Mathematics - 2015-03-14 (page 1 of 3)

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12:03 AM

Is there such a thing as the intended-model of theory of groups ?
It seems that the theoy wants to capture not ONE model ( like zfc or peano arithmetic ) but a CLASS of models
So maybe we use intended-class-of-models there ?
and intended-model in ZFC and PA ?

You could try to create such a thing :-)

But that would be pretty "nerdy" :D

Do you know model theory ? Did what i wrote make sense ?

Its because i'm pretty newbie at it

@nerdy I suppose you could say the intended model was "quotients of free groups"

I don't know abstract algebra too

=/

unfortunately, this is unwieldy to work with, we can't decide if two quotients are even isomorphic

12:13 AM

I was just noticing that it seems like the difference between models of Theory of groups is imensely larger than the difference between models of say PA or ZFC

did what i wrote make sense ?

yes, it makes sense

it is known that there is a (single) isomorphism class of models of PA, under sufficiently strong logic

weaker logic yields more models (i.e. the so called "non-standard models), but we're still not dealing with as much diversity as one sees in groups (there are many "kinds" of groups, the concept is "underdetermined")

category theory is thus "more like group theory" than "set theory", from a model theory perspective, even though one can view set theory as an example of a category, and (perversely), category theory as a subset of set theory (or some conservative extension)

Thats exactly what i perceived

Yeah, that makes a lot of sense, in the classical foundation we have uniformity ( everything is a set ) , the other foundations are trying to break that uniformity ( with types and categories )

right ?

I dont find whats the problem with the uniformity, but i'm too newbie in foundations anyways

Its not like we HAVE to represent everything as a set ( then uniformity would indeed be a problem ) , we just need to know that everything can be thoughth as a set ( which is surprising and interesting )

At least to me

I'm as mad as hell, and I'm not going to take this any more!

2

12:31 AM

y u so mad, meng?

set theory only has one major flaw, the theory itself does not form a set (there is no set of all sets), so it's not "truly comprehensive"

but i don't think a comprehensive logical foundation for math is even possible

for most purposes, it is enough to know that most of the mathematical constructions we encounter can be viewed as sets

12:52 AM

Thats exactly what Godel Incompleteness Theorem is about

As long as your theory is stronger than PA it cant be "comprehensive"

user image

I think it's true in a larger sense: humanity is always going to be in the position of the blind men and the elephant. We'll have a good sense of "certain views", we'll never see the "whole picture".

Does every infinite polycyclic group contain a non-trivial torsion-free abelian subgroup of finite index?

1:09 AM

@Bib look: words ;-)

don't mind me...

@DavidWheeler I've been thinking about, what I feel to be, the excessive closures for no context.

Have you discussed this with the other moderators?

@robjohn Sorry for the lack of response. I have an interest in keeping the chatter to a low level so my question has maximum exposure.

@DavidWheeler I've left some of my rantings, but they are not home right now.

Are there any identifiable trends, such as a certain subset of people being responsible for a disproportionate amount of closures?

@Bib sorry... I have no idea what a polycyclic group is and I've only a vague idea what a torsion-free group is. It's Friday evening, the 13th, and so I blurted the first thing that popped into my head.

The second was "gesundheit" :-)

I didn't mean to be rude. I hope it did not come across that way

@Bib Your question is now on the sidebar, so you can speak

1:20 AM

:)

No offense taken.

@Bib You can also ask it on main. With the proper tags, the proper people will probably see it.

I'd rather not, but thanks for the suggestion.

1:56 AM

@Bib I am pretty sure there are polycyclic groups of exponential growth, so I don't think so

Although I don't know of any examples

@PaulPlummer Do you know if it is the case that every infinite finitely generated nilpotent group has a finite index subgroup with infinite abelianization?

The discrete heisenberg group is an example.

I don't know,

@PaulPlummer Also, does a group having exponential growth rule out the possibility of it having a finite index subgroup of infinite abelianization?

Or just a finite index torsion-free abelian subgroup?

No, the free group on two generators has exponential growth. It rules out finite index abelian subgroups though

Thank you. :)

2:03 AM

No problem

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There is most certainly a certain subset of people being responsible for a disproportionate amount of closures @DavidWheeler

2:18 AM

David Wheeler , i don't know, i don't see a reason to think that it applies to everything

that incompreensiveness we were talking about

It might be possible to have a theory of everything in physics, but in math we already know iots not possible

@Committingtoachallenge I can imagine this, but I have no proof.

