# Prove one to one using rank-nullity theorem

1. Feb 26, 2012

### Hwng10

1. The problem statement, all variables and given/known data

Let V and W be finite dimensional vector spaces.Given dim(V)≤ dim(W)
, prove that there exists a one-to-one linear transformation T : V -> W .
2. Relevant equations

3. The attempt at a solution
What I want to prove here is to show that nullity=0
dim(V) ≤ dim (W)
dim(N(T)) + dim(R(T)) ≤ dim(W)

2. Feb 26, 2012

### HallsofIvy

Staff Emeritus
You can't prove that the nullity of T until after you have T! How are you defining T?

3. Feb 26, 2012

### Hwng10

I really have no idea no how to prove this question . Do I need to prove it for general T or I just need to find an example which satisfies all the properties above ??

4. Feb 26, 2012

### HallsofIvy

Staff Emeritus
I suggest you go back and reread the problem! "prove that there exists"... First define an obvious linear transformation from V to W, then show that it is one-to-one.

Here's a simple special case: suppose dim(V)= 2, dim(W)= 3. Then, choosing some basis for both V and W, you can write any vector in V as <x, y> and any vector in W as <a, b, c>. Okay, what can you say about T(<x, y>)= <x, y, 0>?

5. Feb 26, 2012

### Deveno

hint 2: if dim(V) = n, we have a basis {v1,v2,...,vn}.

since dim(W) ≥ dim(V), what can you say about any basis of W?