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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.
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.

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myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1I think it is false but I need a counterexample to show me that. lol

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1So kerL=0 is the elements v in V such that L(v)=0

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1So 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)

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1So 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 v1v2 = 0, which shows that v1=v2 as desired.

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1@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

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2If the kernel is nonzero 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.

myininaya
 3 years ago
Best ResponseYou've already chosen the best response.1So that is why I couldn't think of a function that fails the surjective part becacuse the statement was true. lol

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0Wouldn't we need to know the dimensions of V and W in order to determine if it was indeed surjective?

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.2I'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.
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