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@DavidWheeler Look at the list of names on the close ribbon and you'll see a trend

"close ribbon"?

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The yellow thing that hangs down saying 'closed for blah blah'

can I view this for all closed questions at once?

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2:24 AM

You can see how many close votes people have over different periods of time by putting your cursor over their identicon here math.stackexchange.com/review

Woodface 122 close vote reviews this week

~3600 all time

math.stackexchange.com/review/close/stats

I guess you can't see their ratios though

I have 0

Me too :P

A question I recently asked got closed but nobody suggested how to improve it

What i see is 3 users with over 100 close votes this week, and then it drops way down

2

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I don't have 3k rep, so I can't see what options they have in there. I imagine they have leave open and close, so that count might include leave open votes

2:31 AM

I think the only time I voted to close was my own question

@ᴇʏᴇs o_O did someone answer already?

@ConorO'Brien No

@ᴇʏᴇs excuse me if I'm missing something obvious couldn't you have deleted it yourself?

@ConorO'Brien There were suggestive comments on it

@ConorO'Brien Also I heard it's bad etiquette to delete questions

@ᴇʏᴇs Ahhhhhh okay ;) just wondering

2:33 AM

@ᴇʏᴇs Which question was it?

@ᴇʏᴇs ... oh. dear. I'm a bad etiquette-er :P

@DavidWheeler Super confusing one

linky?

@DavidWheeler It doesn't really make sense so I don't know if it'd be helpful to look at it

This one I thought the question made sense and I still don't know how to make the question clearer math.stackexchange.com/questions/1179524/…

@PaulPlummer In case you're interested. Yehuda Shalom exhibits a theorem which states that if $G$ is quasi-isometric to a polycyclic group, then $G$ has a finite index subgroup with infinite abelianization. The article is here: link.springer.com/article/10.1007%2FBF02392739

2:40 AM

Here's what I think @ᴇʏᴇs : you are concerned that you cannot show from the definition of "inclusion" that one set includes the other. But to most of the people who commented, this is obvious, so they don't know what your question is.

Suppose that $y$ is such that $d(x,y) < r$.

But just because it's obvious to other people and not to others, they are allowed to do that?

$d(x,y)$ is just a real number, say $t$.

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What is the notation of a change of basis?

@Committingtoachallenge What do you mean?

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If I want to show some column vector of coordinates $v$ in some basis, how do I write this?

2:43 AM

In the real numbers, if $c > 0$ and $t < r$, we have $ct < cr$ (this is a real-number property) @ᴇʏᴇs

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$[v]_b$ where $b$ is the basis?

@Committingtoachallenge I don't know that there is a universally accepted notation, but I use the one you do

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Okay cool, I just guess at that from something I thought I saw in someones book

@Bib Nice. I am pretty sure I saw a question on MO very similar to your question about nilpotent groups and it might have answered it, I will take a look for it again

I tend to write $[v]_B$, to distinguish between "elements" and "sets"

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2:44 AM

And the change of basis has some matrix $T$ and we take the left row vector $\times$ the right transformation matrix, to get it into another basis?

@DavidWheeler Yea, my confusion was that I didn't understand the term "if and only if" well and I couldn't see that $d(x,y) < r$ has the same solutions as $c \cdot d(x,y) < c \cdot r$

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$[v]_B T = [V]_C$

But it would've been nice if someone just explained it that way and not just shut the question down because it's obvious to them

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Or we could take the right column vector against the left Transformation matrix?

@ᴇʏᴇs this site has always had a certain sniff factor

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2:46 AM

$T[v]_B = [v]_C$

Normally, @Committingtoachallenge, we operate on vectors with matrices as left multiplication on column vectors

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Fair enough

However, people get very confused when one of the bases is "the standard basis"

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Hey @abe

Why do they get confused @DavidW?

A Beautiful Mind

Hello @Committingtoachallenge @ᴇʏᴇs.

2:50 AM

Hi @ABeautifulMind

At least you don't insult people who don't understand obvious things @ABeautifulMind

Often one is given a transformation $T$ by $T(x,y,z) = (ax+by+cz,a'x+b'y+c'z,a''x+b''y+c''z)$

A Beautiful Mind

@ᴇʏᴇs Where?

