The trick with this proof is that we aren’t assuming
is finite-dimensional, so we can’t start with a basis of
But we do know that
is finite-dimensional, so we start with a basis
of
Of course, every vector in
is the image of some vector in
so we can write
where
for
Since
is a basis, it is linearly independent. The results of
Exercise 2.1.12 tell us that the set
must therefore be independent.
We now introduce a basis
of
which we also know to be finite-dimensional. If we can show that the set
is a basis for
we’d be done, since the number of vectors in this basis is
We must therefore show that this set is independent, and that it spans
To see that it’s independent, suppose that
Applying to this equation, and noting that for each by definition of the we get
We assumed that the vectors were independent, so all the must be zero. But then we get
and since the are independent, all the must be zero.
To see that these vectors span, choose any Since there exist scalars such that
We’d like to be able to conclude from this that but this would be false, unless was known to be injective (which it isn’t). Failure to be injective involves the kernel -- how do we bring that into the picture?
The trick is to realize that the reason we might have
is that we’re off by something in the kernel. Indeed,
(2.2.1) can be re-written as
so But we have a basis for so we can write
for some scalars and this can be rearanged to give
which completes the proof.