Mathematics - 2015-05-24 (page 1 of 3)

Paul Plummer

12:00 AM

Exercise: what is the contrapositive?

Clarinetist

Well, we have here $\exists $linearly dependent set $\subset X$ $\implies $ $X$ linearly dependent

So the contrapositive would be...

(dang, I wish I actually took a course on predicate logic)

Paul Plummer

Just so you know every non empty set, that is not just the zero element, will contain a subset that is linearly independent.

Clarinetist

$\sim (X $ linearly dependent$)$ $\implies $ $\sim (\exists $ linearly dependent set $\subset X)$

So we have $X$ linearly independent $\implies$ ... $\exists$ changes to $\forall$, that's all I know

I think I'm just going to try a direct proof for the time being @PaulPlummer

Paul Plummer

There exists a linearly dependent subset "changes" to all subsets are linearly independent. Statements like these should become natural (and logical :P) if you end up doing more proof based mathematics. Also a "direct" proof is just as easy @Clarinetist

Samuel Yusim

X linearly independent $\implies$ for all subsets $S$ of $X$, $S$ is linearly independent.

Clarinetist

12:08 AM

@PaulPlummer Yeah, I'm very, very rusty. Just trying to get back to everything after a year of being an Access and Excel slave has been a bit difficult.

Samuel Yusim

the problem that makes the contrapositive non-obvious above, in my mind, is that you didn't give a name to your linearly dependent set $\subset X$

Clarinetist

Okay, let's try a contrapositive proof

Suppose $X$ is linearly independent.

Take an arbitrary $Y \subset X$.

Set $Y = \{x_1, \dots, x_m\}$ WLOG

Then wouldn't this just be the same as what I have in the proof already, starting with "Since $X$ is linearly independent"? @PaulPlummer

Paul Plummer

Ah yah it does, (although what you said above was not the contrapositive) @Clarinetist So your proof was right, I just didn't see it :D

Clarinetist

K so I have some wording to change. Thanks @PaulPlummer @SamuelYusim

New proof of the Lemma @PaulPlummer @SamuelYusim :

If $k = 0$, choose $a_2 \in \mathscr{F} - \{0\}$ and $a_1 = 0$. Then $a_1x + a_2kx = 0$. Hence $X$ is linearly dependent.
If $k \neq 0$, let $a_1 = -a_2k$, $a_2 \in \mathscr{F} - \{0\}$. Thus $a_1x + a_2kx = 0$, hence $X$ is linearly dependent.

user147690

@PaulPlummer I timed my central extension talk and I didn't finish it in 7 minutes lol(and I'll be doing it tomorrow unless he is busy

Paul Plummer

12:19 AM

Oh, I thought you were not going to do it?

user147690

@PaulPlummer I will be in the end(both this and the inverse limit of rings, but that one will be later on)

pjs36

Jeeze, inverse limits of rings? That sounds pretty wild

user147690

@pjs36 It is lol, fortunately balarka gave me some good ideas for the p-adic's

Clarinetist

@PaulPlummer Corollary proof:

In $Y_m$, pair two of the vectors together. By Lemma, this pair of vectors forms a linearly dependent set, and by Theorem, $Y_m$ is linearly dependent. Since $Y_m \subset X$, $X$ is linearly dependent by Theorem.

How do these look?

Paul Plummer

It looks good. Although it is a little indirect on the last part, why not just look at a pair that is linearly dependent and go directly to $X$ is linearly dependent, instead of having this middle man $Y_m$ @Clarinetist

Clarinetist

12:25 AM

K I will do that @PaulPlummer. Thank you!

Soham Chowdhury

12:39 AM

Damn, I'm sick and pissed I can't breathe.
Good morning, guys.

user147690

@SohamChowdhury That's awful. Hopefully it doesn't make studying hard

Soham Chowdhury

It happens whenever my mom leaves the air-conditioner on all night. I'm pretty used to it, and a shower sort of fixes it.

I'm currently on my second pass of chapter 2. Homomorphisms and frees today.

@AlexClark what time is it there?

user147690

@SohamChowdhury 4.5 hours later than where you are I believe(e.g. I am at 10:44, so you are at 6:14 I believe)

Soham Chowdhury

Oh, so quarter to 11?

Yeah.

