too

Ramanewbie

@Sawarnik hippa should be coming now

Hippalectryon

@Sawarnik Ellu

Electrical Engineering

A place to talk with friends from the EE community about vacuu...

1

Ramanewbie

@hippa Oh ok thanks but how did you find the number of the room ? (15)

Hippalectryon

7:49 PM

@Ramanewbie -_-

Daniel Fischer

@fubal Hm, $\frac{1}{2} v$ is the solution of the Dirichlet problem for boundary values $\lvert x\rvert$ on $D$. If we had $u(x_0) = \frac{1}{2} v(x_0) = \frac{1}{2}$ for some $x_0 \in \partial B_{1/2} \setminus \partial C$, then the strictly subharmonic function $w(x) = \lvert x\rvert - u(x)$ would have a local maximum at the interior point $x_0$ of $C$, which cannot be. Something's wrong.

fubal

8:10 PM

@DanielFischer Why do you need $v(x_0) = \frac{1}{2}$ for your argument?

do you mean $v(x_0) = 1$

Daniel Fischer

@fubal Not $v(x_0) = \frac{1}{2}$, it's $\frac{1}{2}v(x_0) = \frac{1}{2}$. The important thing is that we'd have $u(x_0) = \frac{1}{2} = \lvert x_0\rvert$ for an interior point of $C$.

fubal

@DanielFischer yes I get it. Thanks for the help

Daniel Fischer

You're welcome.

Vrouvrou

8:48 PM

someone have an idea please ? math.stackexchange.com/questions/1298594/…

Balarka Sen

9:02 PM

@MikeMiller I read Lefschetz fixed pt thm. Truly appreciating Qiaochu's proof that $S^n$ is a topological group only if $n$ is odd.

Clarinetist

$M$ is a perpendicular projection matrix onto $C(X)$ if and only if:
1) $v \in C(X) \implies Mv = v$
2) $w \in C(X)^{\perp} \implies Mw = 0$

What is the purpose/application of such a matrix? ($C(X)$ denotes the column space of $X$)

Hippalectryon

@Clarinetist purpose as in .. ?

@Clarinetist Are you asking why projectors are useful, or why orthogonal projectors are useful ?

Clarinetist

@Hippalectryon Idk, it just seems to lack any sort of motivation in the textbook I have. I haven't found much online

Orthogonal projector, is that the usual terminology?

Hippalectryon

I don't know (I'm French). That's a quick translation

Well, projectors... project :P a projector takes a given vector and gives you the projection of this vector on a basis

Eg, on R^3, a projector on R^2 would return (x,y,0) if it gets (x,y,z)

Balarka Sen

@Clarinetist The whole point is that you'd want to get a good notion of how the basis of a given vector spaces looks like. If you have a vector space with a hermitian/symmetric form on it, then it's a fact that there is an orthonormal basis of it.

Hippalectryon

9:13 PM

^

Balarka Sen

To prove that fact, you need orthogonal projection operators.

Is that a good motivation for you?

Clarinetist

Yep that's fine. Thanks @BalarkaSen @Hippalectryon

Balarka Sen

@Clarinetist To elaborate a bit more : you can prove that given a vector space with a hermitian/symmetric form on it, there exists an orthogonal basis. However, orthonormal is way way stronger than that, which tells you that not only the basis elements are orthogonal, but $(a_i, a_i)$ for a basis element $a_i$ is either $1$, $-1$ or $0$.

It's actually quite striking : it tells you that a real vector space with a symmetric on it is pretty similar to R^n. so forms are not quite fancy tools like linear operators : they give your vector space a strong inherent structure.

Fargle

@TedShifrin My computational skills have clearly dulled a little bit. That's slowing me down. Otherwise, the notes are intuitive and the exercises frustrating in the best way.