@ABeautifulMind Anywhere

@ABeautifulMind At a Harvard bar, maybe

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@ᴇʏᴇs MIT*

A Beautiful Mind

@ᴇʏᴇs Did someone insult you?

2:51 AM

G'night, everyone! ^^

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@ᴇʏᴇs Assuming you are referencing good will hunting

@Committingtoachallenge I don't know which bar they actually filmed at

A Beautiful Mind

Probably Hollywood bar, lol.

They couldn't secure a real Harvard bar?

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@ᴇʏᴇs Oh maybe it was harvard my bad. The uni with the professor was MIT though[the blackboard and janitorial]

2:52 AM

In the standard basis, elements of the vector space $F^n$ have as coordinates "themselves"

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@DavidWheeler That makes sense

So, often, people want to find the matrix of $T$ in a "new basis"

A Beautiful Mind

I think I am at the worst point in life now. Maybe this means a miracle is about to happen.

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@DavidWheeler So they run all of the column vectors of Basis 1 through basis 2?

@ABeautifulMind If it's really the worst then it can only get better

2:54 AM

To do that, you need to change "new to standard" for the change-of-basis

A Beautiful Mind

@ᴇʏᴇs Do you wear glasses?

And people's intuition says they want "standard to new"

@ABeautifulMind Yes

A Beautiful Mind

@ᴇʏᴇs I wear them when I go out.

@ABeautifulMind But I also have eyebrows

A Beautiful Mind

2:55 AM

@ᴇʏᴇs Me too.

Can eyebrows your computer?

@DavidWheeler It took me like 10 seconds to get that lol

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I am trying to understand the answer here: math.stackexchange.com/questions/1188129/…

@Bib here is the fairly similar question

@Committingtoachallenge What part, in particular, is troubling you?

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2:57 AM

I'll just go through it with what you have told me and find out if I am good now

Yeah, I remember seeing that question. Unfortunately it didn't talk about an abelian torsion-free subgroup of finite index. :P
Thankfully however, Shalom's article has solved my problem.

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$X = [v]_B$ and $X'=[v]_{B'}$ and $P$ equals what?

@DavidWheeler Do you have a lot of math textbooks

@ᴇʏᴇs Not a lot.

@ᴇʏᴇs Libraries do!

2:58 AM

@DavidWheeler Which algebra books do you own

Hopefully...

@Committingtoachallenge The columns of $P$ are the coordinates in the "old" basis of the "new" basis vectors.

@ᴇʏᴇs Herstein, Garrett, Fraleigh, Dummit & Foote, Jacobson BA I

Sounds like a change of basis of a basis

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I don't know how to write that out in 'math' form

so $X = PX'$

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3:01 AM

oh, he says that :\, I couldn't connect it

$P: [v]_{B'} \mapsto [v]_B$

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yep yep

That is so simple argh, I suck haha thanks

So $APX$ is the image of $X$ (under $A$) in the $B$-basis

argh-latex gives me trouble sometimes

$[v]_{B'} \to [v]_B \to [Av]_B$

@DavidWheeler If you were forced to keep only one of your algebra books, which one would you choose

If only one, D&F, it has the most inofrmation

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3:05 AM

I don't understand that last line

Why are there three mappings?

It's like 1,000 pages long innit

There's just 2 mappings so far, starting element $\to$ image under first matrix $\to$ image under composition of both matrices

@ᴇʏᴇs haven't counted

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ohhh I hadn't seen it change to APX

That makes sense, awesome thanks

I just see it like this:

$T' = P^{-1}TP$ so that new->old->T(in old) -> T (in new)

If the "old basis" is the "standard basis" $P$ is extremely easy to write down.

For example, if the old basis is the standard basis for $\Bbb R^3$ and the new basis is $B' = \{(1,0,1),(1,1,0),(0,1,1)\}$ then $P = \begin{bmatrix}1&1&0\\0&1&1\\1&0&1\end{bmatrix}$

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Ahhh very true, the basis vectors become the column vectors, since we are essentially multiplying that with the identity matrix

3:14 AM

Ya, it's the "new basis" in the "old coordinates"

That converts "new coordinates" to "old ones"

Hit that with the "usual" matrix for $T$, and then we convert back.

The most tedious part is computing $P^{-1}$

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Wait I thought it converted old coordinates to new ones?

