Mathematics - 2012-04-08 (page 2 of 3)

t.b.

12:07

hmmm...

Jonas Teuwen

@N3buchadnezzar How do you like this horn?

@tb Hmm, I've tried the same when I was at the level of knowledge...

I quickly decided I better study some other things first 8-).

t.b.

Same here. But notice we didn't look for an in-depth study... I still don't know the first thing about FA

Jonas Teuwen

Oh, like that. I know some very small topics in some reasonable depth I think. Semigroups perhaps.

But only a part of that too, not so much analytic ones.

Benjamin Lim

@tb I am starting to get hold of this base change stuff

t.b.

good. Then you no longer have to ask me about it :)

dupe

@BenjaminLim sorry, I saw your question but I was really not in the mood. I always have to look up what a local ring is...

Benjamin Lim

12:13

@tb It's ok did not need any local ring

t.b.

(I told you several times already that commutative algebra is not at all my cup of tea)

Benjamin Lim

All I needed was the fact that if $M$ is an $R$ - module, $N$ an $S$ - module

then there is an isomorphism

between the $S$ module $M \otimes_R N$ and $(S \otimes_R M) \otimes_S N$

user19161

@BenjaminLim It is not common. Yours is pretty common here.

Benjamin Lim

I know, that's what my dad told me in malaysia and singapore the surname Lim is everywhere

user19161

@tb Really? I thought you knew everything!

Benjamin Lim

12:16

@tb You seem to hate it with a vengeance

avec de la vengence !!!

t.b.

@JasperLoy No, you must confuse me with my Greek namesake θεός

@BenjaminLim I just find it utterly boring.

Benjamin Lim

hahahahahhahahahahahahahaha

@tb what bout algebraic topology?

I will do that next semester

@tb You know I look up to you.

user19161

@BenjaminLim He is taller than you?

Benjamin Lim

No I am quite tall 6 feet 2

user19161

I am quite short, 1.68m.

Benjamin Lim

12:21

@tb Quelle est votre taille?

@tb This checking of maps, bilinearity, S - linear or R - linear or whatever, checking that maps are mutual inverses is tiring me

t.b.

@BenjaminLim You shouldn't! Seriously: I use a lot of algebra but I simply don't like study it for its own sake for some unexplainable reason.

Jonas Teuwen

I'm also tiny 1.99m.

user19161

@JonasTeuwen What? That's huge!

N3buchadnezzar

@tb What algebra do you prefer?

user19161

@N3buchadnezzar prefer

N3buchadnezzar

12:23

@JonasTeuwen width ?

Benjamin Lim

@tb Why not? I miss it when there are no people to talk mathematics to about. But commutative algebra is used for algebraic geometry no?

Jonas Teuwen

@N3buchadnezzar I didn't measure that.

N3buchadnezzar

Oh, I thought 2m was width.

t.b.

@BenjaminLim it is. That's why I never seriously studied it.

Jonas Teuwen

Holy cow, then I would be a monster.

t.b.

12:24

There's too much math that I find more interesting.

Benjamin Lim

@tb Man theo this AM they don't tell you a lot of details do they?

Like automatically something is isomorphic to something else what the?????

N3buchadnezzar

@JonasTeuwen a monster and the coolest professor,

Jonas Teuwen

I'd be satisfied with that 8-).

t.b.

@BenjaminLim no, they don't. But I never actually read the book. When I was trying to learn that stuff a little bit then there was only the too terse AM and the too chatty Eisenbud.

@JonasTeuwen the FA guy wants to know what you would recommend.

Benjamin Lim

AM is very terse indeed

I have told Kannapan to be prepared for a baptism of fire from tensor products in AM

t.b.

12:26

And with Eisenbud you could kill a monster of 2m width by letting it drop onto its head from the height of one meter

N3buchadnezzar

Eisenbud is worse than Rudin? Geesh

user19161

Bourbaki is the most concise.

user19161

I like to write in Bourbaki style myself.

user19161

Or try browsing Federer's geometric measure theory.

t.b.

Oh, that's a wonderful book, too :)

Jonas Teuwen

12:28

@tb Okie.

t.b.

@JasperLoy I once read the assessment: "I don't know if that book was written by a robot or for a robot; probably both."

2

Benjamin Lim

@tb I am told

that if we have finite dimensional vector spaces $V$ and $W$ such that $V \otimes W = 0$

user19161

@ben I realized you still hate using the right arrow.