Ted Shifrin

9:37 PM

Well, some of it's more theoretical than computational, @Fargle, but I believe everyone should be able to compute, too. :)

hi @Clarinet, @Balarka

Clarinetist

Afternoon, @Ted

Ted Shifrin

well, but not $\Bbb R^n$ with the usual inner product, @Balarka.

Fargle

@TedShifrin Yeah. I'm also just not very good at analysis yet.

Ted Shifrin

What analysis, @Fargle?

Fargle

Well, I mean, I guess it's not really analysis. I just meant as opposed to algebra--but geometry features into both, so I dunno

Whatever this is, I'm not good at it yet, but this is how I get good at it, so, I'm not discouraged.

Clarinetist

9:41 PM

This stats book I have has an abuse of notation which it didn't define xP

Balarka Sen

hello @TedShifrin

Ted Shifrin

@Fargle: It's a blend of calculus and linear algebra. Generally, the proofs are much more concrete than in any analysis or algebra course :)

Balarka Sen

interesting question from a fellow highschooler that might be closer to stuff you like to think about : i have a $C^1$ simple closed curve in $\Bbb R^2$, and i know that there is a point $x_0$ in the interior of the curve such that the straightline obtained from joining any point $x$ on the curve with $x_0$ is perpendicular to the tangent of the curve at $x_0$. is the curve necessarily a circle?

Ted Shifrin

Yes. And it generalizes to $\Bbb R^n$. That's a basic lemma for both physics and differential geometry :)

Balarka Sen

interesting!

Ted Shifrin

9:49 PM

One-line proof.

You'll encounter this in Chapter 3. :)

Balarka Sen

nice, ok.

so I guess that's another motivation for learning mult. calc. : I'd get to answer cool questions asked by my fellow classmates. :P

Ted Shifrin

This is actually a single-variable question. :)

Hint: A curve is a $1$-manifold.

Clarinetist

@TedShifrin Any physics text recommendations for when I finally decide to learn physics?

Ted Shifrin

I'm not an expert, @Clarinet. You're talking like a good freshman physics book?

Clarinetist

@TedShifrin Yeah, one with calculus

Ted Shifrin

9:54 PM

Look for Kleppner/Kolenkow. Great book.

Balarka Sen

@TedShifrin upto homeomorphism, not upto conformal things.

Ted Shifrin

Huh? It's parametrized by a $C^1$ function of a single variable.

Balarka Sen

don't know how being perpendicular to something relates to being a manifold

Ted Shifrin

Well, there are different ways of thinking about manifolds. See Chapter 6. :P

Balarka Sen

now you're behaving like RB. :(

(I kid)

Ted Shifrin

9:56 PM

LOL ... All the more reason for me to not have to show up.

Balarka Sen

boo, no. I'll email you anyways.

Mike Miller

Luckily, modern emails all have systems to block unwanted spam.

Ted Shifrin

$\exists$ email filters.

2

high-fives @MikeM :D

Mike Miller

hahaha

Balarka Sen

:(

that's mean of you.

Ted Shifrin

9:58 PM

We've already established that I'm mean.

Mike Miller

Right.

Ted Shifrin

Who's going star-crazy again?

pjs36

Not me, although I'm the one enjoying the existence of email filters.

Ted Shifrin

And I've never even emailed you.

pjs36

That's true, to my knowledge

Mike Miller

10:01 PM

I can't tell you how convenient it is to be able to block all those unwanted emails I get from Ted.

Stan Shunpike

@TedShifrin hey ted! I got a question. So I was given the equation $x^2(x-20)=y^2(y-20)$ and apparently it is irreducible. Why does simply plugging $y$ in for $x$ not work to show it reduces to $y=x$? I gather I am not understanding the difference between iff and if statements.

Ted Shifrin

First of all, you need a polynomial to be irreducible, not an equation?

Stan Shunpike

Okay, yes when you factor, there is an irreducible polynomial that's what I meant

Ted Shifrin

Certainly that curve contains $y=x$. We need to show there's nothing else.

