# Is T an Isomorphism of F^2? Finding Necessary Conditions

• fk378
In summary, the conversation discussed the definition and properties of a homomorphism T defined on F^2. The conversation also addressed the necessary and sufficient conditions for T to be an isomorphism, including being one-to-one and onto, and the importance of the determinant of the transformation matrix in determining one-to-one-ness and invertibility.
fk378

## Homework Statement

Let T be defined on F^2 by (x1,x2)T=(w*x1+y*x2, z*x1+v*x2)
where w,y,z,v are some fixed elements in F.

(a) Prove that T is a homomorphism of F^2 into itself.
(b) Find necessary and sufficient conditions on w,y,z,v so that T is an isomorphism.

## The Attempt at a Solution

Part (b), I'm not sure what it means. For T to be an isomorphism it has to be one-to-one and onto.
To show one-to-one, I need to show that the kernel is 0.
Is showing that T is into F^2 the same thing as saying it is onto F^2? If not, what's the difference?

Last edited:
Into is different from onto. T is onto F^2 provided that for every (u, v) in F^2 there is some (x1, x2) in F^2 such that T(x1, x2) = (u, v).

A simple example of an into function is e^x, which maps the real numbers to a subset of the real numbers.

For the one-to-one part, show that the kernel consists of (0, 0) and nothing else. IOW, if T(x1, x2) = (0, 0), then x1 = 0 and x2 = 0.

Mark44 said:
Into is different from onto. T is onto F^2 provided that for every (u, v) in F^2 there is some (x1, x2) in F^2 such that T(x1, x2) = (u, v).

A simple example of an into function is e^x, which maps the real numbers to a subset of the real numbers.

For the one-to-one part, show that the kernel consists of (0, 0) and nothing else. IOW, if T(x1, x2) = (0, 0), then x1 = 0 and x2 = 0.

Would it have to be that x1 and x2 equal 0 only? What about the fixed elements in F? Also, the question asks about conditions on the elements of F.

fk378 said:
Would it have to be that x1 and x2 equal 0 only?
Yes, and that's exactly what I said.

BTW, I didn't notice it earlier, but I think you have copied the definition of the function incorrectly. You have
(x1,x2)T=(w*x1+y*x2, z*x1,v*x2)
I think it should be
T(x1,x2)=(w*x1+y*x2, z*x1 + v*x2)
For T to be a map from F^2 onto itself, the image vector has to have two coordinates, not three.

fk378 said:

$$\begin{eqnarray} T\left ( \begin{array}{c} \nonumber x_1 \\ x_2 \end{array} \right ) = \left [{\begin{array}{cc} w & y \\ z & v \end{array} \right ] \left ( \begin{array}{c} x_1 \\ x_2 \end{array} \right )$$

What conditions should you place on the elements of the 2 x 2 matrix to make T one-to-one?

I don't see what difference the 2x2 matrix would have for T to be one-to-one if we're already saying that x1=x2=0. Is it just that the entries must be scalars? And how can I show that x1=x2=0, i.e. kerT=(0,0), in order to show that T is one-to-one, anyway?

There's a connection between the determinant of the matrix of a transformation, and the one-to-one-ness of the transformation, and the dimension of the nullspace of the transformation.

If you have a linear transformation, it will always be true that T(0, 0) = (0, 0). (I'm assuming the transformation is from a 2D vector space to a 2D vector space in my notation.) The key is whether there are any vectors (x1, x2) that are nonzero, that also map to (0, 0).

For example, with a different transformation $T: R^2 \rightarrow R^2$, if the matrix of the transformation is [0 1; 0 0], T(1, 0) = (0, 0), so the nullspace does not consist only of {(0, 0)}.

Oh ok, so the vectors have to be linearly independent? And the determinant has to be zero so that it is invertible?

fk378 said:
Oh ok, so the vectors have to be linearly independent?
I don't know what that has to do with it.
fk378 said:
And the determinant has to be zero so that it is invertible?

No, if the determinant is zero, the matrix is NOT invertible, which means that the transformation isn't one-to-one and doesn't have an inverse.

Oops, I meant the determinant cannot be zero. So if the determinant is 0 then this means the transformation is one-to-one since it has an inverse?

fk378 said:
Oops, I meant the determinant cannot be zero. So if the determinant is 0 then this means the transformation is one-to-one since it has an inverse?
No again. Read what I wrote in my previous post.

So if the determinant is not 0 then this means the transformation is one-to-one since it has an inverse?

Yes.

## 1. What is an isomorphism?

An isomorphism is a one-to-one correspondence between two mathematical structures that preserves the structure and relationships between elements.

## 2. What are the conditions for an isomorphism to exist?

The conditions for an isomorphism to exist are that the structures must have the same number of elements, the elements must have the same properties, and the relationships between the elements must be preserved.

## 3. How do you prove that two structures are isomorphic?

To prove that two structures are isomorphic, you must show that there exists a function between the two structures that is both one-to-one and onto, and preserves the structure and relationships between elements.

## 4. Can two structures that are not identical be isomorphic?

Yes, two structures that are not identical can still be isomorphic as long as they meet the conditions for an isomorphism to exist. This means that even if the structures look different, they can still have the same number of elements and relationships between them.

## 5. What is the significance of isomorphisms in science?

Isomorphisms are significant in science because they allow us to make connections and draw conclusions between different structures and systems. They also help us understand complex systems by breaking them down into simpler, isomorphic structures that we can analyze and compare.

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