Question

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.

Answer 1

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

Answer 2

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

Answer 3

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)

Answer 4

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.

Answer 5

@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

Answer 6

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.

Answer 7

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

Answer 8

Thanks JamesJ.

Answer 9

sure

Answer 10

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

Answer 11

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.