Benjamin Lim

Why can we conclude that $V = 0$ or $W = 0$?

@tb forget it

this comes from the fact that

$\dim V \times \dim W = 0$

Hence $\dim V = 0$ or $\dim W = 0$

so that either $V$ or $W$ is the zero space

Jonas Teuwen

So where does geometric measure theory come in handy?

Benjamin Lim

12:31

dimesion is really powerful

Jonas Teuwen

I don't like my dimensions finite.

N3buchadnezzar

@JonasTeuwen Then you really are a monster

Benjamin Lim

@JonasTeuwen Imagine trying to prove $a$ algebraic, $b$ algebraic implies $a + b$ algebraic

by bashing algebra

using dimension everything is so easy

t.b.

@JonasTeuwen For instance compactness theorems. Existence of minimal surfaces is a classic. Read Almgren's booklet. It's a gem!

and not more than 100 pages, if I remember correctly.

Benjamin Lim

Fred Almgren is a legend man

Antonio Ros

user19161

12:35

I wish Bourbaki can write all of mathematics in a new series that is current and relevant.

user19161

Then every math student can just read that and nothing else.

Benjamin Lim

@JasperLoy Prob not a good idea.

user19161

Perhaps a new Bourbaki group will form one day for this purpose.

t.b.

@JasperLoy I do like Bourbaki very much, but I seriously wish that won't happen.

Bourbaki still exists, they just focus on polishing their tomes and organizing the séminaires.

user19161

I am sick and tired of so many textbooks with different definitions, notation and level of generality.

Benjamin Lim

12:37

@JasperLoy what kind of math you are studying now?

our lecturer uses a different definition of what a separable polynomial is, confusing like hell when you want to check with other resources

t.b.

@JasperLoy I never found that particularly distressing. But that's probably because my way of thinking doesn't involve notation that much.

user19161

@BenjaminLim None. I finished my undergrad. Maybe I will apply to grad school in future, maybe.

Benjamin Lim

ok

t.b.

And you're going to have to get used to all kinds of notations. Unless, of course, Mariano becomes the Emperor of Notation.

Jonas Teuwen

@tb Do you perhaps also know the title?

Benjamin Lim

12:39

bye guys

i am going to bed now

all this alcohol is making me tired

t.b.

good night, Ben!

user19161

@BenjaminLim Oh, and then you have to check if the definitions are equivalent or not too.

t.b.

@JonasTeuwen what title?

Jonas Teuwen

@BenjaminLim Maggot?

@tb Of the Almgren's booklet.

user19161

I once saw a blog post where the fridge was infested with maggots as the owner turned it off before a trip.

Benjamin Lim

12:40

@JonasTeuwen not now but before yes

nite guys

t.b.

@JonasTeuwen Oh, Plateau's problem. Cambridge.

Jonas Teuwen

Nite.

@tb This?

user19161

@jonas Are all the people there as tall as you?

Jonas Teuwen

No, certainly not, but there sure are people that are taller.

Well, the average woman is at least 8 cm taller than you 8-).

user19161

It's OK, I have given up on women long ago. :-)

user19161

12:44

I will probably stay single this life...

user19161

No need to remove that, so secretive!

user19161

I also noticed you use a different smile, namely 8-).

robjohn

@JasperLoy he wants to make sure people know he wears glasses :-)

user19161

Why? You are not spending enough time with her?

Jonas Teuwen

"What did you write???????????"

t.b.

12:46

@JonasTeuwen exactly.

user19161

@robjohn Oh!

robjohn

Jonas will leave no further trace of himself on the chat. He seems to be erasing all of his comments :-)

user19161

Matt is infecting everyone with his secrecy! I also have my secrets...

t.b.

@robjohn seen this? I thought the geeky side of you would love it :)

(totally useless, of course)

Jonas Teuwen

@tb Hmm, I can't see how you've done is. Does \eqref do this?

t.b.

12:50

yes. you put a labels inside a \begin{equation} environment or elsewhere, then you can use \ref and \eqref to refer and link automatically to it.

(it's a bit unstable, though). Haven't quite figured out what broke it all the time when I was trying it out.

robjohn

@tb I didn't think that \tag could take math on the inside. I had tried, and failed. Maybe this is new in 2.0

t.b.