Paul Plummer

Free groups ftw

Soham Chowdhury

12:45 AM

Yeah!

user147690

Still haven't mentally mapped the US properly, but that's because they have 4 hours difference across them pretty sure

Soham Chowdhury

@AlexClark I gave up a long while back.

Paul Plummer

Actually just asked a question today about them, sort of

user147690

But it's like 6pm or something

user147690

So 5-7 is my mental image

Soham Chowdhury

12:46 AM

Still working on rings, right?

user147690

Yes, but I have been side tracked worrying about my speech, and doing functional analysis things

Paul Plummer

We passed the School, where Children strove
At Recess -- in the Ring --
We passed the Fields of Gazing Grain --
We passed the Setting Sun -

user147690

And when I say worrying I do mean worrying

Soham Chowdhury

Hmm

Paul Plummer

When I first read that, I totally thought Emily must have been a mathematician secretly

user147690

12:50 AM

@PaulPlummer That is a logical conclusion. Outside of math what even is a ring or a field?

Paul Plummer

(Emily Dickinson)

Soham Chowdhury

My favourite:

"Much Madness is divinest Sense / To a discerning Eye"

(I found it in A Beautiful Mind)

@AlexClark The One Ring, of course.

Paul Plummer

Well what do you think the one ring to rule them all was, it certainly was not some bit of metal that goes around your finger, that would be ridiculous.

Soham Chowdhury

Hahaha

user147690

@PaulPlummer What ring was it though? What ring does rule them all? $\Bbb C[x]$? Is $\Bbb H[x]$ a ring?

Paul Plummer

12:56 AM

Every one knows it's e143ac3dc39cf097879d54543b089d82ae633800

Hmm, I guess it wants to stay a secret...

@AlexClark Yes it is a ring

Soham Chowdhury

@AlexClark we have a little special day here today. I'll be away at my grandma's, dunno about Balarka.

Note that I'm in it for the food mainly, lol.

user147690

@PaulPlummer What is this encoded in? I can't work it out

Paul Plummer

I think it is SHA-1 hex

Soham Chowdhury

md5, sha256, something irreversible

user147690

Hex was the first thing I tried :\

Paul Plummer

1:00 AM

Pain in the ass to reverse @SohamChowdhury, Don't hash anything in md5 if it is suppose to be secure

user147690

What does it say?

Soham Chowdhury

I know, I read HN

:P

Paul Plummer

Don't know what HN is

Soham Chowdhury

Hacker News

Paul Plummer

Oh

user147690

1:01 AM

@PaulPlummer I take it you have also taken crpytography courses?

Paul Plummer

No

user147690

Oh

user147690

But you know so many random things? Do you also know how the enigma worked?

Paul Plummer

It was a series of tubes

:P

user147690

Wut?

user147690

1:02 AM

It was bio tubes

user147690

A green slime transmitted energy through the biotubes

user147690

Combined with ESP use by different work stations

Paul Plummer

It has been explained to me how it worked mechanically before, but I don't remember

user147690

For the record it was a plug board that switched letters in pairs, then through a gear system that works much like a clock(26 turns of the first, turns the second, 26 turns of the second turns the third) and then a reversing drum and back through

Paul Plummer

Yah something like that

user147690

1:06 AM

Where the gears and reversing drum were hardwired

Paul Plummer

And then there was a series of tubes

user147690

Tubes?

user147690

I don't remember any tubes?

user147690

Except the ESP based biotubes

Paul Plummer

Maybe its an American phrase (since it is making fun of a US senator)

user147690

1:08 AM

Oh, I don't know it(nor any US senators)

user147690

Sorry for letting you down :P

Paul Plummer

Series of tubes

It is okay, it will just take me a while to forgive you

Clarinetist

Suppose I have a matrix $A$. What does it mean when we say the columns of $A$ are mutually orthogonal? Does it mean that for all pairs of columns $a, b$ of $A$, $a \cdot b = 0$?

user147690

@PaulPlummer oh lmao

pjs36

That's exactly right, @Clarinetist

Clarinetist

1:12 AM

So for instance, if $A = \begin{pmatrix}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 1 & 1
\end{pmatrix}$, taking the dot product of the first two columns gives $2$, so thus the columns of $A$ are not mutually orthogonal.

