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Linear maps preserve linear combinations. We now learn about special linear maps that preserve generators. These are called epimorphisms.
In the article on monomorphisms we considered linear maps which map linearly independent vectors to linearly independent vectors. There we found out that these maps are exactly injective linear maps.
Injective linear maps therefore "preserve" linear independence.
Using the linear independence, we could express the intuitive dimension notion in mathematical (linear algebra) terms. There, we also encountered generators.
Now: Are there also linear maps that preserve generators?
So let
be two
-vector spaces over the same field
and
a generator.
Now, what properties must a linear map
satisfy, in order for
being a generator of vector space
? For this, we would need to be able to represent any
as a linear combination of
.
That is, we need to find
such that
Since the map
is linear, this is equivalent to
So
must be in the image of
. This is said to hold for every
. Thus
is a necessary condition for
to preserve generators.
Is this also a sufficient condition? Let
. We investigate whether every
can be represented as a linear combination of
. Because
we have for any
a vector
with
. Since
is a generator of
, there are some linear combination factors
with
So we can write
as:
And hence
is within the generated space of the
.
Thus, the linear map
preserves generators if and only if
. Moreover,
satisfies
exactly if
is surjective. Thus, a linear map must be surjective to have the generating property. We call surjective linear maps epimorphisms.
We have already considered in the motivation that surjective linear maps are exactly the maps that preserve generators.
Because the case of finite generators is more important than the general statement, we consider this case first. Then we investigate what we need to change for the general case:
Proof
Proof step: „
is an epimorphism“
„
is a generator“
Let
beliebig.
Then according to precondition there is a vector
with
.
Since
generates the vector space
, there are linear combination factors
with
.
Hence, we have:
So
can be represented as a linear combination of
.
Since
was arbitrary,
is a generator of
.
Proof step: „
is a generator“
„
is an epimorphism“
Let
be arbitrary.
We have to show that there is a vector
with
.
Since
is generated by
, there exist scalars
(for linear combination) with
.
We now set
:
This proves that
is surjective, that is, an epimorphism.
Now we generalize to vector spaces of arbitrary dimension:
Proof
We can almost copy the proof from above: Since
is a generator of
, this means that every vector
has a representation as a linear combination
, where
are scalars and
are from
.
The only thing that changes is that the sums no longer have a fixed number of summands. In the proof above, we could always run the sums of
to
. Here, the number of summands depends on the vectors
and
, respectively. But it is still a finite number of summands. Therefore, the rest of the proof is the same as within the finite case.
We will now be introduced to a second (category-theoretic) characterization of epimorphisms, the possibility of being "right shortened":
Proof (Epimorphisms can be right shortened)
Proof step: 1.
2., by direct proof
Let
be an epimorphism, i.e.,
is surjective.
Let
be a vector space, and
, such that
.
We want to show that
holds.
Since
and
are maps with same domain of definition
and same range
, we need to show that
holds for all
.
So consider any
.
Since
is surjective, there exists a
with
.
Now,
.
Since we have chosen
arbitrary, we obtain
.
Proof step: 2.
1., proof by contradiction
Let
be a homomorphism.
Suppose
is not an epimorphism, i.e., not surjective.
Then there is a
with
.
In particular,
since
.
We extend
to a basis
of
.
now , let us define two homomorphisms
.
First we set
.
Further, we define
using the principle of linear continuation on the basis
:
for all
.
Next we show
:
Consider some
.
Then
, since
.
As
,
we indeed have
.
But
, since
.
This is a contradiction to the assumption, and it follows that
is an epimorphism.
Example
We consider the vector spaces
and
with
, as well as the linear map
For this map, we simply truncate the last
components.
This makes it clear why we must require
(if
, the map is simply the identity).
This map is an epimorphism:
Let
.
Then, we have
.
Example
For a field
and two
-vector spaces
, the following map is an epimorphism:
Here
denotes the Outer direct sum (missing).
We first show for this that the map
is linear.
Consider some
, as well as
.
Then
and
.
This establishes linearity.
Let now
be arbitrary.
Then
for every
.
That is,
is a preimage of
under
.
Thus
is an epimorphism.
If
is not the null space, then there are even multiple (perhaps infinitely many) preimages.
Math for Non-Geeks: Template:Aufgabe
Math for Non-Geeks: Template:Aufgabe
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