Mathematics - 2012-03-15 (page 3 of 4)

Jonas Teuwen

18:06

@Ilya No.

Matt N.

Hi Jonas : )

arete

Is anyone willing to help me out with some linear algebra concepts?

Matt N.

We can try.

arete

Thanks, I just want to discuss Span and Null space.

I would like to solidify my understanding.

So the span is the set of all linear combinations of $v_1\ldots v_r$.

Matt N.

Yes.

arete

18:11

Basically I'm looking at the space that my vectors are capable of taking up.

All the possibilities.

Matt N.

Yes.

arete

So when I'm looking to solve for the given set of vectors's span, what should I be keeping in mind?

Matt N.

I'm not sure what you mean by that.

arete

I understand that when given given a set of linearly independent vectors that it takes up the $\mathbb{R}^n$ however, I'm not quite sure I follow what that means.

So say I'm given the vectors $a=\begin{bmatrix}1\\-1\end{bmatrix}$ and $d=\begin{bmatrix}1\\0\end{bmatrix}$. How should I approach finding the span?

Matt N.

You haven't stated a problem so far.

arete

18:18

Sorry, I'm a little out of it.

Matt N.

The dimension of a vector space is the number of its basis elements and the basis is defined to be a maximal set of linearly independent vectors.

@arete No problem, don't worry : )

arete

We haven't covered basis yet.

Matt N.

Ok.

I'm not sure what they expect you to write down when they ask you to "find the span". The vectors you have there are linearly independent so you might want to prove that.

arete

For the problem above, if I were to go about creating an augmented matrix with some arbitrary vectors $v_1$ and $v_2$. When I solve that augmented matrix, what am I necessarily accomplishing?

Matt N.

What's an augmented matrix?

Hey there, robjohn!

arete

18:22

$\begin{bmatrix}1&1&v_1\\-1&0&v_2\end{bmatrix}$ would be the augmented matrix.

[A|b] Where A={a,d}

Nimza

Guys, do you know some alghoritm that allows to find eigenvectors of a matrix nearest to given ones? Say, given vectors [a1,...,an], i have to find eigenvectors [b1,...,bn] of matrix B such that b1 is the nearest to a1, etc. ?( I wanna receive a continuous time-dependent eigenvector of a time-depedent matrix)

Matt N.

@arete And what does solving it mean?

Godisemo

Hi, do you guys know how I can prove a subset of a metric space is closed?

arete

That's what I'm trying to understand

Why does solving this matrix give me the information that I need?

Matt N.

@Godisemo Show that its complement is open, in most of the cases. If that won't do you need to give more information.

@arete I haven't understood what "solving the matrix" means.

Usually you have something like $Ax = b$.

Then you solve for $x$.

arete

18:24

Sorry reducing to rref.

Matt N.

What's rref?

arete

Reduced Row Echelon Form.

It's kind of overkill to use it for a system of this size

Godisemo

Consider the space l^1(Z_+,R) but with the supremum norm instead of the l^1 norm. My subset is B={a; sum_i^inf abs(a_k) <= 1}

arete

So here's a set of vectors that I was given for the exercises. I'm looking at $a=\begin{bmatrix}1\\-1\end{bmatrix}$$b=\begin{bmatrix}2\\-3\end{bmatrix}$ $c=\begin{bmatrix}-2\\2\end{bmatrix}$d=\begin{bmatrix}1\\0\end{bmatrix}$e=\begin{bmatrix}0\\0\end{bmat‌rix}$

How would you find the Span of S (Sp(S)) if S={a,c,e}?

Matt N.

@arete I'd try to determine whether $a,c,e$ are linearly independent and if they are you have $\mathbb{R}^3$. If they're not then you'd have to see what dimension they span... This might be what they're asking for but I'm not sure.

arete

18:33

Ok, so it's clear that a and c are not linearly independent.

as well as e.

From that point how would I go about finding the span of the system?

Matt N.

