Mathematics - 2012-06-27 (page 1 of 3)

1:33 AM

It's like when you make love for the first time you have to go slow @JasperLoy

user19161

@BenjaLim I am still a virgin...

@PeterTamaroff That's what we call a spanning set

every single element in the given subspace will be the element of some spanning set

user19161

@BenjaLim The members in the spanning set are called generators too.

yes

1:34 AM

@BenjaLim We call it "sistema de generadores" ie "system of generators".

user19161

I think we are all clear about what he wants now.

ah ok

@PeterTamaroff Let me tell you first of all that if you have only one element in the generating set, it will NOT generate $S$

a spanning set is also called a generating set in English. the former is usually tied to vector spaces in particular, whereas the latter refers to groups (even non-abelian ones) more generally.

@JasperLoy Actually the plane is a subspace of $R^3$ so I want a system of generators that spans a subspace of $R^3$

$S$ is a plane

not a line

so you need at least 2

user19161

1:35 AM

@PeterTamaroff I have provided you a complete answer. My two vectors span your plane.

@BenjaLim Yes, I know!

@PeterTamaroff So

@PeterTamaroff Well, you kept saying "a" generator, which was confusing.

@BenjaLim I just couldn't see what to consider to solve the problem.

user19161

@anon Our job is to anticipate his questions!

1:36 AM

Because like I said, every element of the subspace is an element of some generating set.

@anon a system of generator*s*

@PeterTamaroff We only need to choose two vectors that are not scalar multiples of each other to generate the plane span $S$

this question ain't makin any sense to me at all

@BenjaLim Yeah, I was expecting that. I expected I could get two particular solutions and make a linear comination of those, like in ODEs, hehehe....

user19161

@PeterTamaroff Exactly. The two vectors I gave you, take linear combinations and you get the plane!

1:39 AM

@JasperLoy Yay!

quick quiz: what's a direction vector describing the perpendicular line? :)

@anon What are you talking about now?

nevermind

user19161

@anon I know...

Rajesh D

what is given here?

1:40 AM

However, I could have used three vectors, right?

@anon Oh, come on!!!

Rajesh D

a vector and you want perpendicular to it

user19161

@PeterTamaroff One would be redundant then.

user19161

@PeterTamaroff He means the vector perpendicular to the plane.

@JasperLoy Well, thinking about a basis yes, but not for a system of generators, but I get what you mean.

but yes, given any generating set, you can adjoin elements to it arbitrarily and you will still have a generating set. (not true for deleting elements, however)

user19161

1:41 AM

This can in fact be obtained immediately.

@anon Yeah.

@PeterTamaroff Let me give you a few more details

user19161

The perpendicular vector is ... (1,1,-1).

@JasperLoy Can I deal with Peter Tamaroff alone please?

AgainstASicilian

can someone tell me how to rearrange (4x^2+1)*(4y^2+1) = (4z^2+1) into a pell equation?

1:42 AM

@AgainstASicilian Please don't parachute in like that

Rajesh D

cross product?

@JasperLoy Yeah, I know nothing about planes and stuff, so I really don't care about that now.

@PeterTamaroff We have the vectors $(1,0,-1)$ and $(1,-1,0)$ that lie in $S$ and they are not scalar multiples of each other

so I claim that these two span $S$

We need to understand why

@BenjaLim But that's because you know that that subset has dimesion 2

@PeterTamaroff So suppose you take a vector $(a,b,c)$ in $S$

@PeterTamaroff well how do you know a priori? :D

1:44 AM

@BenjaLim since when does 1+0-(-1)=0?

@BenjaLim Because a plane is basically $R^2$

@anon Sorry make that $(1,0,1)$

@anon I think he lost the equation. He prolly means $(1,0,1)$

@PeterTamaroff if the plane contains the origin. otherwise, the plane does not have vector space structure induced from the ambient space

@PeterTamaroff Well stay with me now

1:45 AM

@BenjaLim Ok.

Suppose you take a vector $(a,b,c)$ in $S$

the components must satisfy the relation $a + b - c = 0$

@anon No idea what you're saying there! =)

(though the plane will be isomorphic to R^2 as an affine space)

@BenjaLim Yeah.

@anon Obviously, that's what I meant, dude.

So asking if the vectors $(1,0,1)$ and $(1,-1,0)$ span $S$

is equivalent to asking given any $(a,b,c) \in S$

does there exist $d,e$ such that it satisfies:

AgainstASicilian

1:47 AM

@BenjaLim Sorry

$\left[\begin{array}{c} a \\ b \\ c \end{array}\right] = \left[ \begin{array}{cc} 1 & 1 \\ 0 & -1 \\ 1 & 0 \end{array}\right] \left[\begin{array}{c} d \\ e \end{array}\right]$

@BenjaLim I don't follow that.

@PeterTamaroff

@PeterTamaroff You are asking if the two vectors I gave you span $S$ no?

What I would need to prove is that any $v \in S$ is a linear combination of $v_1=(1,0,1)$ and $v_2=(1,-1,0)$

This is asking if I can write every vector in $S$ as a linear combination of $v_1 = (1,0,1) $ and $v_2 = (1,-1,0)$

1:49 AM

@BenjaLim We concurr then!

@PeterTamaroff Exactly I just put that in a matrix

And now you will see we will always have such a solution because

@BenjaLim AH! OK.

If the coefficients of v1 and v2 are d and e respectively, then in matrix form...

choosing $d = -b$

$e = c$

and then $a$ will be completely determined, it is $e+ d = -b +c$

But then we don't run into trouble

ragib's post is widly popular

1:51 AM

because this is exactly the relation that $a$ had to satisfy in order for $(a,b,c)$ to be in $S$

@PeterTamaroff So we have checked that our vectors span $S$

@PeterTamaroff Is that clear?

@Eugene He thought that the splitting field of $x^3 -5$ had degree 3 over $\Bbb{Q}$ :D :D

@BenjaLim Yeah, I get you. And in turn that means that $S$ has dimesion two.

@PeterTamaroff yes.

@BenjaLim yes i saw

See now? @PeterTamaroff

Any subspace of a space with the same dimension is the whole space.

1:52 AM

@BenjaLim and it's not unreasonable to think that

@PeterTamaroff Right!

SO there you have it

@BenjaLim This is written the other way around!

@PeterTamaroff ah stuff it

@PeterTamaroff Exercise:

australia is a whole bunch of coastal cities right? because most of the land is unlivable?

Prove that two vector spaces are isomorphic iff they have the same dimension (finite dimensional VSs)

1:55 AM

@BenjaLim I haven't reached morphisms yet. I'll let you know then.

@Eugene Well a large chunk of the population is concentrated in Perth, Adelaide, Melbourne, Sydney,Brisbane, Gold Coast - all coastal cities :D

@PeterTamaroff A morphism is one of the most fundamental things in all of algebra

@BenjaLim yeah because it's mostly desert inland right?

@Eugene YOu would have to go easily 500km inland from sydney to get to the desert

@BenjaLim that's really not that much of a distance

@Eugene that's right considering Australia is one big continent

@Eugene I will probably meet ragib for lunch sometime :D