Linear Transformation - Onto

1. Apr 18, 2004

discoverer02

Linear Transformation -- Onto

I'm having trouble with the first part of the following problem:

Let T be a linear transformation from an n-dimensional space V into an m-dimensional space W.

a) If m>n, show that T cannot be a mapping from V onto W.

b) if m<n, show that T cannot be one-to-one.

Part b) I can see. I think. T(v) = Av = w The matrix A will have more columns than rows (more unknowns than equations), so there will be infinitely solutions (more than one mapping from a v in V to a w in W).

I'm stumped by part a). I'm not seeing how m>n guarantees that there are w 's in W that aren't part of R(T).

A nudge in the right direction would be greatly appreciated.

Thanks.

Last edited: Apr 18, 2004
2. Apr 18, 2004

Hurkyl

Staff Emeritus
Why doesn't the same approach work?

3. Apr 18, 2004

discoverer02

I'm wondering the same thing, so there must be something that I'm not seeing.

I'll think about it some more.

Thanks.

4. Apr 19, 2004

Stevo

Well, for a linear map: $$T:V^n \rightarrow W^m$$ where $$n<m$$

There is a useful formula which describes subspaces of V and W in terms of conditions they satisfy with respect to T. Then have a look at the dimension of these subspaces, the dimension of V, and the dimension of W. You should be able to establish that there are certain elements in W that aren't the image of any element in V. Hint: Consider a basis of W.

Last edited: Apr 19, 2004
5. Apr 19, 2004

NateTG

Hint: Do you think that T inverse is defined for all of W?

6. Apr 19, 2004

discoverer02

OK, let's see what I've got so far:

A basis of W would consist of 3 elements, A basis of V would consist of 2 elements.

T(x + y ) = T(x ) + T(y )
T(kx ) = kT(x )

I also have the equation dim(ker(T)) + dim(R(T)) = dim(R2) = 2

Anyway you look at it dim(R(T)) <= 2. This means that any other element in R(T) is a linear combination of these two elements. But W has a basis of 3 elements meaning that R(T) could not possibly contain at least one of W's basis elements.

I think this makes sense finally. I'll think about it some more just to make sure it's solid.

Thanks very much for the hints.

discoverer02