all bases for a vector space have the same cardinality
All bases for a vector space have the same cardinality.
Let be a basis for ( exists, see this link (http://planetmath.org/ZornsLemmaAndBasesForVectorSpaces)). If is infinite, then all bases for have the same cardinality as that of (proof (http://planetmath.org/CardinalitiesOfBasesForModules)). So all we really need to show is where has a finite basis.
Before proving this important property, we want to prove something that is almost as important:
If and are subsets of a vector space such that is linearly independent and spans , then .
If is finite and is infinite, then we are done. Suppose now that is infinite. Since is linearly independent, there is a superset of that is a basis for . Since is infinite, so is , and therefore all bases for are infinite, and have the same cardinality as that of . Since spans , there is a subset of that is a basis for . As a result, we have .
Now, we suppose that and are both finite. The case where is clear. So assume . As spans , . Let and
and assume . So for all . Since spans , can be expressed as a linear combination of elements of . In this expression, at least one of the coefficients (in the field ) can not be (or else ). Rename the elements if possible, so that has a non-zero coefficient in the expression of . This means that can be written as a linear combination of and the remaining ’s. Set
As every element in is a linear combination of elements of , it is therefore a linear combination of elements of . Thus, spans . Next, express as a linear combination of elements in . In this expression, if the only non-zero coefficient is in front of , then and would be linearly dependent, a contradiction! Therefore, there must be a non-zero coefficient in front of one of the ’s, and after some renaming once more, we have that is the one with a non-zero coefficient. Therefore, , likewise, can be expressed as a linear combination of and the remaining ’s. It is easy to see that
spans as well. Continue this process until all of the ’s have been replaced, which is possible since . We have finally arrived at the set
Suppose and are bases for . We apply the lemma. Then since is linearly independent and spans . Similarly, since is linearly independent and spans . An application of Schroeder-Bernstein theorem completes the proof. ∎
|Title||all bases for a vector space have the same cardinality|
|Date of creation||2013-03-22 18:06:51|
|Last modified on||2013-03-22 18:06:51|
|Last modified by||CWoo (3771)|