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...generating set as
, so
***(explanation needed, why? [or your question])*** since
is a some linearly independent...
Theorems & Proofs
Theorem: Let
be a subspace of a finite dimensional vector space
. Then
is finite dimensional and
Proof: Let
be a basis for
. Then we know that
is a finite set since
is a finite dimensional. Then, for given a subspace
, let us construct a linearly independent set
by adding vectors from
such that
is maximally linearly independent. In other words, adding any other vector from
would make
linearly dependent. Here, L has to be a finite set by the Replacement Theorem, if we choose the generating set as
, so
since
is a some linearly independent subset of
. Now we want to show that
is a basis for
. Since
is linearly independent, it suffices to show that
. Suppose not:
. (We know that
since
is made of vectors from
.) Then
But this means
is linearly independent, which contradicts with maximally linearly independence of
. Therefore
and hence,
is a basis for
Replacement Theorem: Let
be a vector space generated by
(perhaps linearly dependent) where
and let
be a linearly independent subset of
such that
. Then
and there exists a subset
of
with
and
.
Proof: We will prove by induction hypothesis on
:
For
:
,
and
so,
Now, suppose true for
: