Prove one to one using rank-nullity theorem

[Hwng10]

Homework Statement

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 .

Homework Equations

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)

 

[HallsofIvy]

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

 

[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 ??

 

[HallsofIvy]

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>?

 

[Deveno]

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>?

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

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