$P^{-1}=\frac{1}{|Det{P}} blah blah$ still?

Lol, I TOLD you people get confused

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xD yep I am

What is $(1,0,1)$'s coordinates in the basis $B'$?

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Where it is $(1,0,1)$ in $B$?

at the moment?

(2,1,1)?

3:17 AM

yes, (1,0,1) is in "standard coordinates" = $e_1 + e_3$

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$(1,0,1)_B = (2,1,1)_{B'}$

No..let's name the vectors in $B'$: $v_1 = (1,0,1), v_2 = (1,1,0),v_3 = (0,1,1)$

Now, by definition $(a,b,c)_{B'} = av_1 + bv_2 + cv_3$

Isn't it true that $(1,0,1)_B = 1v_1 + 0v_2 + 0v_3$?

So, $(1,0,1)_B = (1,0,0)_{B'}$

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Ohhh I see

That is $P[v_j]_{B'} = [v_j]_B$

But people want to do it "backwards", their intuition says, find $P^{-1}$.

so they tell themselves...hmm, how do I describe $e_1,e_2,e_3$ in the new basis?

But that's not going to be helpful, because we trying to FIND $T$ in the new basis, so if we convert "old to new", we don't know what to hit it with.

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So $P[v_1]_{B'} \ne P\times [v_1]_{B'}$?

3:31 AM

I have no idea what you are trying to say, there

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${v_1}_{B'} = (1,0,1)$ so $P[v_1]_{B'}=(1,1,2)$

I can see why that is wrong from what you have said

But I don't know how to do the matrix multiplication behind that

Isn't$ P(1,0,0)^T = (1,0,1)$?

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Yep that works, but I thought "That converts "new coordinates" to "old ones""

meant that we put in some vector in the basis $B'$ and get out a vector from $B$

We do-but the "coordinates" in the new basis, don't look like we expect.

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We are treating $B$ here as the standard basis right?

3:38 AM

0

A: Matrix for a linear transformation for a non-standard basis

If that were true, one would expect the matrix of $T$ in the non-standard basis to take the coordinates of $v$ in that basis to $Tv$ in that basis. However: $\begin{bmatrix}5&3\\1&1 \end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}5\\1\end{bmatrix}$ So your matrix takes $1(2,3)^T ...

How can I prove that if $\theta \in \mathbb{R}$ is such that $0 \leq \theta < 2\pi$, and $\cos(n\theta) \geq 0$ for every positive integer $n,$ then $\theta = 0?$

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I'll read that now thanks

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3:49 AM

Yep I don't understand it: we put standard basis vectors into it and you say we get 'standard basis coordinates', but rather to me it looks like we put the standard basis coordinates in and get the new basis coordinates out

you'll have to be more specific about what you mean by "it"

@DavidWheeler yes, what condition is necessary for compact = closed and bounded (sorry I answer so late, I went swinming :) )

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So you say we have $P_{AS}$ going from basis $A$ to the standard basis, but it looks like we put in standard basis stuff and get out new basis stuff $P_{SA}$

After here: The matrix which sends [u,v]T to its "standard basis coordinates" is:

@Anonymous First off, $\theta$ cannot be between $\pi/2$ and $3\pi/2$ (or else we fail for $n = 1$)

@Committingtoachallenge Here's the thing-by itself (1,0,0) is just an array of 3 numbers.

Normally, what we MEAN is $1e_1 + 0e_2 + 0e_3$, where $e_1 = (1,0,0),\ e_2 = (0,1,0), e_3 = (0,0,1)$

But just "as an array"it might mean $1v_1 + 0v_2 + 0v_3$ for ANY basis $\{v_1,v_2,v_3\}$

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That makes sense yep

3:55 AM

In the "standard basis", $B$, we have: $(1,0,0)_B = (1,0,0)$

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Yep

But in the basis $B' = \{(1,0,1),(1,1,0),(0,1,1)\}$, we have: $(1,0,0)_{B'} = (1,0,1)_{B}$

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Yep that makes sense

(assuming it is an ordered basis I imagine)

you can verify that: $\begin{bmatrix}1&1&0\\0&1&1\\1&0&1\end{bmatrix} \begin{bmatrix}1\\0\\0 \end{bmatrix} = \begin{bmatrix}1\\0\\1\end{bmatrix}$

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So $P[v]_B = [v]_{B'}$

4:00 AM

No, $P[v]_{B'} = [v]_B$

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oh

ohh

I am retarded

$P$ takes "new coordinates" to "old coordinates"

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I am so sorry haha

I see it now thank you so much holy crap

Suppose I take a "typical" new coordinate vector $2v_1 + 3v_2 - v_3 = (2,3,-1)_{B'}$

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We can use $P$ to turn it back to $B$

4:03 AM

In "old coordinates" that is $2(1,0,1) + 3(1,1,0) + (-1)(0,1,1) = (2,0,2) + (3,3,0) - (0,1,1) = (5,2,1)$.