Soham Chowdhury

@PaulPlummer "an Internet was sent"

lol

Clarinetist

But if we had removed one of the first two columns of $A$, that wouldn't be the case.

pjs36

Don't the last two columns also give you problems? They're at a $45^\circ$ angle, if you think in terms of $\Bbb R^2$, not a right angle

Clarinetist

Oh yeah

But, for example, $B = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}$ would have mutually orthogonal columns

pjs36

Yep, it definitely would. It's even better than orthogonal, you'll see it's also orthonormal, meaning all columns have magnitude 1

Clarinetist

1:18 AM

Oh, of course - take $1^2 + 0^2 + 0^2 = 1$, square root it... I see

Thanks @pjs36

pjs36

Any time :)

Clarinetist

@pjs36 When trying to find orthogonal bases (of a column basis for a matrix), do you just use Gran-Schmidt or is there an easier way? I know that Gran-Schmidt gives an orthonormal basis, so I thought that might've been perhaps overkill

Oh wait

Duh

Never mind, 'twas a stupid question

The Substitute

off topic - i am trying to understand an example from dummit/foote. Let $G=S_n$ act on the set ${1,2,...,n}$ by $f*i=f(i)$. Why is the permutation representation of this function the identity map on $S_n$?

user147690

@TheSubstitute Because it permutes $i$ to $i$

user147690

@TheSubstitute it gives you $(f(i),f(i+1),f(i+2),\cdots, f(n))$ in matrix representation is the identity matrix

The Substitute

1:32 AM

@AlexClark doesn't it permute $i$ to $f(i)$?

user147690

Do you know matrix represenation?

The Substitute

Yea, wouldn't the matrix be a $2xn$ matrix with first row $1,2,...,n$ and 2nd row $f(1),f(2),...,f(n)$?

user147690

No

user147690

Where is the example in D&F

The Substitute

1 second.

user147690

1:35 AM

Just so I don't give you incorrect advice without context

The Substitute

Example 1 of Section 4.1. Pg 113 of 3rd ed.

Paul Plummer

You have it worded strangely, but I think what it is getting at is that you have an action $S_n$ on $[n]$, which is the same thing as a group homomorphism $S_n \to \mathrm{Aut}([n])=S_n$, which is obviously the identity

@TheSubstitute

Where $[n]=\{1,...,n\}$

Clarinetist

On orthogonal bases: I am to find an orthogonal basis for the space spanned by the columns of six matrices. Is there an easy way to do this?

user147690

Usually we take a permutation matrix(a single one in each row and each column), and we construct a permutation by placing a $1$ in the $i^{th}$ row and $f(i)^{th}$ column

Paul Plummer

Does that sound right from the context @TheSubstitute, because I feel like that is right, but from the way you worded things, it sounds like you are saying something else

The Substitute

1:41 AM

I think I got it - the permutation representation is sending the permutation $f$ to itself?

Paul Plummer

Yes

The Substitute

ah, thanks fellas.

Clarinetist

Anyone wanna help me with my question? :P

pjs36

That sounds really terrible, @Clarinetist! I don't know any easy way, off-hand...

anon

I would call this the permutation action, not the permutation representation. The latter more typically refers the linear action of $S_n$ on the vector space $F^n$, which is not the topic of discussion.

Clarinetist

1:43 AM

@pjs36 Yeah, this stats book... I just want to finish the exercises in this section quickly so I can get to the generalized inverses section (completely new material to me)

pjs36

Probably because you know representation theory, @anon :P Much like algebra teachers love to call $\frac{b^m}{b^n} = b^{m - n}$ the quotient rule...

anon

there are many more than two quotient rules

quotient rules for exponents, radicals, logarithms, derivatives, ...

Paul Plummer

This is interesting, there is a question on the front page with over 500 views but 2 upvotes

pjs36

@Clarinetist Yeah, that was what I hated about stats. It was so computational, I just couldn't stand doing very computationally-demanding similar work, several times over

Clarinetist

I can write it out here... Give an **orthogonal** basis for the space spanned by the columns of each of the following matrices:

$$A = \begin{pmatrix}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 1 & 1
\end{pmatrix}, D = \begin{pmatrix}
1 & 0 \\
1 & 0 \\
2 & 5 \\
0 & 0
\end{pmatrix}, N = \begin{pmatrix}
1 \\
2 \\
3 \\
4
\end{pmatrix}, K = \begin{pmatrix}
1 & 0 & 0 \\
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 0 & 1
\end{pmatrix}, H = \begin{pmatrix}
1 & 0 & 2 & 2 & 6 \\
1 & 0 & 2 & 2 & 6 \\
7 & 9 & 3 & 9 & -1 \\

@pjs36 Why is that people only know extremes? I was really spoiled as an undergraduate... had a really nice mix of computational and theoretical classes.

I was doing actuarial and pure math classes simultaneously until I decided to switch to statistics, and... I'd like to study stochastic processes, and of course, I can't make up my mind if I want to do applied or theoretical.

If anyone has any suggestions for that orthogonal basis question, I would love to hear them :P

Paul Plummer

1:52 AM

Well I would start off first by finding a basis for each of those...

And then make them orthogonal :D

Clarinetist

@PaulPlummer Wait, I need an orthogonal basis for the space spanned by the columns of ALL of those matrices, not each respective matrix...? [or is my thinking you're wrong wrong? xP ]

user147690

It's sad when bounties are close to ending with no answers especailly at 300 bounty(more than the remaining rep of the user

Clarinetist

Because as far as I can tell, the space spanned by the columns of all of those matrices is $\mathbf{R}^4$.

Paul Plummer

No, find one for each of them

Clarinetist

Find one for each of them, make them orthogonal, and then take the union of those and remove any redundancies (i.e., vectors that are linear combinations of each other)?

Now the question is, how to make them orthogonal? Do I just use Gram-Schmidt?

Paul Plummer

1:58 AM

Well that is how you find an orthonormal basis, normally.... I think (I never really did to much with such things)

pjs36

The first isn't so bad: The first two columns span a single subspace, $c_1(1, 1, 0,0)^T$. The second two span the $x_3 x_4$ plane, so you can really pick $c_2(0, 0, 1, 0)^T$ and $c_3(0, 0, 0, 1)^T$

Scattered thoughts above, take with several grains salt

Clarinetist

@pjs36 Yeah, it's when you have to add all of those columns together when I get confused

Because as far as I know, the span of all of those columns is $\mathbf{R}^4$.

pjs36

Well, you can throw away one of the first two, and the column-space won't change

Clarinetist

Indeed

pjs36

So it really will be the span of those vectors above, with constants $c_1, c_2$, and $c_3$

Clarinetist

2:02 AM

Yep, I follow you there

I know that a vector in the column space of $A$ has to be in the form $(a, a, b, c)^{T}$.

Paul Plummer

I should really do some work from my Linear Algebra and Geometry by Kaplansky.... (they definetly do not span $\mathbb{R}^4$

pjs36

Ah, right! So you can just split that up into $(a, a,0,0) + (0,0,b,0) + (0, 0, 0, c)$, ignoring the transposes, as a base for vectors of that form

Clarinetist

@PaulPlummer IDK, maybe I'm wrong. My thought is if you were to take a linear combination of the columns of all of those matrices, you would get 4 distinct components, neither of which are pairwise scalar multiples of each other, so anything in the space spanned by the columns of all of those matrices would be in the form $(a, b, c, d)^{T}$.

pjs36

Well, that's not the base, scattered again. But that decomposition leads to a base

Clarinetist

Or dang, maybe I should just try computing the components. Ugh :P

19 columns.

This will not be fun :P

Paul Plummer

2:06 AM

Is $(1,2,0,0)^T$ in the space spanned by the columns? @Clarinetist

Clarinetist

@PaulPlummer Without doing (some long) computations, I'm not sure

pjs36

Wouldn't you have to get it as a span of the first two columns alone?

Paul Plummer

Well you said above that the vector has to be of the form $(a,a,b,c)^T$, is $1=2$? :P

(so not that long of a computation)

It is a three dimensional subspace

Clarinetist

Lol, yes, that wouldn't be in the space spanned by the columns of $A$, I agree. @PaulPlummer But when you take ALL of the columns of EVERY matrix into account....

Paul Plummer

Who is talking about that...

Clarinetist

2:10 AM

@PaulPlummer It says to give an orthogonal basis for the space spanned by the columns of each of the following matrices

pjs36

I think they mean, "For each matrix below, find an orthogonal basis for the space spanned by its columns"

Clarinetist

@pjs36 If that were the case, it's a much easier problem (just a bit long)

Paul Plummer

Not really it is easier the way you have it

pjs36

I personally have never heard of column spacing multiple matrices, so I'm assuming that's what they mean

Paul Plummer

"basis for the space spanned by the columns of each of the following matrices:" That is calculate the column space for each matrix individually, and an orthogonal basis

Clarinetist

2:13 AM

@pjs36 Yeah, they do talk about that in this book. I'm not sure how useful it is given my limited knowledge of linear algebra in statistics. They've even invented a notation for it. If $C(A)$ is the column space of $A$, you can think of $C(A,B)$ as the union of the column spaces of $A$ and $B$.

pjs36

Wow, maybe that is what they mean... weird!

Paul Plummer

And if you want to interpret it your way, then you can just look at the three you have already calculated, and find a column that is independent from the other three (which should be easy), then just use the standard basis

Clarinetist

@pjs36 Yeah, any idea why that would be useful? Beats me.

Paul Plummer

Are you sure they want you to calulate $C(A,B,D,E,F)?$

Clarinetist

@PaulPlummer That's how I interpret the question. The text is as I gave above. There was a similar question 8 questions ago

It gave me a matrix $A$ and asked if the space spanned by its columns it was equal to the spaces spanned by the columns of these 5 other matrices

Paul Plummer

2:16 AM

Which does not even make sense since the last matrix does not have columns of the same dimension

Maybe you interpreted that question wrong too

:P

pjs36

To be safe, you'd better do both then :P but the "spanned by each" signifies that you do it individually. I'm not sure.

Clarinetist

Idk :P

This isn't homework, thankfully. Just linear algebra review before I start the M.S. in October

I'm debating just skipping some of these

I think this is going to make me miss pure math somewhat, given these problems. Ugh. Wish I had the time or money to pursue a Ph.D. right now

Paul Plummer

Well if it is your interpretation, you can find a fourth vector that is not in the subspace generated by the ones you found, so you would have 4 linearly independent vectors

And if not, then they all would have to be in the subspace of the one you already calculated, hence you can just use the basis for the one you calculated.

And you already described what all the elements look like in the space you calculated, so it should not be too difficult from there.

@AlexClark Interesting question, I normally don't upvote. but I did this time, might even give it some thought

pjs36

While we're on the subject of interesting question, if anyone enjoys combinatorics, I've been thinking about this one

With pictures from this page

Soham Chowdhury

2:33 AM

That's really cool, @pjs36.

Try this one I like: $n$ children sit in a line. They can change places, but each child can only move by one place at most. How many possible configurations can they reach?
(Harder version: Kids in a circle.)

Don't look at my MSE question history, it has spoilers.

pjs36

The general case is very hard, but the $k = 2$ case is interesting enough, the parallels with Catalon numbers breaks down a bit, or at least isn't completely bijective when we discount symmetries

I'll have to think about that one too

Clarinetist

Here's a not-so-interesting question. I need to find the orthogonal complement of $$\left\{\begin{pmatrix}
a \\
a \\
b \\
c \end{pmatrix}\right\}$$ with components in $\mathbf{R}$. If I have an arbitrary vector $(a_1, a_2, a_3, a_4)^{\prime}$, taking the dot product, I have $a_1a + a_2a + a_3b + a_4c = (a_1 + a_2)a + a_3b + a_4c = 0$ to try to find the components.

Soham Chowdhury

I'm reminded of hypergraphs somehow @pjs36.

@Clarinetist use $V^{T}$ for heavens' sake

Paul Plummer

Solution, push them into a volcano, so there is only one configuration

Clarinetist

@SohamChowdhury Just trying to adjust to statistics notation, unfortunately. :/

Soham Chowdhury

2:36 AM

@PaulPlummer That's just . . .

@Clarinetist Don't mind.

Paul Plummer

Get comfortable with statistics notation in the privacy of your home

2

Clarinetist

I hope my professors don't mind if I use $T$, but just in case...

Well, I guess I could set $a_3 = a_4 = 0$ and $a_1 = -a_2$

Soham Chowdhury

@pjs36 look at any one of the children at the two extremes, find an extremely familiar recurrence :)

Clarinetist

So $$\left\{\begin{pmatrix}
a \\
a \\
b \\
c \end{pmatrix}\right\}^{\perp} = \left\{\begin{pmatrix}
t \\
-t \\
0 \\
0 \end{pmatrix}\right\}$$

?