Well as soon as you have the zero vector in your set your set is going to be linearly dependent.

arete

Yes

Matt N.

You observe that you have only one vector so you get one dimension. That's a line in $R^2$ and it looks like $x = -y$.

arete

However my book has the answers $Sp(S)=\{\bf{x}:x_1+x_2=0\}$.

Matt N.

That's the same. Right?

arete

18:37

yeah

I'm just confused by why that's my span I guess.

Oh wait, so that's saying that my span of vectors is only capable of living on that line right?

So no matter what combination of vectors that I have, I'm only capable of having my vectors take up values that are equivalent to that line.

@MattN right?

Matt N.

@arete Yes! : )

arete

So it's the space that my set of vectors is allowed to "live" on.

Matt N.

Yes, exactly.

arete

Thanks for bearing with me there.

It's hard to articulate what you're trying to learn sometimes.

Matt N.

I know : )

arete

18:43

So, on to null space.

Null space consists of all the vectors $x$ such that $Ax$ is the zero vector.

Matt N.

Yes.

arete

Or more formally $N(A)=\{x:Ax=\mathbb{0}, x\in \mathbb{R}^n\}.$

that 0 is not a scalar.

I don't quite know how to properly notate the zero vector in latex.

So why is the null space important?

A system is linearly independent iff x=zero vector.

So would this be telling me whether or not my system is linearly independent?

Matt N.

If your zero space is only the one element set consisting of the zero vector you know that your $A$ is an injective map.

@arete Yes. Because that's basically plugging in the definition of linearly independent.

arete

Hmm I know what an injective map is, we just haven't even discussed it in my lin alg class.

Matt N.

If the zero space is only the zero vector this means that the only coefficients you can pick to make your vectors into a linear combination equalling zero is the zero coefficients.

arete

18:50

Ahh

What if my null space were something like $\mathbb{R}^2$?

Matt N.

OTOH, if your vectors are linearly dependent you know that you can find coefficients not all zero such that your vectors multiplied with these coefficients added up yield zero (even though not all coefficients are zero).

jay wendt

If $K$ is a Galois extension over $\mathbb{Q}$ and we are assured somehow that $K$ has a unique subfield $F$ of degree $2$, can we conclude that $F=\mathbb{Q}(\sqrt{D_f})$, where $D_f$ is the discriminant of the splitting polynomial?

Matt N.

@arete Then your $A$ maps all of $R^2$ to zero which is saying it's the null map.

arete

Hmm I like to see this map speak. It makes you think about the systems in question, instead of just manipulating a matrix or vector.

Asaf Karagila

What the... where the hell is there a place for a downvote here except perhaps someone who's angry at me for some reason?

jay wendt

18:54

i countered the passive aggressive downvote, asaf

Asaf Karagila

That wasn't needed, but thanks anyway!

jay wendt

your answer is fine

Matt N.

@Godisemo Sorry for taking so long: You are in a metric space $M$. So you know that a set $A$ is closed iff every sequence in $A$ that converges to a limit in $M$ has its limit in $A$. Here.

Asaf Karagila

I know it is. :-)

jay wendt

Asaf, do you have a comment about my simple question? i was led to believe i should be able to conclude this, but i don't see why exactly.

@Asaf

Matt N.

18:56

@AsafKaragila Who do you have in mind?

Asaf Karagila

@MattN I have no idea. I pissed a few people in the past two days... :-)

Matt N.

@AsafKaragila Heh. : )

Asaf Karagila

@jaywendt I haven't read your question. I'm struggling to bring myself to grade more exams.

jay wendt

@AsafKaragila, i understand

Asaf Karagila

@MattN I've had a hard time coping with life and I've been quite prickly lately.

Daniil

18:58

Hi, is anyone here interested in an invite to computerscience.SE private beta?

Godisemo

@MattN No problem. I know this property about limit points and read that it is probably simpler to use this property. The problem is that I don't understand how to use it.

Asaf Karagila

@Daniil Maybe Henning is. Ping him.

Matt N.

