Sarahisme Posted August 14, 2005 Share Posted August 14, 2005 umm here is the question "Here V is a nonzero finite-dimensional vector space: If dim V = p and if S is a linearly dependent subset of V, then S contains more than p vectors. True or False?" Now i am not sure , i could argue either way, although the answers says that it is false. i think it is false because it doesnt say that S is a basis, and so S can have less than p vectors and still be a subset of V. on another note, can a z-dimensional vector space V have a basis with less than z vectors in it? (or is this a related note? ) Link to comment Share on other sites More sharing options...
Yggdrasil Posted August 14, 2005 Share Posted August 14, 2005 The answer is false because linear dependence doesn't really depend on the number of vectors in a set. You can prove it's false just by showing a trivial counterexample. For instance, in R3 (a 3-D vector space), you can show that the subset {(1,1,0),(2,2,0)} contains less than three vectors and is linearly dependent. On a related note, the converse of this statement is true: given a finite-dimensional vector space, V, with dim V = p, any subset containing more than p vectors must be linearly dependent. To answer your related note, the number of vectors in a basis for your subspace will always be equal to the dimension of the subspace. Link to comment Share on other sites More sharing options...
Sarahisme Posted August 14, 2005 Author Share Posted August 14, 2005 right ok, yep that makes sense i think, thanks Yggdrasil Link to comment Share on other sites More sharing options...
matt grime Posted August 14, 2005 Share Posted August 14, 2005 do you understand the idea of 'negation of a proposition'? because if this were true then its converse would be false, and the converse is that all sets of fewer than p vectors in a space of dimension p are linearly independent, which is obviosuly completely, utterly, and trivially false. (eg v, 2v for any nonzero vector v in a space of dimension 3 or more. and even more trivally, any set with the zero vector in is linearly dependent.)) Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now