spanning sets stuff

[Sarahisme]

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? )

[Yggdrasil]

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.

[Sarahisme]

right ok, yep that makes sense i think, thanks Yggdrasil

[matt grime]

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.))