Soham Chowdhury

@Clarinetist pls

'sbetter

Clarinetist

2:40 AM

or is that just a subset?

:P

Paul Plummer

Well there is only one other dimension to fill, so if that works, then that is the whole thing

Soham Chowdhury

2:53 AM

Damn, this Polya enumeration theorem is great.

Clarinetist

Okay, this is bugging me more than it should. So I have this vector in $\mathbf{R}^3$: $(a, b, b)^{\prime}$. The book says that $\{(a, b, b)^{\prime}\}^{\perp} = \{(0, c, -c)^{\prime}\}$. I'm not getting how they get this.

Take, for example, a vector in $\mathbf{R}^3$: $(c_1, c_2, c_3)$. Dot it with $(a, b, b)$ and you get $ac_1 + bc_2 + bc_3 = 0$, so that $c_1 = \dfrac{-b}{a}c_2 - \dfrac{-b}{a}c_3$. So we have $$\{(a, b, b)^{\prime}\}^{\perp} = \text{Span}\{(-b/a, 1, 0)^{\prime},(-b/a, 0, 1)^{\prime}\}$$, not matching with what I have

Soham Chowdhury

What's that bottom for?

$\perp \leftarrow$ this

Clarinetist

That is the orthogonal complement

Soham Chowdhury

Oh

I can see one thing: $(a,b,b)\cdot(0,c,-c) = 0$ if that helps

Clarinetist

Yes, that's clear. But how do you derive it?

or, a better question would be, is my answer wrong?

Soham Chowdhury

3:08 AM

Derive what? (Note that I don't know any lin-alg)

Clarinetist

It is obvious that $(a, b, b) \cdot (0, c, -c) = 0$. But how do I know that only vectors in the form $(0, c, -c)$, when dotted with vectors in the form $(a, b, b)$, would give a dot product of 0?

Soham Chowdhury

Oh.

Clarinetist

In other words, is the orthogonal complement unique?

Soham Chowdhury

Oh, so that "orthogonal complement" is the set/space of vectors which dot to zero with it?

Clarinetist

Yes

@SohamChowdhury If you are familiar with set notation, assuming we're working in $\mathbf{R}^n$, for $A \subset \mathbf{R}^n$, $$A^{\perp} = \{x \in \mathbf{R}^n \mid x \cdot a = 0, a \in A\}$$

[it's not 100% correct, but just for our purposes]

Soham Chowdhury

3:16 AM

Yeah, I got it

Mike Miller

@Clarinetist: You're assuming $a$ is nonzero in your discussion above, so that you can divide by it. Remember that the orthogonal complement is the set of vectors that are orthogonal to everything in your subspace.

So if $(c_1, c_2, c_3)$ is in the orthogonal complement of $\{(a,b,b) \mid a, b \in \Bbb R\}$, it had better be perpendicular to $(0,1,1)$ (aka, $c_2 = -c_3$) and it had better be perpendicular to $(1,0,0)$ (aka, $c_1 = 0$).

Clarinetist

@MikeMiller Probably a dumb question, but is a way to generate $A^{\perp} \subset \mathbf{R}^n$ algorithmically? I can see why my method wouldn't work since no matter what you do, division would end up excluding cases where at least one of the components is equal to zero.

Mike Miller

I can't parse the second part of your question. For the first part, yes. Pick an orthogonal basis of your subspace and extend this to a basis of $\Bbb R^n$. Now apply Gram-Schmidt. The last $n-m$ terms, if $m$ is the dimension of $A$, are a basis for $A^\perp$.

This is almost certainly rather inefficient. There's probably a better method.

Clarinetist

@MikeMiller Thanks, that's all I needed. What I meant in the second part was that if I were to solve for (in my work above) either $c_1$, $c_2$, or $c_3$, it would require making $a$ or $b$ equal to $0$, so when dividing by either $a$ or $b$, only the cases when $a \neq 0$ or $b \neq 0$ respectively would be taken into account given the way I'm trying to compute $A^{\perp}$. If I'm being too wordy, sorry

So I should really try to learn the Gran-Schmidt process for this

Mike Miller

Oh, I'm looking at your question again, and I think you misread it. I think you're reading it as: $a$ and $b$ are fixed real numbers. Find the orthogonal complement to $\text{span}\{(a,b,b)'\}$. What it means to say is: consider the subspace $A = \{(a,b,b)' : a,b \in \Bbb R\}$. Find $A^\perp$. (Note that here, $a$ and $b$ are not fixed! $(1,0,0) \in A$ and $(0,1,1) \in A$.)

