## myininaya 3 years ago True/False?: Let V and W be vector spaces. Let L:V->W be a linear map. If kerL=0, then L is a bijection.

1. myininaya

I think it is false but I need a counterexample to show me that. lol

2. myininaya

So kerL=0 is the elements v in V such that L(v)=0

3. myininaya

So we need to show that for some mapping L that L is not a bijection Remember a bijection has the following properties: A )x,y in V, L(x)=L(y) implies x=y (injective) B )if for every w in W there exists at least one v in V such that L(v)=w (surjective)

4. myininaya

So we know that L is injective. So L fails the surjective part of the definition of bijective. Assume that ker(L) = 0. If L(v1) = L(v2), then linearity of L tells that L(v1 - v2) = 0. Then ker(L) = 0 implies v1-v2 = 0, which shows that v1=v2 as desired.

5. myininaya

@jamesj @Zarkon I hope I don't bother you guys too much...I can't figure out such a function for my counterexample. You don't have to answer if you don't want to but of course you already knew that. lol

6. JamesJ

If the kernel is non-zero the the dimensionality of the image of V, L(V) is equal to V. This tells us intuitively that the map is 1:1. To formalize it, we need two things, as you observe a) surjectivity b) injectivity a) is trivial, and notice that L(V) need NOT equal W Now using your argument above, you've shown the L is injective, i.e., 1:1, because if L(v1) = L(v2), then v1 = v2. Hence indeed L is a bijection.

7. myininaya

So that is why I couldn't think of a function that fails the surjective part becacuse the statement was true. lol

8. myininaya

Thanks JamesJ.

9. JamesJ

sure

10. Zarkon

Wouldn't we need to know the dimensions of V and W in order to determine if it was indeed surjective?

11. JamesJ

I'm just saying it surjective onto its image and the sense of a function being a bijection being just one to one and onto. If the question is: is the map L a bijection *between V and W*, yes, then clearly we need that dim(V) = dim(W) in the finite case.