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$\begin{bmatrix}1&1&0\\0&1&1\\1&0&1\end{bmatrix}\begin{bmatrix}2\\3\\-1\end{bmatrix}‌=\begin{bmatrix}5\\2\\1\end{bmatrix}$

and that is $(5,2,1)_{B}$

Exactly.

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I get it :)))

Thank you so much again!

Another thing people get hung up on is: "what basis is $P$ in?"

It's not in any basis, it's a change-of-basis calculator

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Yep cool

4:06 AM

See, $\text{Lin}(V,V)$ is "basis-less", to write things as "coordinates" (arrays, column vectors) we have to "choose a basis"

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Yep

It's like "picking where we draw our coordinate axes", the plane itself doesn't care.

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Yeah that makes sense

But to do "arithmetic", we need to turn "geometry" into "numbers"

People understand how numbers work-points in an arbitrary space aren't so easy to get a handle on

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Thank you so much!

I can understand that answer now fully

4:17 AM

Honestly, bases are kind of a pain-in-the-(deleted). I try to avoid them until the last possible minute.

But linear algebra teachers often INSIST on using matrices....

4:34 AM

changes of bases are part of what makes bra-ket notation in QM rather handy (though of course it really is nothing but notation)

4:51 AM

@Sawarnik Life goes the same. It's like the stock market, goes up a few points, goes down a few points, sometimes it crashes but the market cycle always trends in growth. How about you?

2 hours later…

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6:44 AM

http://math.stackexchange.com/questions/1188129/square-matrices-of-same-linear-transformation-has-a-and-b-similar

Why does $Y' = BX'$?

Because Y is the coordinates in B and Y' is the coordinates in B'

Y=AX \implies Y' = BX'

?

7:05 AM

because $B = P^{-1}AP$

It makes more sense if you write $[w]_{B'} = [Tv]_{B'}= [T]_{B'}[v]_{B'}$

The $X$'s and $Y$'s are vectors, the $P,A,B$ are matrices

$B$ is the matrix of a linear transformation $T$ in the basis $\mathcal{B}'$

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@DavidWheeler That relation is the only reason I knew why $Y' = BX'$ haha

@DavidWheeler That is what I suspected, thanks

This is an example of a problem-solving procedure called "the by-pass"

let's say you want to know how to get from point $A$ to point $B$, but you don't know how

But you do know how to translate your problem to another arena, in which you know how to get anywhere.

so you: translate, go where you want, and translate back

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I am not sure if you are referring to the changing of the notation, or my trying to find the step that was missing by finding their difference(I suspect the latter, with my slight misinterpretation)

For example, suppose you want to go 1 block north, but there's a car crash.

you go one block east, no car crash there, go one block north, and go one block west. Voila! problem solved.

This technique is common throughout mathematics

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Where did I use this here?

7:18 AM

The change-of-basis matrix is the "translation"-it takes you from your new basis (where you don't know what matrix you want) to your old basis (where you're super-confident in doing everything)

Another name for a change-of-basis matrix is: linear isomorphism.

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Ahhhh I see

Fair enough

then you do your matrix thing in your old basis (which is where you know everything already), and translate back

that solves your problem

And this sort of stuff actually happens. For example, you probably know that the derivative of a quadratic is: $(at^2 + bt + c)' = 2at +b$

In the basis $\{t^2,t,1\}$ for the space of all quadratic and lower degree polynomials, we have the derivation matrix: $D = \begin{bmatrix}0&0&0\\2&0&0\\0&1&0\end{bmatrix}$

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Very true, good point

Now, we might (for some reason) want to consider the basis $\{t^2-2t+1,t-1,1\}$

That is, powers of $t-1$ instead of $t$.