@AsafKaragila I know others... : )

Daniil

Ah, right. But he is not online

Asaf Karagila

@Daniil Pinging him will leave him a notification on the main site too.

@MattN I also know others. What's your angle?

Matt N.

19:00

@AsafKaragila I was hinting at grumpy people that aren't here atm.

Daniil

But I can't ping him. When I start typing @H nothing comes up

Asaf Karagila

@MattN Oh.

jay wendt

i think if you type his user name it will work

Asaf Karagila

@Daniil Just write him name.

Also, whatever you do: Do note ping Candyman three times in a row while looking in a mirror.

Matt N.

@AsafKaragila well I don't know about the "hard time coping with life"...

Daniil

19:01

Henning Makholm, @HenningMakholm

Matt N.

@Godisemo Now you need to either pick an arbitrary Cauchy sequence in your set $B$ and show that it converges in $B$ (that is, that its limit is again an absolutely converging series or you find an explicit sequence that is Cauchy but has a limit that is not an absolutely convergent sequence.

What a crappy day.

Time to have wine.

jay wendt

May I ask why it's crappy? It looks pretty good from this angle.

Asaf Karagila

Well said, @MattN. I like how you quit quitting alcohol! :-D

@jaywendt I have to grade tons and tons of exams if I want to get hammered on St. Patrick's Day. Also despite the junior staff strike there's a threat to cut my salary if I won't finish grading by Sunday. So I have to sit and work really hard today and tomorrow. Ergo, crappy day.

Matt N.

If you need me ping me, I'll be reading afk but within hearing distance of ping.

Asaf Karagila

@MattN This is a test. Can you hear me now?

Godisemo

19:06

@MattN Since $\sum^\infty_{k=1}|a_k|<\infty$ all cauchy series converge to 0, right?

jay wendt

@Asaf, that does sound crappy!

Asaf Karagila

Well, I shall return to force myself to grade... only 50 more. :|

Matt N.

@Godisemo Well I think you're confusing two things: yes, $a_k \to 0$ but your space is not the space of $a_k$s but rather the space of $(a_0, a_1, a_2, \dots )$ of sequences! So what you have are sequences of sequences.

Godisemo

@MattN that is confusing... :/

Matt N.

@Godisemo Yes it is but you will get used to it quite quickly. Fiddle with it for some time.

Jeff

19:21

Hey anyone, does @Henning mean "random" when he says "Brownian motion" in the third line of this answer: math.stackexchange.com/a/120274/22544 ?

Godisemo

Does this proof hold, for showing the complement is open?
The complement to $B$ is $\{x; ||x||_1>1\}$. Now choose an arbitrary point $a$ in the complement. Then create the open set $\{x; abs(||x||_1-||a||_1)<\epsilon\}$. Now $||a||>1$ and we can choose $\epsilon$ arbitrarily small so we get $||x||>1$. This x is then a part of the complement to $B$, so the complement is open.

Note that I used that $||x-a||_\infty < ||x-a||_1 < abs(||x||_1-||a||_1)<\epsilon$ when I created the open ball around $a$.

Matt N.

I misremembered $B$! Ignore what I just wrote.

Godisemo

@MattN $B=\{x; ||x||_1 <= 1 \}$

Matt N.

Yes, thanks.

Godisemo

So the complement is $B=\{x; ||x||_1 > 1 \}$

@robjohn What are you doing? :P

Matt N.

19:28

@Godisemo And did you prove $\|x-a\|_\infty < \|x -a \|_1 < abs(\|x\|_1 - \|a\|_1) $?

And why is the set you describe there open? I'm not sure I see that.

Godisemo

@MattN You have a point there

@MattN This is just using $||b||_\infty <= ||b||_1$ together with the triangle inequalities

Matt N.

Nitpick but wouldn't you have $\|x-a\|_\infty <= \|x -a \|_1 < abs(\|x\|_1 - \|a\|_1)$?

Godisemo

yes, true

I've been stuck on this problem for a couple of hours now...

