Does isomorphic imply

[WiFO215]

Does isomorphic imply...

I was reading Hoffman and Kunze where I came across the following definition:

Let us call a linear transformation T non-singular if T(a) = 0 implies that a = 0, i.e., if the null space of T is {0}. Evidently, T is 1:1 iff T is non-singular.

So what I gathered was, Saying a linear transformation is non-singular is the same as saying it is 1:1. But a few pages later they define the following:

If V and W are vector spaces over the field F, any one-one linear transformation T of V onto W is called an isomorphism of V onto W. If there exists an isomorphism of V onto W, we say that V is isomorphic to W.

So don't isomorphism and non-singular mean the same thing? Why would we want to give it two names?

 

[dx]

Your first quote from Hoffman and Kunze is talking about linear transformations V → V. A general linear map V → W can be non-singular (i.e. null space = {0}) and not one-to-one at the same time.

Example:

The linear map T from R² to R³ represented by the matrix

[1 0]
[0 1]
[0 0]

The only element that goes to zero here is (0,0), but T is not one-to-one.

EDIT: Ignore this, or replace 'one-to-one' everywhere in the post with 'onto'.

 

[Office_Shredder]

Your first quote from Hoffman and Kunze is talking about linear transformations V → V. A general linear map V → W can be non-singular (i.e. null space = {0}) and not one-to-one at the same time.

Example:

The linear map T from R² to R³ represented by the matrix

[1 0]
[0 1]
[0 0]

The only element that goes to zero here is (0,0), but T is not one-to-one.

You mean T is not onto.

In fact, given any linear map T with a kernel of only {0} it must be 1-1 as if Tv = Tw, then Tv - Tw = 0 so as T is linear, T(v-w) = 0. But it has a trivial kernel, so v-w=0 i.e. v=w

In this way, if V is a vector space, T:V->W a 1-1 linear map, then T(V) the image of T is always isomorphic to V. This can be pretty handy. Of course, T(V) is in general NOT equal to W

 

[dx]

Whoops! Yes I did mix up one-one and onto. Sorry.

 

[jbunniii]

I was reading Hoffman and Kunze where I came across the following definition:

Let us call a linear transformation T non-singular if T(a) = 0 implies that a = 0, i.e., if the null space of T is {0}. Evidently, T is 1:1 iff T is non-singular.

This is a nonstandard definition.

If T(a) = 0 implies that a = 0, i.e., if null(T) = {0}, then T is usually called INJECTIVE (or one-to-one).

If T:V->W, and if for every w in W there exists v in V such that Tv = w, then T is called SURJECTIVE (or onto).

If T is both injective and surjective, then we call it NONSINGULAR or INVERTIBLE or an ISOMORPHISM. Note that if T is an isomorphism, this implies that dim(V) = dim(W).

If dim(V) = dim(W) < infinity, then the three conditions are actually equivalent: T injective <==> T surjective <==> T invertible. This follows from the rank-nullity theorem:

dim(range(T)) + dim(null(T)) = dim(V)

If T is nonsingular, then it has an inverse. If T is an isomorphism, then not only does it have an inverse, but the inverse is also a linear map. (The inverse "preserves the structure" of the vector space.)

It happily turns out that if T has an inverse, then the inverse is automatically linear, but this is a theorem that has to be proved.

An arbitrary (not necessarily linear) map can have an inverse without being an isomophism, which is why the two terms exist. Fortunately, for linear maps, the two notions are equivalent.

 

[WiFO215]

A general linear map V → W can be non-singular (i.e. null space = {0}) and not one-to-one at the same time.

Okay. So that clears it up somewhat. Can you give me an example of null(T) = {0} and not 1:1? Is it possible that the T can be one-one and singular?

 

[Tibarn]

If V and W are vector spaces over the field F, any one-one linear transformation T of V onto W is called an isomorphism of V onto W. If there exists an isomorphism of V onto W, we say that V is isomorphic to W.

So don't isomorphism and non-singular mean the same thing? Why would we want to give it two names?

Key word in the definition is "onto". By the definitions you gave, a non-singular linear transformation need not be onto, so it wouldn't be an isomorphism.

 

[WiFO215]

OK. Now I got that. Thanks Tibarn. But I still don't understand how something can be non-singular and not 1:1. If it is 1:1, then it would also be onto if dim V = dim W. Then it is definitely isomorphic right? I seem to be going in circles.

