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…