Matt N.

(But your proof is sound. : ))

Godisemo

does it hold?

Matt N.

19:35

I meant how you argued that these inequalities hold.

Godisemo

ah

Matt N.

But your other proof looks good to me too, I hope I'm not missing anything.

Godisemo

The only thing I'm not 100% sure of it what you said earlier

that the set I created really is open

(but I'm quite sure it is)

N3buchadnezzar

Yarr mates!

Matt N.

@Godisemo It's open because it contains $x$ such that $\|a - x\|_1< \varepsilon$. So it's an open ball around $a$ with respect to the $1$-norm.

N3buchadnezzar

19:41

Anyone mind helping me with some trigonometry ? gosh I feel so embarressed =(

Godisemo

@MattN that is true.

@MattN Thanks for the help, just have to write it all down...

Matt N.

@Godisemo Pleasure : )

N3buchadnezzar

I need to prove that $$\sin(\pi - \arctan(c)) = \frac{1}{\sqrt{c+1}} $$

Nimza

Any real symmetric square matrix is O(n,R)-similar to the real diagonal. And what can we say about matrices, SO(n,R)-similar to the real diagonal?

is it a set of all real symmetric matrices?

N3buchadnezzar

HAH

I figured it out

anon

20:39

@Nimza: Yes. If the matrix of the similarity transformation A has determinant 1, then it's in SO and we're done, otherwise it has determinant -1 and the similarity transformation associated to -A is exactly the same, and -A is in SO.

Skullpatrol

@N3buchadnezzar How did you prove it?

anon

@N3bu, @Skull: It's actually incorrect.

Skullpatrol

@anon Why?

anon

@Skull: The actual answer is $c/\sqrt{c^2+1}$. To derive this, first use sin(pi-x)=sin(x), then construct a triangle with side c, base 1, and thus hypotenuse $\sqrt{c^2+1}$.

Skullpatrol

21:02

@robjohn may interrupt you to ask your opinion about something Sir?

robjohn

@Skullpatrol what's up?

Skullpatrol

@robjohn What do you think of this video.google.com/videoplay?docid=-5382609196671955257

robjohn

I'll have to get back to you in an hour and twenty minutes...

Skullpatrol

Np enjoy 1:19:42

Kannappan Sampath

I am sad that people don't understand hints here on the site.

Another user writes an answer totally derived from hints three hours later and (s)he gets three upvotes. This is holy crazy!

Asaf Karagila

21:35

All aboard the train to Hammersvile!

Matt N.

@AsafKaragila Done.

Teddy's absence makes me wonder whether he had an attempt at reinstalling his socket...

Kannappan Sampath

Well, this is my last guys and I will not return. I am bidding a fare well to all of you. Bye and take care.

Matt N.

@Profiletobedeleted Wot?

Skullpatrol

@Profiletobedeleted Why?

Matt N.

Oy.

@Profiletobedeleted What's happened?

Skullpatrol

21:44

@Profiletobedeleted Hope you all best in the future.

Benjamin Lim

@Profiletobedeleted Kannappan what is happening?

Skullpatrol

@MattN I think he's gone.

Asaf Karagila

Finally. Done with the exams. One more push like this tomorrow, and I'm done with that course altogether. Now: home, food, sleep.

Matt N.

@AsafKaragila No drink?

N3buchadnezzar

@Skullpatrol \sin +\arctan x = x * ( 1 + x^2 ) ^ -0.5

I just made a simple triangle

Skullpatrol

21:52

@N3buchadnezzar Thanks

N3buchadnezzar

And ofcourse there should be no + on the left handside

Skullpatrol

@MattN I wonder what happened to Kannappan?

Matt N.

@Skullpatrol Looks as if he's annoyed about how people vote and write answers.

Skullpatrol

Hi @BillDubuque

Bill Dubuque

It looks like Kananappan was upset about this thread. But there must be more to it than that. Does anyone know more?