 

[vandanak]

like if there is some linear transformation between 3d and 2d it can be onto but it cannot be 1-1

 

[HallsofIvy]

No one here (except dx by mistake and he corrected it) has said that there are linear transformations that are "non-singular" but not "one-to-one". Every non-singular linear transformation is one-to-one as you said initially. There exist non-singular linear transformation that are not "onto" and so not an isomorphism.

If T is a linear transformation from vector space U to vector space V, and V has higher dimension than U, T will map U onto a subspace of V.

 

[de_brook]

like if there is some linear transformation between 3d and 2d it can be onto but it cannot be 1-1

Yes it can be onto. But the linear transformation can be 1-1 to a subspace of the space with higher dimension. Thus when speaking of isomorphism or nonsingular transformation, we must be conscious of the dimension of the space we are working with. If the spaces are not of the same dimension, there can't be an isomorphism between the spaces. We must note that there could be an isomorphism between a space with dimension 2 and a subspace of a space with dim 3. Isomorphism of a linear transformation on vector spaces is equivalent to its non singularity.

 

[JG89]

"There exist non-singular linear transformation that are not "onto" and so not an isomorphism."

Sorry to interrupt here Halls, but (assuming non-singular is synonymous with invertible) my textbook (Linear Algebra by Friedberg, Insel and Spence) says on PG. 100 "We often use the fact that a function is invertible if and only if it is both one-to-one and onto"

1) Can't a linear transformation be thought of as a function, and so must also be onto (as well as one-to-one) in order for it to be invertible?

2) My textbook on page 102 has the following definition: "Let V and W be vector spaces. We say that V is isomorphic to W if there exists a linear transformation
T: V->W that is invertible. Such a linear trasformation is called an isomorphism from V onto W". You said that a there exist invertible linear transformations that are invertible but not an isomorphism because they aren't onto. According to my definition an isomorphism from V to W is T:V->W such that T is invertible, and if you were right in saying that all linear transformations are one-to-one, and that is a sufficient condition for the transformation to be invertible, then shouldn't it be an isomorphism according to my definition?

 

[dx]

"There exist non-singular linear transformation that are not "onto" and so not an isomorphism."

Sorry to interrupt here Halls, but (assuming non-singular is synonymous with invertible)

Non-singular is not synonymous with invertible.

1) Can't a linear transformation be thought of as a function, and so must also be onto (as well as one-to-one) in order for it to be invertible?

Yes, an invertible linear transformation must be both onto and one-one.

2) My textbook on page 102 has the following definition: "Let V and W be vector spaces. We say that V is isomorphic to W if there exists a linear transformation
T: V->W that is invertible. Such a linear trasformation is called an isomorphism from V onto W". You said that a there exist invertible linear transformations that are invertible but not an isomorphism because they aren't onto.

No, there are no linear transformations that are invertible but not onto. All invertible functions must be onto.

According to my definition an isomorphism from V to W is T:V->W such that T is invertible, and if you were right in saying that all linear transformations are one-to-one, and that is a sufficient condition for the transformation to be invertible, then shouldn't it be an isomorphism according to my definition?

one-one is not a sufficient condition for a linear transformation transformation to be invertible. It must also be onto.

 

[WiFO215]

Every non-singular linear transformation is one-to-one as you said initially. There exist non-singular linear transformation that are not "onto" and so not an isomorphism.

Okay. That was somewhat clear. Here's what I've gathered:

1) All non-singular functions are 1:1
2) An isomorphism is a 1:1 and onto function i.e. an invertible function

So obviously, a 1:1 function need not be onto. Which implies that a non-singular/ 1:1 function need not necessarily be invertible.

 

[JG89]

Thanks for clearing that up dx. What does non-singular mean then? According to wikipedia: "In linear algebra, an n-by-n (square) matrix A is called invertible or non-singular". Other definitions I find are ones that say something like a square matrix A whose determinant is not 0 is said to be non-singular, implying the matrix is invertible, since the determinant is 0.EDIT: I guess we call a function one-to-one if it is non-singular?

 

[dx]

EDIT: I guess we call a function one-to-one if it is non-singular?

A linear transformation is one-one if it is non-singular.