@robjohn It is a 2.0 feature. It didn't work under 1.0. J.M. and I spent quite some time to convince it to accept an asterisk as a tag.

robjohn

@tb That is handy.

@tb also I didn't know that \label picked up the contents of the \tag before it

I guess I am used to the auto numbering from the environment I usually use offline.

t.b.

@robjohn heh, you noticed that :) I was somewhat surprised, too.

robjohn

@tb is that documented somewhere?

t.b.

13:03

@robjohn I played around with it myself, but see here

robjohn

It seems to be hinted at here since our equation environment is really equation*

David Wheeler

@tb will you do me a favor?

t.b.

@DavidWheeler what's up?

robjohn

Autmatic Equation Numering?

David Wheeler

look at this: and tell me if what i said makes sense? math.stackexchange.com/questions/128788/…

t.b.

13:06

@robjohn a further improvement is that you no longer seem to need to do this ugly enclosing of everything into double dollar signs.

@DavidWheeler Just a sec.

@DavidWheeler Looks okay. I wouldn't argue that the projections are epimorphisms though: I would use the universal property of the quotient map a few time to cancel $p_M$, $p_{M/N}$ away. Else you'd need to argue why they are epimorphisms (which categorically comes down to the same). Of course it's obvious because they're surjections of the underlying sets, but I'm nitpicking from a puristic viewpoint.

David Wheeler

the "why" is what we talked about yesterday, right?

t.b.

no, yesterday we talked about the converse: every epimorphism in groups is a surjection of the underlying sets.

David Wheeler

ok, well that's easy

Chris

Hello all!

How can we type an array here?

Thank you

David Wheeler

but i'm interested in how you would use the universal property to cancel the projections away

t.b.

13:16

@Chris en.wikibooks.org/wiki/LaTeX/Mathematics#Matrices_and_arrays

\begin{array} ... \end{array}

Chris

@tb Thank you! Question related to linear algebra coming up :P

t.b.

good luck!

@DavidWheeler Let me state the universal property in detail: You have an inclusion $i_N: N \to G$ and the quotient $p_N: G \to G/N$. If $f: G \to H$ is such that $f i_N = 1$ (the map sending every element of $N$ to $1$ in $H$ then there exists a unique morphism $f': G/N \to H$ such that $f'p_n = f$.

David Wheeler

right, if f kills N (which is what $fi_N = 1$ means) f factors through $p_N$

t.b.

exactly.

David Wheeler

so how can you "cancel" the p's"?

t.b.

13:22

To verify that two morphisms $h,k: G/N \to H$ are the same you check if $hp_N = k p_N$ and the universal property tells you then that $hp_N = kp_N$ factors over both $h$ and $k$ and by uniqueness it follows that $h=k$.

(so this is actually the proof that $p_N$ is an epimorphism).

David Wheeler

ok, so i didn't need to invoke cancellation, uniqueness does it for me

t.b.

yes, exactly. I just noticed that in the universal property I forgot to state that $p_N i_N = 1$...

The fun of this is that you don't need to mention any elements at all. Everything is just massaging arrows.

David Wheeler

the thing is, i didn't have anything factoring though the projections, so i would have had to invoke the identity morphism at some point, but i see what you're saying

yeah...the subgroup N can be replaced by the injection monomorphism, right?

t.b.

give me a second...

David Wheeler

i'll give you 10, i'm in a generous mood :)

N3buchadnezzar

13:27

I tried using the euler substitution $\sqrt{x^2-2x}=x \cdot u$ to integrate $\int \sqrt{x^2-2x} \, \mathrm{d}x$, and now my head really hurts. Great..

t.b.

@DavidWheeler The relevant diagram is this:

Forget about the words.

David Wheeler

so "C" is "G", "A" is "M", "B" is "N", "X" is "N/M" and "Y" is "G/N" yes?

what does "bicartesian" mean?

t.b.

yes and Z is G/M

@DavidWheeler I told you to forget about the words :) It means both a push-out and a pull-back (this isn't true for groups)

This is from a text on additive categories.

Chris

Hello again,

In linear algebra, given 2 linear transformations, $\theta : R^{3x1} \rightarrow R^{3x1} $ and $\varphi : R^{3x1} \rightarrow R^{3x1} $.

