# Pproving (g ° f)* = f* ° g* - is this correct?

1. Jun 29, 2013

### Emspak

pproving (g ° f)* = f* ° g* -- is this correct?

1. The problem statement, all variables and given/known data

assume g: F→G is a linear map

prove that (g ° f)* = f* ° g*

My solution

(g ° f)* = g* ° f* by the properties of associativity in linear maps.

if we assume that g* ° f* = f* ° g* then g and f are inverse functions of each other.

by the properties of linear maps the only way that g* ° f* = f* ° g* is if they are inverse functions.

therefore (g ° f)* = f* ° g*

is there anything wildly wrong with my reasoning?

2. Jun 29, 2013

### Dick

Certainly looks wildly wrong to me. Is '*' supposed to be a transpose or adjoint or what?

3. Jun 29, 2013

### Emspak

i honestly don't know if you don't, i am writing the problem exactly as it appeared on the assignment.

4. Jun 29, 2013

### Dick

I don't have the rest of your course to look at. If they asked you to prove this then they certainly should have told you what '*' means. Look back. You'll need to use the definition.

5. Jun 29, 2013

### Emspak

I got two definitions: one is "any value" as it appears in some summation problems, the other is a map from G the dual space. Which one should be there I would have no way of figuring out since I am not the math PhD. My prof seems to think we are telepathic.

6. Jun 29, 2013

### Dick

I don't either. It is a sloppy statement, I'll give you that. They said g:F->G but they didn't say anything about f. My mind reading says it's about dual spaces. It's basically the same as showing $(AB)^T=B^TA^T$ where T means transpose.

7. Jun 29, 2013

### Emspak

Well that helps, at least it tells me that I am approaching this the wrong way. I really wish there was a better method of teaching this stuff; we're using the Freidberg text but I don't find it all that helpful because (of course) my prof decides he wants to use a different notation. I tried using Linear Algebra Done Right but the people who wrote it have never heard of an index. I'm on a third text now hoping it will help.

8. Jun 29, 2013

### Dick

Try thinking of it in terms of matrix representation. If the matrix representation of g is G and the matrix representation of f is F, then matrix representation of (g ° f) is GF. Now try to figure out what '*' does to matrices. It's really a form of 'transpose'.

9. Jun 30, 2013

### Emspak

Dick --
OK, I went to the prof. He says "the *" is defined in the problem which was about as helpful as a bullet to the brain, which is what I am contemplating right about now.

Here's the whole assignment problem:

Let f: E --> F be a linear map. Let f* : HomK(F,K) --> HomK(E,K) denote the function such that f*(u) = u composed of f (I'm too tired right now to mess with itex).

first he wants us to prove f* is a linear map. then the question I gave you.

Frankly the way this is set up might as well be written in Madarin as far as I am concerned.

10. Jun 30, 2013

### Dick

It makes a lot more sense when you state the whole problem. I would do that in the future. Your first job is to figure out what all the maps are. Then it won't seem like Mandarin. $Hom_K(F,K)$ is the dual space of $F$. Do you know what that is? Can you describe it in simple words?

11. Jun 30, 2013

### Emspak

i know that a homomorphism is by definition a linear map, or at least that is the definition I found online. (neither of the three books I used -- Freidberg, Linear Algebra Done Right -- deigns to define it, though Schaum's seems to and says it is the space of all linear transformations). But no, the concept of a dual space is completely foreign to me right now.

12. Jun 30, 2013

### Dick

Well, time to start learning that foreign concept. You just said what it is. F* (the dual space to F) is the set of all linear maps from F to K. In fancier language, $F^*=Hom_K(F,K)$. That is a vector space, right? Do you see why? Now to first part of the exercise is to show f* defines a linear map from G* to F*. Try it.

13. Jun 30, 2013

### Emspak

ye, thanks a lot, rethinking it helped quite a bit.

14. Jun 30, 2013

### Dick

So the whole thing isn't that bad once you figure out what the parts are, right?

15. Jun 30, 2013

### Emspak

yeah, I think I was just venting a bit too. I can't be the only one who feel the frustration.