the dimension of a vector space is the number of elements in one of its bases (the cardinality) we write where is a basis of
number of free variables
let be a vector space of dimension , for every subset that has elements the following statements are equal
- is linearly independant
we prove given is linearly independant, we need to prove let , a basis of , let we know according to previous lemmas that therefore is linearly dependant since is linearly dependant then we have the following non-trivial linear combination that gives us : we know because otherwise another would be equal to 0 and wouldnt be linearly independant by manipulating the expression we get: therefore therefore
we prove given assume in contradiction that is linearly dependant according to a previous lemma we there exists that is linearly independant such that therefore is a basis therefore we arrived at a contradiction becacuse
if and , if or is linearly independant then is a basis of
given is finitely spanned, if then
according to the previous lemma, since is finitely spanned then is finitely spanned too which means has a basis and a dimension
we assume in contradiction that assume is a basis of , then we know assume where is a basis of , then we know which means is linearly independant but it cant be because we have a set where with less elements that is linearly independant so a set with cant be linearly independant
given is finitely spanned, and then
let be a basis of and so and is linearly independant therefore according to this lemma and since it spans and is of size then it is a basis of
if is finitely spanned and then for every basis of there exists a basis of such that we call this operation as complementing the basis to the basis
we prove using induction of
the first step of an induction proof is a check we check if meaning we pick
to be continued…
let be a finitely spanned vector space, let then:
with previous lemmas in mind, we know let be a basis of since then we can complement to a basis of such that is a basis of and since then we can complement to a basis of such that is a basis of so far we know to be continued…