Find the linear transformation

That is why, in this problem, you do not need to do anything with the kernel to determine that f is an isomorphism- if it is injective, it is automatically surjective.
  • #1
Physicsissuef
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0

Homework Statement



Check if the linear transformation [tex]f : \mathbb{R}^2 \rightarrow \mathbb{R}^2[/tex], defined with [tex]f(x,y)=(x+y,y)[/tex] is isomorphism? If so, find the linear transformation [tex]f^-^1[/tex]

Homework Equations



V and U are vector sets. The linear copying [tex]F:V \rightarrow U[/tex] which is bijection is isomorphism.

The Attempt at a Solution

 
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  • #2
Linear 'map', not 'copying' right? Pick a general point in the image R^2, say (u,v). Can you solve f(x,y)=(x+y,y)=(u,v) for x and y uniquely in terms of u and v? I think you can. If so then the expression for x and y in terms of u and v is f^(-1) and the map is an isomorphism.
 
  • #3
It is also true that if a linear map is a surjection, it is a bijection and an isomorphism. I say that because showing that a linear map is a surjection is essentially the same as finding the inverse function which you are also asked to do.

Suppose (u,v) is any member of R2. You want to show that there exist (x,y) in R2 so that f(x,y)= (u, v). If that is always true, then f is a surjection and so an isomorphism. f(x,y)= (x+y,y)= (u, v) so we have x+ y= u, y= v. Can you solve that for x and y no matter what u and v are? If so then f is an isomorphism. The best way to show that is to actually solve for x= ..., y= ... And once you have done that the function f(u,v)= ( x, y) is the inverse function.
 
  • #4
And can you please give me some example, where it wouldn't be bijection, but it will be linear map?
 
  • #5
f(x,y)=(x,0) is a linear map that's not a surjection (onto). g:R^2->R^1, g(x,y)=x+y is a surjection but not a injection (1-1). Neither is a bijection for two different reasons.
 
  • #6
Dick said:
f(x,y)=(x,0) is a linear map that's not a surjection (onto). g:R^2->R^1, g(x,y)=x+y is a surjection but not a injection (1-1). Neither is a bijection for two different reasons.

Why f(x,y)=(x,0), it is not surjection?

why g(x,y)=x+y is is a surjection, but not injection?
 
  • #7
1) Because there isn't any vector (x,y) such that f(x,y)=(0,1). It's not onto.
2) Because g(1,-1)=g(2,-2)=0. It's not 1-1.
 
  • #8
Dick said:
1) Because there isn't any vector (x,y) such that f(x,y)=(0,1). It's not onto.
2) Because g(1,-1)=g(2,-2)=0. It's not 1-1.

We prove surjection for the first f(x,y)=(0,1)=(p,t), so p=0, and t=1, why it is not surjection?
 
  • #9
Physicsissuef said:
We prove surjection for the first f(x,y)=(0,1)=(p,t), so p=0, and t=1, why it is not surjection?

No. You don't. You want to find X and Y, such that f(X,Y)=(0,1). f(X,Y)=(X,0). (X,0) can NEVER be equal to (0,1).
 
  • #10
Where did you get f(X,Y)=(X,0) from?
 
  • #11
Physicsissuef said:
Where did you get f(X,Y)=(X,0) from?

I DEFINED IT. You asked for an example of a linear map that was not a surjection. I suggested f(x,y)=(x,0). What f have you been talking about? You're not paying attention.
 
  • #12
Dick said:
I DEFINED IT. You asked for an example of a linear map that was not a surjection. I suggested f(x,y)=(x,0). What f have you been talking about? You're not paying attention.

The f(x,y)=x+y is not injection, because of [tex](x_1,y_1) \neq (x_2,y_2) , g(x_1,y_1)=g(x_2,y_2)[/tex], right?
 
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  • #13
I have no idea what that sentence means! What do you mean by "Do we get arbitrary (0, 1)"?

Physicsissuef, you asked for an example of a linear map that is NOT an isomorphism. Since an isomorphism must have an inverse, it must be both injective and surjective. Since a linear map is surjective if and only if it is injective it is enough to give an example of a linear map that is not injective- that has non-trivial kernel. Surely you have posted enough "kernel" problems here to be familiar with that!

f(x,y)= (x, 0) is obviously not injective because any vector of the form (0, y) is mapped to (0, 0), f(0, y)= (0, 0), and so is in the kernel. So it cannot be an isomorphism and is an example of a linear map that is not an isomorphism.
 