Then our matrix $P = \begin{bmatrix}1&0&0\\-2&1&0\\1&-1&1\end{bmatrix}$

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Ahhh I see

7:30 AM

We find $P^{-1} = \begin{bmatrix}1&0&0\\2&1&0\\1&1&1\end{bmatrix}$

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Good demonstration :)

So $P^{-1}DP = \begin{bmatrix}0&0&0\\-2&0&0\\0&1&0\end{bmatrix}$

oops, did it wrong

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Is that the only matrix similar to?

I mean is $P^{-1} D P$ if it were correct above the only matrix similar to $D$?

Next is trying to find out why the conjugate transpose of a complex valued linear operator is represented by a symmetric matrix

It should be $P^{-1}DP = \begin{bmatrix}0&0&0\\2&0&0\\0&1&0\end{bmatrix}$

So we would expect $D(a(t-1)^2 + b(t-1) + c) = 2a(t-1) + b$

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D is similar to itself?

7:42 AM

We can check this:

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Oh of course lol(similarity is an equivalence relation)

$(a(t-1)^2 + b(t-1) + c = at^2 - 2at + a + bt -b + c = at^2 + (b - 2a)t + a-b+c$

which has derivative: $2at + b-2a = 2a(t-1) + b$.

The reason that there's "no difference here", of course, is that $D(t - 1) = Dt$.

But if we chose a wildly different second basis (like Legendre polynomials), it would give a different matrix

Now, here is another interesting aspect-suppose $A$ is diagonalizable, so there exists an invertible $P$, such that:

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Ahhh yes of course I should have seen that

$P^{-1}AP = D$

Then $AP = PD$

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And thus we have a basis of e-vectors?

e.g. semi simple?

7:48 AM

Yes, which can be seen by looking at any column of $P$

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Can you help me understand why the conjugate transpose of a complex-valued linear operator is represented as a symmetric matrix? I can't see it from the definition(or maybe it isn't true, another student claimed it is)

I don't think it is true.

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What is the linear operator equivelent of a symmetric matrix in $\Bbb C$?

Do you know it?

They are called Hermitian matrices

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I'll go read it now

7:54 AM

They're not symmetric, they're equal to their conjugate transposes

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Which is a sort of conjugate symmetry

Does that sort of symmetry have a name?

An example would be: $\begin{bmatrix}2&1-i\\1+i&3\end{bmatrix}$

Basically we want $v^Hv$ to be real, so that it can be used to induce a metric.

This, in turn means that $v^HAv$ is real, for such a matrix.

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How old are you?(If you don't mind telling me)

Just as symmetric matrices can be diagonalized by an orthogonal change-of-basis, Hermitian matrics can be diagonalized by unitary matrices.

I am 53.

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That is a little beyond me, but I will read these terms, thanks

8:10 AM

It's better to deal with the real case first (symmetric matrices), before moving to complex inner-product spaces.

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Yeah for sure haha

I have done almost nothing with inner product spaces, and that which I have done was a year ago

Well, typically, people first learn the "dot product" which is ONE possible inner product out of many.

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Yeah I have used dot product a bunch before when I was doing actual linear algebra

Inner products are intimately related to symmetric matrices, since they are bilinear functionals.

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Inner products are bilinear functionals?

8:17 AM

They are linear in each argument, ya?

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Yes, I see, I had never seen the term before was all

The symmetric matrix associated with the "standard" inner product, is the identity matrix: $\langle x,y\rangle = x^TIy$

replace that with a different symmetric matrix $A$, and you get a bilinear functional $x^TAy$

If $A$ is positive-definite, you have an inner product.

Often such things are called "weighted" inner products, as they assign different "weights" to the scalar coordinates.

Such things first arose in considering how to balance a region with different weights in different positions at the center of gravity

Committing to a challenge

Interesting stuff

I believe I will be learning this stuff over the next few weeks

(I hope so)

They're still used in statistical analysis to compensate for errors in measurement

Committing to a challenge

I have to go now, I will be back later. You have been overwhelmingly helpful @DavidW Thank you very very much!

8:30 AM

bye!

9:18 AM

Assume we have a semidefinite matrix inequality, $ A \circ B >= C$. Can we divide both sides of inequality and get $ A>= C \oslash B$. i.e. divide both sides of a semidifinite inequality by a semidefinite matrix. Division is elementwise.

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