Over $\Bbb R$, I assume?

Stan Shunpike

Yes exactly

Ted Shifrin

10:05 PM

So, write $f(x,y) = x^2(x-20)-y^2(y-20)$, divide by $x-y$, and show that what's left has no real solution.

(When you set it $=0$.)

Stan Shunpike

Okay, that's what we ended up doing. I just wanted to check and see why that made sense

Is that a common way to handle when you get an irreducible polynomial?

Ted Shifrin

In what context are you getting this?

This seems unusual for an undergraduate question.

Mike Miller

10:27 PM

@TedShifrin: The rest of the paper fell pretty quickly after that. So I went and wrote up a sketch of the talk... there's way too much here, and as far as I can tell it's all fairly essential to what I want to say :(

ᴇʏᴇs

Aww John Nash passed

Stan Shunpike

@TedShifrin Sry! Something came up didn't mean to dash off. Yeah, its really weird. Its for econ

@TedShifrin the TA just picked some functions I think because they looked nice. But he never does the problems before assigning them, so this one looked realllllly ugly.

Its been a pretty frustrating experience. We basically do the set. Make mistakes. And then have to figure out if we screwed up or the TA is wrong. We asked the prof but he just differed to the TA

robjohn

10:45 PM

@Guesswhoitis. I was gone all weekend. That is indeed sad news.

Ted Shifrin

Wb, @robjohn ... Yes, unexpected and sad.

Hi mr eyeglasses

ᴇʏᴇs

Hi @TedShifrin

I got a 'B' in complex analysis

Ted Shifrin

@Stan: The polynomial is not irreducible, as it factors nontrivially. But the variety is, as the other factor has no real solutions. Over $\Bbb C$ that would be very false.

Did that bad question show up again, mr eyeglasses?

ᴇʏᴇs

@TedShifrin No

Ted Shifrin

Good. But Still ...

ᴇʏᴇs

10:56 PM

I didn't understand most of the material anyway

I just hope I won't have to encounter too much complex analysis in the future

Gridley Quayle

Given sin^2(x)=1/8(5^0.5 + 3) how would I go about finding sin(x) in the form p+q(5^0.5)?
One method leads to the quadratic:
1280q^4-96q^2+1=0 which gives q=+_ 1/4 but I was wondering if there was a neater method.

Clarinetist

Okay, this is really bugging me. They used $C(X)$ earlier in the book to denote the column space of $X$ and now it means an "arbitrary space.... for some matrix $X$." What the heck.

ᴇʏᴇs

Which book @Clarinetist

Clarinetist

@ᴇʏᴇs It is a statistics book, Plane Answers for Complex Questions. I am using it mainly to just get the essentials of linear algebra I need before I start studying multiple regressions for statistics

ᴇʏᴇs

oh

Clarinetist

11:06 PM

I've had to correct a few things in this chapter already because I've seen a lot of this stuff before

Ted Shifrin

That makes no sense, @Clarinetist

robjohn

@TedShifrin Hey there... I am back in town, but I think we may have family plans for the Holiday.

Ted Shifrin

Write $(a+b\sqrt 5)^2= ...$, @GridleyQuayle

Oh, I thought the holiday was over. Have a good time, @robjohn

Gridley Quayle

@TedShifrin That's what I did, then 2ab=1/8 and a^2+5b^2=3/8. Eliminating a gives the quadratic/quartic that I wrote. This seems a bit long winded though especially since I am not allowed a calculator.

Clarinetist

@TedShifrin Yeah, the definition makes no sense... it says here:

>We begin by defining a perpendicular projection operator (ppo) onto an arbitrary space. To be consistent with later usage, we denote the arbitrary space $C(X)$ for some matrix $X$.

>**Definition**. $M$ is a perpendicular projection operator (matrix) onto $C(X)$ if and only if: (i) $v \in C(X)$ implies $Mv = v$ and (ii) $w \perp C(X)$ (abuse of notation here, FYI) implies $Mw = 0$.