  • #14
HallsofIvy said:
I have no idea what that sentence means! What do you mean by "Do we get arbitrary (0, 1)"?

Physicsissuef, you asked for an example of a linear map that is NOT an isomorphism. Since an isomorphism must have an inverse, it must be both injective and surjective. Since a linear map is surjective if and only if it is injective it is enough to give an example of a linear map that is not injective- that has non-trivial kernel. Surely you have posted enough "kernel" problems here to be familiar with that!

f(x,y)= (x, 0) is obviously not injective because any vector of the form (0, y) is mapped to (0, 0), f(0, y)= (0, 0), and so is in the kernel. So it cannot be an isomorphism and is an example of a linear map that is not an isomorphism.

Wait, wait... Let's clear some things. If some linear map is injective, it is not necessarily surjective or vice versa. What does the kernel has to do now?
 
  • #15
What the "kernel" has to do with that is that "injective" means "one to one"- If f:U->V, one and only one member of U is mapped into one member of V. In particular, only one member of U is mapped into 0 (the only thing that must be in every V). Since any linear map must take 0 to 0, that means that only 0 is mapped into 0- a linear map is "injective" if and only if its kernel is the trivial subspace, {0}.

For this problem, or any problem in which "U" and "V" (f:U->V, remember. In this problem U= V= R2) have the same dimension, if f is injective, so the kernel has dimension 0, by the "dimension theorem" I have mentioned before, the image of f must be the same as the dimension of U which is the same as the dimension of V: the image of f is all of V so f is surjective.

If f:U->V with dim(U)< dim(V), since the dimension of the image of f cannot be greater than the dimension of U so f is not surjective. If f:U->V with dim(U)> dim(V), since the dimension of the image of f cannot be greater than the dimension of U, we must have dimension of kernel of f larger than 0 and f is not injective. In either case, f is not an isomorphism which is what we would expect: if two subspaces are "isomorphic" then they are essentially the same- in particular, they have the same dimension.
 
  • #16
And how will explain with kernel and dimension with my problem?
[tex]f : \mathbb{R}^2 \rightarrow \mathbb{R}^2[/tex]

[tex]f(x,y)=(x+y,y)[/tex]
 
  • #17
The kernel is zero dimensional. So it's injective. The range is two dimensional. So it's onto. All of these are just fancy ways of saying (x+y,y)=(x'+y',y') only if (x,y)=(x',y') (injective) and for any (x,y), f(x-y,y)=(x,y) (surjective).
 
  • #18
f(x,y)= (x, 0)
Here f(x,0)= (x,0)
and f(y,0)= (0,0), right?
Why then it is not injective?
Lets say we will get arbitrary value for x and y two times.
[itex]x_1=1[/itex]
[itex]y_1=2[/itex]
f(1,0)=(1,0)
f(2,0)=(0,0)
[itex]x_2=1[/itex]
[itex]y_2=-1[/itex]
f(1,0)=(1,0)
f(-1,0)=(0,0)
So [tex]f(2,0)=f(-1,0)[/tex] but [tex]y_1 \neq y_2[/tex], right?
Now f:U -> V
U=V=R2
dimU=2
dimV=1
So dim(U)>dim(V), it is also not surjective, right?
 
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  • #19
Physicsissuef said:
f(x,y)= (x, 0)
Here f(x,0)= (x,0)
and f(y,0)= (0,0), right?
Why then it is not injective?
Lets say we will get arbitrary value for x and y two times.
[itex]x_1=1[/itex]
[itex]y_1=2[/itex]
f(1,0)=(1,0)
f(2,0)=(0,0)
[itex]x_2=1[/itex]
[itex]y_2=-1[/itex]
f(1,0)=(1,0)
f(-1,0)=(0,0)
So [tex]f(2,0)=f(-1,0)[/tex] but [tex]x_1 \neq x_2[/tex], right?
Now f:U -> V
U=V=R2
dimU=2
dimV=1
So dim(U)>dim(V), it is also not surjective, right?

If f is defined by f(x,y)=(x,0), then f(x,0)=(x,0) and f(y,0)=(y,0). Your problem may be less with linear maps and more with not understanding function notation, because you are posting nonsense.
 