Kaj Hansen

3:28 AM

@BalarkaSen, Gram-Schmidt on the face of it is kind of a mess. Draw a picture and look at the geometry, however, and there is hardly any "learning" involved. It's pretty intuitive.

Mike Miller

No, probably not. That was the first algorithm that came to mind as someone who doesn't think about this very much. There are almost certainly better ways to algorithmically find the orthogonal complement of a set.

Clarinetist

Yep, so let's take an arbitrary vector in $\mathbf{R}^3$, say $(c_1, c_2, c_3)^{\prime}$. Dot it to $(a, b, b)^{\prime}$, you get $ac_1 + bc_2 + bc_3 = 0$. I thought of solving for $c_1, c_2, c_3$, but this doesn't work so well since I would have to assume either $a$ or $b \neq 0$. @MikeMiller

Mike Miller

@Clarinetist: Here's the trick here. Set $a=1, b=0$ and see what criterion this gives you. Set $a=0, b=1$, and see what criterion this gives you. These together will determine the entire orthogonal complement.

i.e., $a=1, b=0$ gives $c_1 = 0$. $a=0, b=1$ gives $c_2 = -c_3$. This is as desired.

Clarinetist

@MikeMiller Thank you, perfect explanation! Now I imagine I would have to do something similar for $(a, a, b, c)^{\prime}$ in $\mathbf{R}^4$.

Kaj Hansen

@pjs36, in regards to the starred message, one of the problems given to us in homework when I took combinatorics was to establish the relationship between $Q(n,k)$ and the Catalan numbers. Interesting question indeed.

Clarinetist

3:35 AM

@MikeMiller I have a feeling the solution isn't unique...

Mike Miller

I don't understand what you mean by 'the solution isn't unique'. Doing the same strategy gives $x_1+x_2 = 0$, $x_3 = 0$, $x_4 = 0$. The orthogonal complement is $\text{span}(1,-1,0,0)$.

Clarinetist

@MikeMiller Okay, I'm definitely wrong. So let me gather this... the idea is set one of the arbitrary values equal to $1$ and set all others equal to $0$.

Mike Miller

Yes.

Clarinetist

@MikeMiller Thank you!

Clarinetist

4:08 AM

Question: is the space spanned by the columns of $$X = \begin{pmatrix}
1 & 1 & 4 \\
1 & 2 & 1 \\
1 & 3 & 0 \\
1 & 4 & 0 \\
1 & 5 & 1 \\
1 & 6 & 4 \end{pmatrix}$$ just $\mathbf{R}^6$, since no two components inside a vector inside the span of the columns are the same?

Samuel Yusim

if that were true then $\mathbb{R}^6$ would have a 3-element basis

Clarinetist

Ah dang, that's right @SamuelYusim

So I need to find an orthogonal basis for that space.

It's clear that the columns create a basis for that space.

So now...

Does anyone have tips on finding an orthogonal basis?

Balarka Sen

@KajHansen Dunno how I got pinged but TOTALLY NOT.

Gram-Schmidt is pretty intuitive and not a mess.

Kaj Hansen

What @BalarkaSen ?

Oh absolutely. That was my point @BalarkaSen

But if you look at just the algebra without drawing any pictures, it looks a bit messy

Balarka Sen

Coincidentally, I was doing Gram-Schmidt to show that $GL_n(\Bbb R)$ deformation retracts to $U(n)$ at about the same time :P

How can you work with inner products and not draw pictures?

Kaj Hansen

4:21 AM

haha, I don't even know what the process is off the top of my head, but I know I can derive it in a few minutes with some sketching

Soham Chowdhury

@BalarkaSen No jamai shoshti stuff going on at your place today?

Kaj Hansen

Did I tell you I was taking another algebra course this fall @BalarkaSen ? I'm very excited.

Balarka Sen

Yes, but that doesn't involve me, @Soham. :P

Cool, @Kaj. Yes, you told me.

Are you going to learn anything new?