$\theta : \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \rightarrow A\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} $

And for $\varphi$ is given this: $\varphi : R^{3x1} \rightarrow R^{3x1} $.
Is there any $\varphi$ that would make this* statement true?

* : $ Ker(\varphi) = Image(\varphi) = Ker(\theta) $

What I did was this:
We have this for $\varphi$ : $dim(R^{3x1}) = dimKer(\varphi) + dimImage(\varphi) \Rightarrow

David Wheeler

oh, i think i have it wrong: "Y" is G/M (the second row is exact), and "Z" is G/N (we "cancel the M's")

t.b.

13:34

Some people call pull-backs Cartesian squares (because they generalize the Cartesian product) and push-outs Cocartesian squares (by daulity).

@Chris use bmatrix instead of matrix

David Wheeler

so a product is a pull-back?

t.b.

yes, with the terminal object in the lower right corner.

Chris

@tb Thank you :)

David Wheeler

terminal...thinking...singleton set?

t.b.

yes.

@Chris this looks good :)

David Wheeler

13:39

the first time i ever heard of pull-backs was with pre-composition in manifolds

t.b.

that's a different kind of pull-back

(same word but no strict relation)

You know how products are defined, right?

You're given $M$ and $N$ and a product is an object $P$, equipped with two projections $p_M: P \to M$ and $p_N : P \to N$ such that....

David Wheeler

products are defined as universals that factor through the projection pair?

Chris

@tb: Really? Great! Thank you!

t.b.

yes
....for every object $Q$ together with two maps $q_M: Q \to M$ and $q_N: Q \to N$ you have a unique map $f: Q \to P$ such that $q_M = p_M f$ and $q_N = p_N f$.

David Wheeler

it's like the group quotient, but with a pair that factors

which is why, no doubt, AxB/Ax{0} is iso to B

t.b.

13:43

@DavidWheeler No, it's somewhat different in that $f$ goes into the product. In the quotient you have a map coming from the quotient (dual situation)

David Wheeler

ok, existence of target, rather than existence of source?

t.b.

Yes. Kernels and products have universal "target" property, while cokernels (quotients) and coproducts (disjoint unions for sets) have a universal "source" property.

The technical term for "target" property is "limit" while "source" is "colimit".

David Wheeler

cokernels aren't the same as quotients in groups, though

i mean, we have to do some funky thing with the normal closure, right?

abelian groups are "nice" coker(f) = H/f(G)

t.b.

True. But let's look at normal subgroups :)

The quotient by a non-normal subgroup isn't a group in the first place...

David Wheeler

well, i mean we'd have to look at something like $H/\langle f(G)^H \rangle$

could get messy

robjohn

13:53

too much algebra :-)

t.b.

@DavidWheeler Yes. That's because every homomorphism annihilating a non-normal subgroup $H$ annihilates a normal subgroup containing $H$ and conversely for every normal subgroup N containing H the quotient map $G \to G/N$ annihilates N, so the smallest possible quotient is what you just wrote.

David Wheeler

if f is epi, though, the cokernel should be nice :)

t.b.

Or if $f$ is the inclusion of a normal subgroup :)

David Wheeler

that's a more interesting situation, because the cokernel isn't trivial then

t.b.

Well, it's nice whenever the image of $f$ is a normal subgroup...

David Wheeler

13:55

like the inclusion of the rotations in the dihedral group

t.b.

@robjohn that's probably my punishment for having complained about algebra just an hour ago :)

David Wheeler

i'm starting to really like the dihedral groups...they're the next-best thing to cyclic

robjohn

@tb :-) I just go back and work some questions. I would only get lost in the diagrams :-)

right now, I am trying to decipher an answer by Didier that seems hard to follow. I answered much later, but I think it is easier to follow.

David Wheeler

i think diagrams are to make things clearer, not more complicated

t.b.

@DavidWheeler yes, they're nice. Ever looked into reflection groups?

robjohn

13:59

@DavidWheeler I'm sure they do, but I really need to brush up on things first.

Chris

Let A be a matrix: $ \begin{bmatrix}
7 & 2 & 3 \\
10 & 5 & 6 \\
17 & 7 & 9
\end{bmatrix} $
are there any P and Q matrices that make the $PAQ$ matrix invertible?

In addition, $rank(A) = 2 $ and A is not an invertible matrix.
So due to the fact that A is not invertible, the product of PAQ is not an invertible matrix?

