Isomorphism only if it is a linear transformation?

[McLaren Rulez]

Hi,

I've come across this result which says that if there are two isomorphic vector spaces with a transformation between them, then that transformation must be linear. Can anyone help me prove this?

For instance, if I have a transformation T: Z -> Z where Z is the set of integers, T(z) = z+1 which is non linear, I still get an isomorphism right? The elements in the codomain and domain have a one to one relation with each other.

What is my misconception here? Thank you.

 

[chiro]

From what I remember about linear transformations, you have to have the following:

let a,b be an element of some set. Linearity obeys the axioms:

T(a+b) = T(a) + T(b) and
T(d.a) = d T(a)

where d is some field.

So your sets are just integers. So using the fact that T(z) = z + 1, let a,b be integers.

T(a + b) = (a + b) + 1
T(a) + T(b) = (a + 1) + (b + 1) = (a + b) + 2

So linearity is broken from the above example.

I'm thinking the reason why they have to be linear is because of the following:

Lets say we calculate T(a), T(b) and T(a+b).

Now if we wanted to get a from T(a) we could, and similarly get b from T(b) with regard to the inverse transformation.

But what about T(a+b) or T(a + b + c) or so on? In this situations we can't get our parameter back (the [a+b] or [a+b+c]) since the answer is not unique, and hence there is no unique inverse and hence no isomorphism.

 

[McLaren Rulez]

Sorry, I don't see why the example above is not an isomorphism. For any element z, T(z) is z+1. Similarly, for any element in the codomain, say z', it gets mapped back to z'-1 under the inverse. So it is a one to one relation, isn't it? And yet, it is non linear.

 

[HallsofIvy]

Please state exactly what your original question was. The title of the thread was "Isomorphism only if it is a linear transformation?" but in your first post you talk only about a "transformation" between two isomorphic vector spaces. Those are two very different questions. Certainly, two isomophic space can has a non-linear transformation between them that is one-to-one and onto and not linear, as your example shows. But that transformation is not itself an isomophism. So are you talking about general transformations between two vector spaces that happen to be isomorphic or are you talking specifically about isomorphisms between two vector spaces?

(In fact, any homomorphism from one vector space to another must be linear.)

 

[McLaren Rulez]

Well this actually came up in a special relativity text. The author considers two frames S and S'. S has x y z and t and S' has x' y' z' and t'. The two frames coincide at t=t'=0. Now, in order to derive the relations (eventually turn out to be the Lorentz transformations) between the primed and unprimed coordinates, the author says: "In order that there be a 1-1 correspondence between the points of S and S' i.e. a single event in one inertial frame may transform into a single event in another inertial frame, the connection between x' y' z' and t' and x y z and t must be linear"

So, I'm not sure why this must be so. And HallsOfIvy, what is the distinction between a transformation between isomorphic vector spaces and an isomorphism? Thank you.

 

[HallsofIvy]

An isomorphism between two "agebraic structure" (group, ring, vector space, etc.) is an invertible (one-to-one and onto) that "preserves the operations"- that is, f(a*b)= f(a)*f(b) where * represents an operation in each structure. For a vector space, the operations are addition and scalar multiplication- an isomorphism from one vector space to another is one to one function such that f(u+ v)= f(u)+ f(v) and f(av)= af(v).
Those are precisely the conditions for a linear transformation on a vector space.

The reason I made a distinction between "isomorphic vector spaces" and an "isomorphism" is that you can have two spaces that are isomorphic and still talk about some function that is not an isomophism. The example you give, f(x)= x+ 1 from the set of real numbers to itself, (I just noticed you say Z- the set of integers is not even a vector space), the set of real numbers is, of course, isomorphic to itself, but f is NOT an isomorphism.

I notice that the text you quote does not use the word "isomorphism". Where did you get that?

 

[McLaren Rulez]

Ah okay, my mistake with the title and opening post. Sorry about that. I confused isomorphisms with 1-1.

So now, why must the transformation be linear between the two coordinate frames?

EDIT: In particular, why is my previous example incorrect? The mapping is one to one but not linear. Thank you

 

[Fredrik]

"In order that there be a 1-1 correspondence between the points of S and S' i.e. a single event in one inertial frame may transform into a single event in another inertial frame, the connection between x' y' z' and t' and x y z and t must be linear"

Not true. There are certainly bijections from [itex]\mathbb R^4[/itex] onto [itex]\mathbb R^4[/itex] that take 0 to 0 and still aren't linear. For example, the map that multiplies every vector by 2.

I assume that what you're doing is to "derive" the Lorentz transformation from Einstein's postulates. I've had a lot to say about such derivations in the past, so I'll quote myself from one of my earlier posts:

Einstein's postulates are loosely stated ideas, not mathematical axioms, so they can't be used as the starting point of a proof. There are however mathematical statements that seem to capture Einstein's ideas (or aspects of them) quite well. One of them is the Lorentz transformation. If you want to, you can start with another collection of mathematical statements that can be thought of as expressing different aspects of Einstein's ideas, and derive the Lorentz transformation from them. My favorite derivation of that sort is the one I posted here. (Start reading at "The explicit..."). This is just for the 1+1 dimensional case. Note that I'm not using Einstein's postulates (which my posts refer to as "the numbered list in my previous post") as axioms, but as an inspiration for a number of mathematical assumptions that are needed along the way.