  • #20
What was wrong with my explanation?
If f(x,y)=(x,0)
that means that:
f(x,0)=(x,0)
f(0,y)=(0,y)=(0,0)
I still can't undestand what you want to say with the kernel and images. Please tell me, when its/its not surjection, its/its not injection and please give me example.
 
  • #21
This is getting repetitive. If f(x,y)=(x,0) then f(1,1)=(1,0), f(1,2)=(1,0). Not injective. There are no values of x and y such that f(x,y)=(0,1). Not surjective.
 
  • #22
Dick said:
This is getting repetitive. If f(x,y)=(x,0) then f(1,1)=(1,0), f(1,2)=(1,0). Not injective. There are no values of x and y such that f(x,y)=(0,1). Not surjective.

And how will you represent that with dim(U) and dim(V)?
 
  • #23
Dimension of kernel(f)=1. Not injective. Dimension of domain R^2=2. So dimension of range=2-1=1. So the dimension of the range<2 (dim(V)). Not surjective.
 
  • #24
And why HallsofIvy wrote:
HallsofIvy said:
If f:U->V with dim(U)< dim(V), since the dimension of the image of f cannot be greater than the dimension of U so f is not surjective. If f:U->V with dim(U)> dim(V)
In your post, dim(U)=dimension of the range. Is the U domain?
 
  • #25
I don't know what would give you the impression that he said that dim(U) is the dimension of the range. He is clearly contrasting the dimension of U with the dimension of the image of f. And yes, U is the domain. Have you not seen the notation for functions that goes like this?

f: U -->V

That means f is a function that maps U into V. So U is the domain, and V is the codomain.
 
  • #26
Physicsissuef said:
And why HallsofIvy wrote:

In your post, dim(U)=dimension of the range. Is the U domain?

Why are you asking that? I didn't have a dim(U) in my post. By 'range' I mean f(U). Maybe I should be calling that 'image' to distinguish it from V, or call V the codomain like Tom suggests.
 
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  • #27
Ok, let's say f:U ->V
where f(x,y)=(x,0)
so the dimU=dim(x,y)=2
the dimV=dim(x,0)=1
so dimU > dimV, that means that it is not injective, right?
What want to say HallsofIvy with the post that I quote?
If f:U->V with dim(U)< dim(V), since the dimension of the image of f cannot be greater than the dimension of U so f is not surjective. If f:U->V with dim(U)> dim(V), since the dimension of the image of f cannot be greater than the dimension of U, we must have dimension of kernel of f larger than 0 and f is not injective. In either case, f is not an isomorphism which is what we would expect: if two subspaces are "isomorphic" then they are essentially the same- in particular, they have the same dimension.
 
  • #28
Don't write dim(x,0). Write dim(f(U)). It's more standard. I would say U=V=R^2. So dim(U)=dim(V)=2. For f(U), the (probably bad word 'range') or 'image', dim(f(U))=1. dim(f(U))<dim(U) implies it's not injective. dim(f(U))<dim(V) implies it's not surjective.
 
  • #29
Ok. I understand now, thanks. And what about your post:
Dick said:
Dimension of kernel(f)=1. Not injective. Dimension of domain R^2=2. So dimension of range=2-1=1. So the dimension of the range<2 (dim(V)). Not surjective.
Why you wrote dimf(u) (dimension of range)=2-1. What is 2, what is 1? Is this definition, different from the HallofIvy's?
Why you said kerf=1?
 
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  • #30
dim(f(U))=dim(U)-dim(kernel(f)). No, that's not different from what Halls is saying. It's not a definition, it's a theorem.
 
  • #31
Dick said:
dim(f(U))=dim(U)-dim(kernel(f)). No, that's not different from what Halls is saying. It's not a definition, it's a theorem.

kernel(f)?
Is it kernel(f(U))?
Why dim(kerF)=1
f(x,y)=(x,0)
(x,0)=(0,0)
x=0
so kerF=0
 
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  • #32
(0,1) is in the kernel of f.
 
  • #33
Dick said:
(0,1) is in the kernel of f.

How did you find it?
 
  • #34
It found me. f(0,1)=(0,0). I just looked at the definition of f.
 
  • #35
Dick said:
It found me. f(0,1)=(0,0). I just looked at the definition of f.

Sorry, and how will you find it for my example?
f(x,y)=(x+y,y)??
(x+y,y)=(0,0)
or?
 

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