Have I done this proof correctly and how do I finish it?

Question

1.	Let V and W be finite dimensional vector spaces. Show that:
a. If dim(V) ≤ dim(W), there exists a one-to-one linear transformation V→W.
b. If a one-to-one linear transformation V→W exists, then dimV ≤ dimW

I wrote:
Let T : V-->W
The dimension theorem gives that dimV=nullity(T) + rank (T) 
and rank(T)≤dimW because Im(T) is a subspace of W.
Therefore, dimV - dimW ≤ nullity(T)
If dimV is less than or equal to dimW then, the nullspace is trivial and T must be one-to-one.

I'm not sure whether this is correct and I don't know how to prove the converse. I would appreciate any help that you can offer.

arthur · Accepted Answer

You seem to imply that EVERY linear transformation is injective, which is certainly not the case. I think your error stems from your claim that b <= a and b <= 0 implies a = 0, which is absolutely false.

Part (a) can be done by direct construction: map different basis elements of V to different basis elements of W.

For part (b), your idea to use the rank & nullity theorem is brilliant. Now we can start by assuming null(T) = 0.

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Have I done this proof correctly and how do I finish it?

1 Answer