One of those assumptions is linearity. My post contains a partial motivation for it: If a function that represents a coordinate change from one inertial frame to another doesn't take straight lines to straight lines, an object that's moving with a constant velocity forever in one inertial frame would be accelerating at some point in the other. If we supplement this with the requirement that these functions are smooth (that their partial derivatives up to order n exists, for all integers n), and that they preserve the origin, we can show quite easily that the only possibility is a linear function. These assumptions are stronger than they need to be. (They're sufficient to prove linearity, but not necessary). I have never cared enough to find out what a set of necessary assumptions look like.​

 

[Anonymous217]

In my upper div linear algebra, I was taught that an isomorphism is defined as a linear bijection (among vector spaces). In abstract alg, I was taught that an isomorphism is defined as a homomorphic/structure-preserving bijection (among groups).

I'm assuming that the definition I was taught in abstract alg. was a more general definition, since linearity would certainly be a structure-preserving requirement.

 

[Fredrik]

Yes, for vector spaces, "structure preserving" is equivalent to "linear". For rings, groups, etc, it means something else. It's actually kind of hard to properly define the terms "structure" and "structure preserving", so I think the traditional way of teaching these things is to not even mention them as long as the students are only familiar with one kind of structure, and then introduce them informally when they've seen a few more examples of structures. I think the formal definition isn't even discussed in courses on abstract algebra.

And yes, isomorphisms are structure-preserving bijections, no matter what type of structures we're talking about.

 

[McLaren Rulez]

Not true. There are certainly bijections from [itex]\mathbb R^4[/itex] onto [itex]\mathbb R^4[/itex] that take 0 to 0 and still aren't linear. For example, the map that multiplies every vector by 2.

Sorry if this is a bit stupid but why is a multiplication of every vector by 2 not a linear function? Also, would my previous example which takes a real number z to z+1 be a correct counterexample then, i.e. it is non linear but it gives a one to one correspondence between R and R?

I assume that what you're doing is to "derive" the Lorentz transformation from Einstein's postulates. I've had a lot to say about such derivations in the past, so I'll quote myself from one of my earlier posts:

Einstein's postulates are loosely stated ideas, not mathematical axioms, so they can't be used as the starting point of a proof. There are however mathematical statements that seem to capture Einstein's ideas (or aspects of them) quite well. One of them is the Lorentz transformation. If you want to, you can start with another collection of mathematical statements that can be thought of as expressing different aspects of Einstein's ideas, and derive the Lorentz transformation from them. My favorite derivation of that sort is the one I posted here. (Start reading at "The explicit..."). This is just for the 1+1 dimensional case. Note that I'm not using Einstein's postulates (which my posts refer to as "the numbered list in my previous post") as axioms, but as an inspiration for a number of mathematical assumptions that are needed along the way.

One of those assumptions is linearity. My post contains a partial motivation for it: If a function that represents a coordinate change from one inertial frame to another doesn't take straight lines to straight lines, an object that's moving with a constant velocity forever in one inertial frame would be accelerating at some point in the other. If we supplement this with the requirement that these functions are smooth (that their partial derivatives up to order n exists, for all integers n), and that they preserve the origin, we can show quite easily that the only possibility is a linear function. These assumptions are stronger than they need to be. (They're sufficient to prove linearity, but not necessary). I have never cared enough to find out what a set of necessary assumptions look like.​

Okay, so you are saying that it isn't really possible to prove linearity just because the correspondence is one to one. But the author of this book (Special Relativity by Mohammed Saleem and Muhammad Rafique) says that one to one correspondence implies linearity. Could you give me an example of a transformation which takes 0 to 0 and is non linear but still one to one? As I mentioned, I don't quite get the multiplication by 2 example.

Anyway, your argument about constant velocity in one frame changing to accelerated motion in another does make sense. Thank you.

 

[Fredrik]

Sorry if this is a bit stupid but why is a multiplication of every vector by 2 not a linear function?

Lol, I can't believe I picked an example that's very obviously linear. I obviously had some sort of brain malfunction. Except for the identity map, there's no function that's more obviously linear than that one. I'll reply to your other questions in a while. Right now I need a minute to get over the embarrassment.

Edit:

OK, how about these examples:

1. The function that multiplies every vector by 1+sin θ, where θ is the angle it makes with the t axis.

2. The function f defined by f(t,x,y,z)=(t³,y³,x³,z³) for all (t,x,y,z).

3. Functions like the f defined by f(t,x,y,z)=(t, x cos θ-y sin θ, x sin θ+y cos θ, z), where θ isn't a constant, but depends on e.g. the vector's distance from the x,z plane, its distance from the t axis, or the angle it makes with the t axis.

Also, would my previous example which takes a real number z to z+1 be a correct counterexample then, i.e. it is non linear but it gives a one to one correspondence between R and R?

Yes, translations are good examples of non-linear bijections. I would have used that as my first example, if the book hadn't been talking about transformations that take 0 to 0.

 

[McLaren Rulez]

Thank you, Fredrik! You've helped me a lot