Kaj Hansen

Definitely. This time around, it's a course for graduate students preparing for their quals.

Balarka Sen

ooh, nice.

Kaj Hansen

4:24 AM

It's my first of this kind, so I'm anxious, but also excited since I really enjoy the subject.

Balarka Sen

Hey, did you think about the Galois theory problem I gave you?

Algebra is very hard, but once you get used to it, it gets addictive.

Kaj Hansen

About extensions of $\mathbb{C}(t)$? I can't say I have. I guess I just haven't been doing that much math over the past week or two. About the extent of what I've done mathematically is introducing my girlfriend how to reading and writing proofs.

I'm actually putting a lot of energy into learning how to drive right now and preparing for getting my license. I've put that off for much too long.

Balarka Sen

noise of impatience

Kaj Hansen

Yeah, I get super lazy every May. Very bad habit of mine.

Balarka Sen

I am always lazy, but you know why I am showing impatience. :P Anyway, that one is a good problem.

Kaj Hansen

4:29 AM

Great response on that covers question yesterday. I enjoyed reading it, and ended up deleting mine :P

Balarka Sen

yes, I noticed. you shouldn't have, it was fine.

I just expanded on your answer and comments, that's all I did

Kaj Hansen

Eh, I felt like you did a better job conveying everything I did, and your response was more organized and so forth. I often have trouble responding to the softer questions since I don't know the OP's level of mathematical maturity. So I feel like I end up fumbling around a bit.

Also, I felt a little uncomfortable not being able to speak at all to why they are looking at covers in PDEs. I know literally nothing about PDEs.

Balarka Sen

@Kaj just in case you're interested : galois extensions of $\Bbb C(t)$ has a completely geometric description. look up covering spaces, you'll be able to understand the definition now that you know topology. it's a provable fact that category of galois extensions of $\Bbb C(t)$ corresponds to covers of $S^2$ minus finitely many points.

Kaj Hansen

Oh boy. More algebraic topology :P

If I do well in my course this fall, I could take algebraic topology in the spring if I choose.

Balarka Sen

I think it's pretty fascinating if every extension of $\Bbb C(t)$ of the same degree are isomorphic, if it's true.

Kaj Hansen

4:34 AM

No doubt! That's certainly not true everywhere.

Balarka Sen

that's nice. but you don't need algebraic topology to understand this geometric description of galois groups, though

Kaj Hansen

I guess it's true over finite fields, but blatantly false over $\mathbb{Q}$.

Balarka Sen

right.

which year of undegrad are you in, now?

user147690

What is $\Bbb C(t)$?

Kaj Hansen

Oh @BalarkaSen - I think I might've talked to you about it in the past, but have you ever thought about the group theory behind perfect shuffles on decks of cards?

@AlexClark, the field of rational functions with coefficients in $\mathbb{C}$.

Balarka Sen

4:38 AM

That's symmetric group, isn't it?

Oh, fixed-point-free shuffles.

Kaj Hansen

I'm going to be a 4th year undergrad this fall.

And yeah. Cut a deck in half, weave the cards together one at a time from each side

It's one of my favorite problems I've thought about as an undergrad, mostly because I discovered the pattern independently.

So if you have a deck of an even number of cards, say $N=2n$ for some $n$, this shuffle will be a permutation in $S_N$, and so doing it over and over will eventually return the deck to its original position. But its order depends heavily on what $N$ is.

Balarka Sen

Right, yes, because every elt of $S_N$ has finite order.

Kaj Hansen

mhmm

Balarka Sen

that's interesting. I don't know if there's an easy formula for the number of fixed-point-free elts in $S_n$.

Kaj Hansen

You are indeed correct. However, for this specific problem, I can tell you what the order will be in terms of the order of an element in a better-understood group.

Balarka Sen

4:49 AM

i'm all ears.

Kaj Hansen

So it turns out that a deck of $2n$ cards will return to its original position after $y$ "perfect shuffles", where $y$ is the order of $2$ in the multiplicative group $\mathbb{Z}_{2n+1}$. And the problem is, of course, there is no general formula for computing that.

@BalarkaSen

Balarka Sen

ahh

Kaj Hansen

Here's a sketch of a proof: math.stackexchange.com/questions/886083/…

There's a bit of confusion at the beginning since my definition and the OP's definition of a "perfect shuffle" are slightly off.

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