Thank you!

t.b.

@Chris are P and Q supposed to be 3 x 3 matrices?

David Wheeler

@tb, like the symmetry group of the cube?

t.b.

@DavidWheeler exactly. There's a nice book by Humphreys on them.

robjohn

@Chris If A is not invertible then PAQ won't be either. unless P is 2x3 and Q is 3x2

Chris

14:01

@tb : We are not given any more information. :/
@robjohn: Is there any link that I can use as a source?

robjohn

@Chris the same question that tb asked. Are P and Q supposed to be square matrices?

if so, then $\det(PAQ)=\det(P)\det(A)\det(Q)$

Chris

@robjohn: We aren't given any more information about P and Q matrices :S

t.b.

Most probably they are, but if you're not given any more information then the correct answer is "it depends". For instance if P = [1 0 0] and Q = [1 0 0]^T then PAQ = [7] is invertible.

robjohn

However, if $P$ is 2x3 and $Q$ is 3x2, then $PAQ$ would be 2x2 and might be invertible.

David Wheeler

well, for an easy counter-example if P,Q don't have to be square, let Q be 3x1 and P be 1x3, with Q = $e_1$ and P = $e_1^T$

t.b.

14:05

@robjohn I leave that to you, robjohn. You said you need to brush up on your algebra :)

David Wheeler

then PAQ(x) = 7x

robjohn

@tb Linear Algebra is fine :-p

David Wheeler

if P,Q are 3x3 however, no le-go-go

robjohn

@DavidWheeler Since $\det(A)=0$

David Wheeler

or, put another way, rank(A) = 2, means the null-space has dimension 1, so some non-zero vector gets zapped.

it's rather hard to invert a zapped vector, we have too many pre-images to choose from

t.b.

14:10

so... let's have mercy on Chris :)

David Wheeler

should 0 go back to 0, or the zapped vector v? i can't decide...i'm....ill-defined...

@Chris if that's all the information you're given, you'll have to write out TWO answers: one saying it's impossible if P and Q have to be 3x3 (and saying WHY), and another saying it IS possible if P,Q don't have to be 3x3 (and give an example).

t.b.

@Chris So, you determined that A has rank two (I haven't checked). Since $\operatorname{rank}(AB) \leq \min{\{\operatorname{rank}(A), \operatorname{rank}(B)\}}$ you can conclude from this that $\operatorname{rank}{(PAQ)} \leq 2$. This tells you that $PAQ$ can't be invertible if $P$ and $Q$ are $3\times 3$-matrices. You're absolutely right about that.

David Wheeler

yeah it has rank 2, row 3 is row1 + row 2

t.b.

right.

David Wheeler

and row1, row 2 are LI

t.b.

14:14

I tend to forget about numbers as soon as I'm given more information :)

David Wheeler

well the exact numbers aren't important right? just the dimension of the row space/or column space

because we could fiddle with the numbers by multiplying a row by something (except 0, because that's just...not...fair!)

N3buchadnezzar

heya

Chris

@tb: The only choices that we are given to answer and prove are " a. There are P and Q matrices and b. There aren't any " . I think that P and Q matrices should be square and that this $\det(PAQ)=\det(P)\det(A)\det(Q)$ would be a good proof that PAQ matrix is not invertible.

@DavidWheeler: What does "no le-go-go" mean? :P
I got what you said here " let Q be 3x1 and P be 1x3, with $Q = e_1$ and $e_1^T$" :)

Why is that " rank(A) = 2, means the null-space has dimension 1, so some non-zero vector gets zapped" ?

t.b.

@Chris see en.wikipedia.org/wiki/Rank_(linear_algebra)

N3buchadnezzar

Does anyone know how to make a list in matlab with all the positive integers from 1 to n ?

David Wheeler

14:19

a mathematician and his buddies were playing poker. when the night ended, they started arguing over how to split the pot on the table. the mathematican pushed all the bills on the the floor. "well," he said, "the homogeneous case has a solution."

N3buchadnezzar

I know I can use a for loop, but I would like to avoid doing this.