And then they show an example

Samuel Yusim

11:14 PM

they mean $C(X)$ for arbitrary $X$, I think

Clarinetist

>For example, consider the subspace of $\mathbf{R}^2$ determined by vectors of the form $(2a, a)^{\prime}$. It is not difficult to see that the orthogonal complement of this subspace consists of vectors of the form $(b, 2b)^{\prime}$. The perpendicular projection operator onto the $(2a, a)^{\prime}$ subspace is $$M = \begin{bmatrix}
0.8 & 0.4 \\
0.4 & 0.2\end{bmatrix}$$

I was able to indeed prove that $M$ is the perpendicular projection operator, but this whole $C(X)$ business bugs me. What the heck is $X$?

I tried to fix the definition up and thought it's maybe supposed to be the following:

> $M$ is a perpendicular projection matrix (or operator) onto $V$ if and only if: (i) $v \in V$ implies $Mv = v$ and (ii) $v^{\perp} \in V^{\perp}$ implies $Mv^{\perp} = 0$.

But then they have this theorem which says:

> If $M$ is a perpendicular projection operator onto $C(X)$, then $C(M) = C(X)$.

So is $C$ supposed to be an arbitrary space now, or a column space?

Samuel Yusim

again, $C(X)$ is the column space of an arbitrary matrix $X$

Clarinetist

@SamuelYusim So what you're saying is that it's not the space that is arbitrary, but $X$ which is?

Samuel Yusim

also any space is the column space of the matrix formed by column vectors from a basis

yes, that's what I'm saying

Clarinetist

Hopefully that will clear things up... thanks @SamuelYusim

Ted Shifrin

11:28 PM

@Clarinetist: They're saying that if you're given a subspace, you can choose a matrix (how?) whose column. Space is that subspace.

@gridley: I would try $a=A/4$ and $b=B/4$ and look for integers $A,B$.

Yeah, Samuel said it ....

Clarinetist

**Theorem.** $M$ is a perpendicular projection operator on $C(M)$ if and only if $M^2 = M$ and $M^{\prime} = M$.

*Proof*. Write $v = v_1 + v_2$, where $v_1 \in C(M)$ and $v_2 \perp C(M)$.

What is the vector space $v$ comes from?

Ted Shifrin

Huh? $v$ is an arbitrary vector in $\Bbb R^n$.

Clarinetist

facepalm

Clearly I've been doing this for too long :P

Ted Shifrin

Or not long enough :D

Clarinetist

Or that

:P

And if I recall, if $M^2 = M$... from way back in my linear algebra class, I think such a matrix is called idempotent

Ted Shifrin

11:40 PM

Not that the term matters, but yes.

Clarinetist

They claim here that $(I - M)v = (I-M)v_2$. This is because $(I-M)v = (I-M)(v_1 + v_2) = (I-M)v_1 + (I-M)v_2 = Iv_1 - Mv_1 + (I-M)v_2 = v_1 - v_1 + (I-M)v_2 = (I-M)v_2$.

since $M$ is a perpendicular projection operator

Mike Miller

@TedShifrin: How comfortably can you talk about spectral sequences?

Clarinetist

Now I know that $(A+B)C = AC + BC$ for appropriately-sized matrices... to convince myself that FOIL applies to matrices, I have $(A+B)(C+D) = A(C+D) + B(C+D) = AC + AD + BC + BD$

so my work should be right

Ted Shifrin

11:57 PM

I knew them when I was a grad student, @Mike, but ...

Mike Miller

yeah, I asked for a reference in the homotopy theory chatroom because i figured i'd get a better reception there

Ted Shifrin

Sure, @Clarinetist

It's funny how the internet has replaced talking to one's colleagues or going to the library @Mike.

Mike Miller

@Ted: It's a holiday, or else I'd be asking people I know...

Ted Shifrin