Affine images of convex sets.

[bloodhound]

Let [math]F\colon \mathcal{A}\to \mathcal{A}'[/math] be an affine linear map, and let [math]\mathcal{C}[/math] be a convex subset of [math]\mathcal{A}[/math]. Show that its image [math]\mathcal{C}'=F(\mathcal{C})[/math] is a convex subset of [math]\mathcal{A}'[/math]

 

Any Hints on how to proceed?

[matt grime]

pick two points in F© say F(x), F(y) Then as F is affine the line x+t(y-x) t in[0,1] is...

 

if it helps F(x) = L(x)+v for some linear map L and some fixed vector v, though I presume you know that (since the other definition of affine is that i preserves straight lines, making the question easier).

[bloodhound]

ok pick two points in F©,say

F(X)=X'

F(Y)=Y'

 

then since F is affine, the line

[math](1-t)\vect{x}+t\vect{y}[/math] is mapped into

[math](1-t)\vect{x'}+t\vect{y'}[/math]

 

ARGH... then is that line a subset of F©?

[matt grime]

Let me clarify what your definition of an affine map is: is it one that preserves lines or is L(x) + v for some linear map L? (v=F(0) incidentally)

Since you've said that the line is mapped to the other line then since the original line is in C its image must be in F©, agreed? end of proof.

[bloodhound]

 

if it helps F(x) = L(x)+v for some linear map L and some fixed vector v' date=' though I presume you know that (since the other definition of affine is that i preserves straight lines, making the question easier).[/quote']

its confusing the notation ur using, cos it doesnt differentiate between points and vectors. i assume by little x u mean position vector of point X. and ins't v a point instead of a fixed vector.

[matt grime]

Erm, I'm not sure what you're getting at - this is linear algebra: a vector isn't "something with magnitude and direction", it's an element of a vector space. v is an element of the vector space, it is the image of 0 under the affine transformation. Point and vector are intrchangable, it just means an element of the vector space.

[bloodhound]

I meant since F is an affine map, its takes points to points. so F can't act on vector.

 

but you said at the beginning " F(x) = L(x)+v for some linear map L and some fixed vector v" which sort of doesnt make sense as F is defined from A->A' and not from the respective vector space V->V'.

 

although i do understand what you are saying.

[matt grime]

Ok, just subsitute the word point every where where you see the word vector then - was taking as an example R^n which is affine and thinking in those terms. However since an affine map is simply one that preserves lines in some affine space then you see how the proof was completed above?

 

As I said I'd like to clarify what you mean by affine here. Spec(k[x,y,...,z])? Actually forget that, that was silly. Been a while since I thought about these things. Anyway, essentially an affine space is just a vector subspace of a vector space translated along. In essence we can pick an origin and a basis, however for any choice of point as origin in the space we must always get the same results, is that about right?

[bloodhound]

huh? whats Spec(k[x,y,...,z])??

 

Where did we use the fact that C is a convex subset of A in the proff above?

[matt grime]

we used convexity of C because we required the line segment from x to y to be in C, and hence its image to be in F©

 

Forget spec, that was me being stupid.

[bloodhound]

so since the line segment from x to y is in C then the line segment from F(x) to F(y) is in F©. is that all the proof?

[matt grime]

yes, that's it - an affine map sends straight lines to straight lines - the end points are correct and it is a line. I can't see anything else you need to show, though like I say, this (version of affine) isn't my stuff (hence the stupid reference to something else completely useless).

[bloodhound]

still dont get it I thought the statement in my last post WAS what we wanted to prove... thats a bit like saying its true because it is.

 

Well ok. it sends st. lines to st.lines. and the endpoints are in F©, but i still dont get how the line between F(x) and F(y) HAS to be in F©. Can't it be that some portions are in A'\F©

[matt grime]

No.

 

Let F(x) and F(y) be any two points in F©. each point in the segment l(t)=x+t(y-x) is in C, so l'(t)=F(l(t)) is in F©, and is a straight line, by the assumption F is affine, and l'(0)=F(x), l'(1) = F(y) so it is a straight line segment lying wholly in F© with end points as required.

[bloodhound]

OHHHHH,,, how stupid of me, i should have thought of each point , rather than just the endpoints. I get it now.

 

Couple more Affine Spaces question to do for tmr.

 

Cheers