David Wheeler

@Chris "no le-go-go" is just slang for "that will not work out" (or as we say here in texas: "that dog won't hunt").

robjohn

$\left[
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0
\end{array}
\right]
\left[
\begin{array}{ccc}
7 & 2 & 3 \\
10 & 5 & 6 \\
17 & 7 & 9
\end{array}
\right]
\left[
\begin{array}{cc}
1 & 0 \\
0 & 1 \\
0 & 0
\end{array}
\right]
=
\left[
\begin{array}{cc}
7 & 2 \\
10 & 5
\end{array}
\right]
$

David Wheeler

@Chris there is an important result in linear algebra, called the rank-nullity theorem

t.b.

@N3buchadnezzar something like x = 1:1:n? or look up linspace

David Wheeler

14:22

it says that the rank of a matrix, plus its nullity equals the total number of columns

the nullity of a matrix is the dimension of its null space

t.b.

^ which is the same as the $\dim{\ker{A}} + \dim{\operatorname{im}{A}} = \cdots$ equality Chris used earlier on.

David Wheeler

we have 3 columns, and rank 2, so that leaves 1 left over for the null space.

yes rank(A) = dim(im(A)) as well

t.b.

I was taught this identity under the name "dimension formula for linear transformations".

but that's just a translation from German.

Chris

@DavidWheeler: Oh, thank you for clearing this out.
Isn't the nullity of a matrix equal to 0? (I will have to look for this at our book)

@tb: Will do now!

@robjohn: Which is invertible!

David Wheeler

it's actually the first isomorphism theorem in the arena of vector spaces

t.b.

14:26

I wouldn't have understood "rank-nullity theorem" a year ago.

@Chris the nullity of a matrix is the dimension of its kernel

David Wheeler

@Chris while 0 is IN the null space, it's not always the ENTIRE null space

null space = kernel

t.b.

okay, gotta go bbl

Chris

@tb and @DavidWheeler: Thank you for clarifying this " null space = kernel ", I will keep that in mind!
Bb t.b.! Thank you for your help!

David Wheeler

for your example matrix A, we have A(1,4,-5) = (0,0,0) even though (1,4,-5) isn't the 0-vector.

by linearity, every vector of the form (x,4x,-5x) also gets sent by A to (0,0,0).

and since rank(A) = 2, those are the only vectors A sends to 0. {(1,4,-5)} is a basis for the kernel.

linear algebra is cool like that, everything works out so clean

Chris

14:43

How did you find the (1,4,-5) ? //Got what you mean though!

So all in all, the answer would be something like this:
* if $P, Q$ are $2 \times 3$ and $3 \times 2$ respectively then $PAQ$ matrix is invertible

* if $P, Q $ belong in $R^{3x3}$, then $det(PAQ) = det(P) \times det(A) \times det(Q) = 0 $ as $det(A) = 0$.
Then $PAQ$ is not invertible.

David Wheeler

15:07

@Chris the way i found the vector (1,4,-5) was solving the 3 simultaneous equations: 7x+2y+3z = 0, 10x+5y+6z=0 and 17x+7y+9z = 0, just like you normally would.

@Chris by the way, PAQ is not necessarily invertible for ANY 2x3 and 3x2 matrices (for example, the 0-matrices would be poor choices)

Chris

@DavidWheeler: OK, this means I have to take a break :P
So after all I believe that P and Q matrices belong in $ R^{3x3}$. Is there any source for this $det(PAQ) = det(P) \times det(A) \times det(Q) = 0$ ? (like wikipedia?)

David Wheeler

sure..look up "determinants"

Tyler Bailey

planetmath.org/encyclopedia/Determinant2.html

See under remarks. the first one.

Chris

15:26

@DavidWheeler and @TylerBailey Thank you :)

Tyler Bailey

Whoa, that scared me

haha, i may have to turn the ping sound off ;p

Chris

hahah :p

robjohn

15:46

@Chris Yes, which was the point :-)

Chris

@robjohn:
For this $det(PAQ) = det(P) \times det(A) \times det(Q)$ to be valid/ legal, should matrices P, A and Q be of the same dimension?

robjohn

@Chris For the $\det$ to make sense, the matrices must be square

@Chris For $PAQ$ to make sense, $P$ and $A$, and $A$ and $Q$ must have at least one dimension in common

The end result is that all have to be square matrices of the same dimension

Chris

@robjohn: Got it, so for the det to make sense it should be:
$ 2 \times 3, 3 \times 3, 3 \times 2$ ?

robjohn

@Chris $\det(PAQ)$ makes sense, but you can't take the determinant of $P$ or $Q$

since they aren't square.

