Linear copying

1. Feb 29, 2008

Physicsissuef

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

Let's say that $$f: \mathbb {R}^3 \rightarrow \mathbb {R}^2$$, $$g: \mathbb {R}^3 \rightarrow \mathbb {R}^2$$, and $$h: \mathbb {R}^2 \rightarrow \mathbb {R}^2$$ are linear copying defined with $$f(x,y,z)=(y,x+z)$$, $$g(x,y,z)=(2z,x-y)$$ and $$h(x,y)=(y,2x)$$. Find the linear copying:

h o f and h o g

2. Relevant equations

3. The attempt at a solution

Sorry, but again, I don't know the principle of solving this task. My book is not good at all. Thanks from the start.

2. Feb 29, 2008

CompuChip

What does $h \circ f$ mean?
What is its domain and co-domain (e.g. what are A and B in $h: A \to B$)? How is it defined ($h(\cdots)= \cdots$ )?

3. Feb 29, 2008

Physicsissuef

probably h, f, g are the linear copying

4. Feb 29, 2008

CompuChip

I don't know what language you translated it from, but I think you mean "linear map" instead of "linear copying".
Also I think you are a little confused as to what $h \circ f$ means. f is a function that takes something in R3 and gives you something in R2. Into h you put something in R2 and get something (else) in R2 back. Now what you can do, is use the output of f as input for h. So
$$h \circ f: \mathbb{R}^3 \to \mathbb{R}^2$$
is a function defined by
$$(h \circ f)(x, y, z) = h(f(x, y, z))$$.
So to calculate (h o f)(x, y, z) you first calculate (u, v) = f(x, y, z) and then h(u, v).
Does that clarify ?

5. Feb 29, 2008

Physicsissuef

Yes, sir, that's what it means. Sorry for my mistranslation.

6. Feb 29, 2008

HallsofIvy

Staff Emeritus
Okay, so do it! If h(u,v)= (v, 2u), and (u,v)= f(x,y,z)= (2z, x- y), what is $h\circle f(x,y,z)$. What is $f\circle g(u,v)$?

7. Feb 29, 2008

Physicsissuef

How do you know that (x,y)=f(x,y,z)=(2z,x-y)? I truly don't know what is $h \circ f(x,y,z)[/tex] and [itex]f \circ g(u,v)$

Last edited: Feb 29, 2008
8. Feb 29, 2008

HallsofIvy

Staff Emeritus
Because that was what YOU told us:
$h\circ f(x,y,z)= h(f(x,y,z))= h(y, x+ z)$. If h is defined by h(a,b)= (b, 2a), what is h(y, x+z).

9. Mar 1, 2008

Physicsissuef

Can you tell me just this, please, there is one line more, I guess. Thanks.
Probably $h(y, x+ z)=(b,2a)$
But, what next?

Last edited: Mar 1, 2008
10. Mar 1, 2008

CompuChip

I think you are having some troubles with basic calculus concepts.
Suppose we have a function f(x) = 2 x + 3.
Then can you tell me what are f(3), f(a) and f(y + 1) ?

11. Mar 1, 2008

Physicsissuef

f(3)=2*3+3
f(a)=2a + 3
f(y+1)=2(y+1)+3

No, I don't have troubles with basic calculations.

12. Mar 1, 2008

CompuChip

Well, this is exactly the same, but there are just two values to substitute. So
h(y, x + z) is just (b, 2a) with a = y, b = x + z, i.e. h(y, x + z) = (x + z, 2y).

13. Mar 1, 2008

Physicsissuef

But it should be h(y, x+z) and (y,2x), so y=y and 2x=x+z. Hmm...

14. Mar 1, 2008

HallsofIvy

Staff Emeritus
No, you should have learned long ago that the letters used as variables in defining functions are just "place holders"- you can replace them with anything you want. If f is defined by f(x, y)= 3x+ y then f(a, b)= 3a+ b, f(y, x)= 3y+ b, etc. $h\circ f(x,y)= h(f(x,y,z))= h(y, x+z)= (x+ y, 2y) as I said. 15. Mar 1, 2008 Physicsissuef I know that, they are just 'place holders'. But can't understand the whole process. $$h \circ f$$ is what we need to find. $$(h \circ f)(x,y,z)=h(f(x,y,z))=h(y,x+z)$$ Now if $$h(x,y)=(y,2x)$$ then $$h(y,x+z)=(x+z,2y)$$, right? I think I understand now. Now for $$h \circ g$$ is the 2nd think that we need to find. $$(h \circ g)(x,y,z)=h(g(x,y,z))=h(2z,x-y)$$ Now if $$h(x,y)=(y,2x)$$ then $$h(2z,x-y)=(x-y,4z)$$, right? I think, I understand now. 16. Mar 1, 2008 Physicsissuef And if we get $$f \circ h$$? $$(f \circ h)(x,y)=f(h(x,y))=f(y,2x)$$ Now if $$f(x,y,z)=(y,x+z)$$ then $$f(y,2x)=??$$ In my book, it says that it is impossible, why? 17. Mar 1, 2008 CompuChip What could it be? [itex](y, 2x)$ is an element of $\mathbb{R}^2$, whereas $f$ is a function on $\mathbb{R}^3$. In other words, h produces a pair of numbers, whereas f expects a triple as input. So you cannot plug the output of h into f, in other words: you can't compose them in that order.

18. Mar 1, 2008

Physicsissuef

Ok, and what for $$h \circ (f+g)$$ and $$h \circ f + h \circ g$$.
First, are they same?
Other thing, how will I solve it?

19. Mar 1, 2008

CompuChip

Can you write down the definitions of (f + g) and h o (f + g) and (h o f) + (h o g)?
In particular, write down from what space into what space they map, like I did in my last post. Then you can already see if there will be a problem in the composition (like: if the space (f + g) maps to is different than the space h maps from, you cannot compose them, as you have just seen).

Last edited: Mar 1, 2008