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

[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

I already proved (a).
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?

 

[Mark44]

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.

 

[fk378]

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.

 

[Mark44]

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.

Also, the question asks about conditions on the elements of F.

\begin{eqnarray}<br /> T\left (<br /> \begin{array}{c} \nonumber<br /> x_1 \\<br /> x_2<br /> \end{array}<br /> \right ) = \left [{\begin{array}{cc}<br /> w &amp; y \\<br /> z &amp; v<br /> \end{array}<br /> \right ]<br /> <br /> \left ( \begin{array}{c}<br /> x_1 \\<br /> x_2<br /> \end{array}<br /> \right )\end{equation}

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

 

[fk378]

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?

 

[Mark44]

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)}.

 

[fk378]

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

 

[Mark44]

Oh ok, so the vectors have to be linearly independent?

I don't know what that has to do with it.

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.

 

[fk378]

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?

 

[Mark44]

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.

 

[fk378]

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

 

[Mark44]

Yes.