Chris

@robjohn: the $det(PA)$ and $det(AQ)$ make sense?

robjohn

15:57

@Chris No, because $PA$ is $2\times3$ and $AQ$ is $3\times2$

Chris

@robjohn: So the matrix of the product must be square!

Also, something irrelevant, is there any way to add the username recommendations without having to click on it? In example when I click "@rob" I get the "robjohn" above the textbox, is there any way to select this using only the keyboard?

robjohn

@Chris to take the determinant, a matrix must be square

@Chris I am not sure what you mean by "username recommendations".

Chris

@robjohn : indeed!

About the irrelevant question, just found it...using tab

Matt N.

I did it and it seems to be $$ (1 + (x + x^{-1}))^n = \sum_{i = 0}^n \sum_{j = 0}^i {n \choose i} {i \choose j} x^{2j - i}$$
So the coefficient of $x^k$ (where $k = 2j - i)$ turns out to be $\frac{n ! }{(n-i)! j! (i - j)!}$
But if I can't write down the coefficient as a function of $n$ and $k$ my answer seems rather useless.

Chris

Username recommendation is that when I type "@" and start typing your username, a small box with your username above the textbox appears..

robjohn

16:04

@MattN You are working on $(x+1+x^{-1})^n$?

Matt N.

@robjohn Yes, I'm thinking about my answer here.

robjohn

@MattN I am not sure that is the answer to the question being asked...

Since his question seems to view 1+1+0 the same as 1+0+1

"By distinct I mean that the number of 0s, 1s, and -1s of two solutions must be different."

The two I gave above have the same number of 0s, 1s, and -1s.

Matt N.

Oh. Grrr. Thank you. When I read n-tuple I automatically assumed (1,1,0) \neq (1,0,1).

@robjohn You're right. And of course I can still think about the other case : )

robjohn

@MattN Indeed

anon

yes. why do you think it was deleted?

Matt N.

16:16

Ignore me. : )

I assume you're correcting and then undeleting?

It seems she's happy with the comment.

anon

nah, I'm reading wikipedia and listening to youtube

robjohn

I've answered two moderately hard questions in two days, and gotten two upvotes in those two days. One of the upvotes was for an old answer. Weird weekend.

Matt N.

Must be your scary avatar that scares them away before they get to vote.

Chris

@robjohn: Is there any way to upvote in chat? :P

anon

I can't be sure, but I think while I was robjohn I got fewer votes.

robjohn

16:20

@anon wouldn't surprise me :-)

Matt N.

@Chris You can upvote him on the site.

robjohn

@Chris there are no upvotes in chat. You can star a comment

Matt N.

One more answer and you've contributed 500.

robjohn

@MattN I know. I was looking for a good question to be number 500 :-)

Chris

@MattN I can upvote at the site, only through an answer he gave, right?

@robjohn: Will have to star a lot of comments then ;)

Matt N.

16:23

@Chris No you can also upvote his question.

robjohn

@MattN I think he did. I just got two upvotes on some pretty old answers :-)

Matt N.

That was me : )

robjohn

@MattN Oh :-) Thanks!

Matt N.

Welcome : ) It's a thank you for when you helped me with PDOs.

(a second thank you : ))

Bbl : )

t.b.

geee: one more for my own personal hall of blame

On a completely different note: can we please not suggest to people to star answers to questions they asked here?

3

while the stars rarely make much sense it would be nice to see at least one message that is either funny or remarkable otherwise...

Chris

16:43

Sorry for that. I thought it was something like upvoting

t.b.

@Chris: no need to apologize :) It is nice of you to show your appreciation!

David Wheeler

i star random comments by tb, just for the sheer hell of it

3

i also lie a lot

t.b.

oh, so you were one of those smiley voters?

Chris

I am using chat for about a month and still learning :)

David Wheeler

if i were to ask the other knave which message i starred, what would he say?

t.b.

16:48

^ sphinx

David Wheeler

can't we just have robjohn arbitrarily wipe stuff periodically for no good reason?

@tb on 3 legs! bingo!

Jonas Teuwen

Are it just the students at my institution or is complex analysis hard to grasp? They need to compute an integral over the real line and they compute and integral with a large circular contour around the poles.

Many of them do.

Voila, you have your real integral for some reason.

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$Mathematics$ Mathematics