N3buchadnezzar

21:55

Mind giving me a few tips on the following problem?
$\text{Find} \iiint_R x \, \mathrm{d}V \ \text{and} \ \iiint_R z \, \mathrm{d}V \\
\\ \text{over that part of the cone} \\
0 \, \geq \, z \, \geq \, h \left( 1 - \frac{\sqrt{x^2+y^2}}{a} \, \right ) \\
\text{that lies in the first quadrant}$
Sorry for nt asking more about Kananappan, I am sure he will be back soon =)

robjohn

@Profiletobedeleted This is just something that happens.

@BillDubuque I just returned, so I know no more.

I don't see much in the log.

Benjamin Lim

@robjohn He is not online on skype

Bill Dubuque

It's always a sad day when we lose a big contributor. I don't understand why some folks take votes so seriously.

Benjamin Lim

@BillDubuque Recall Kannappan is only 17. I will try to persuade him to return.

Matt N.

@BillDubuque Unfortunately I was afk and saw his previous message too late.

Otherwise I would've talked him out of it. It looks like a reaction made without a cool head to me.

Benjamin Lim

21:58

People downvote on this site. I have been downvoted at least 10 times. So what? It's the nature of things....

Matt N.

I think he was upset because of a full answer being upvoted.

@BenjaminLim Well good job you have him on skype.

Benjamin Lim

@MattN Add me on skype, we can have a teleconference to persuade him

N3buchadnezzar

Hahah =)

Great community

Bill Dubuque

Voting seems to be more and more erratic recently. Often good answers get little or no votes and trivial answers get most of the votes. This probably will drive some folks away from the site, esp. those who take votes too seriously. How can we fix this? Vote up good answers (and vote down poor answers if need be)

N3buchadnezzar

As long as everyone gets a vote, this will always happen

Matt N.

22:01

@BenjaminLim Not now. He's gone to bed probably, it's quite late there.

N3buchadnezzar

People are stupid, it is that easy

Benjamin Lim

Kannappan should know by now that there are questions on batman that get 200+ upvotes, very good questions that get at most 10 upvotes...

@MattN What is your skype?

N3buchadnezzar

I saw an answer that got 38 upvotes or something, and all it said was "W"

Matt N.

I'm not posting that here. Let me see if I can find you.

: D

Benjamin Lim

search that

Matt N.

22:03

k.

Benjamin Lim

Done.

Matt N.

See you in a bit. Going afk for a while.

Skullpatrol

Bye

Asaf Karagila

22:30

@MattN I did say food. Beer is food.

Skullpatrol

@robjohn Were you just being polite when you said, "I'll have to get back to you in an hour and twenty minutes...?"

@N3buchadnezzar Could you explain to me why you use a "3" in N3buchadn?

N3buchadnezzar

22:46

@Skullpatrol Nebuchadnezzar was taken

Skullpatrol

@N3buchadnezzar Thanks for clarifying that for me ;-)

N3buchadnezzar

@MattN haha

Jonas Teuwen

@MattN Not allowed?

t.b.

:3840282 hey there, I'm not sure. I haven't slept in 23 hours...

Skullpatrol

@MattN (removed)?

Asaf Karagila

22:50

@tb At least you have slept since 1945! :-)

Skullpatrol

@tb Sleep deprivation is not good for you pal.

t.b.

@Profiletobedeleted I have no clue what happened, but I sincerely hope you reconsider...

In any case: I had fun talking to you and it would be a pity if you left...

Skullpatrol

Dear Kannappan, please come back. (star this message if you agree)

9

Godisemo

Who is Kannappan?

Skullpatrol

Who are you?

Godisemo

22:56

fair question ;)

Skullpatrol

►

Matt N.

@tb 23 hours? But I thought you'd gone to bed early.

t.b.

Okay, that's it for me guys. I might come back a bit later, but not sure. I declare March 15 official drama day....

2

Asaf Karagila

@tb Oh noes. What happened?

Matt N.

No!

Transcript for

$Mathematics$ Mathematics