Can a Linear Transformation Be Onto If the Codomain's Dimension Is Higher?

[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.

 

[Hurkyl]

Why doesn't the same approach work?

 

[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.

 

[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.

 

[NateTG]

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

 

[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