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Topic: Failing Linear Algebra:
Replies: 91   Last Post: Jan 10, 2007 12:56 PM

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Marc Olschok

Posts: 409
Registered: 12/6/04
Re: Failing Linear Algebra:
Posted: May 3, 2004 8:10 AM
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Anonymous wrote:
> Marc:
>

>>The first time, I spent in a linear algebra course (Kaiserslautern, 1986)
>>introduced sets, groups, rings and fields, modules and then vector spaces.
>>(in that order)

>
> Odd. We did sets and then groups in algebraic structures.
> We touched upon rings, I think, but I don't remember what they are.
> Have no idea about fields, if they're not vector fields,
> which someone already said they aren't. Never heard of modules.


As I wrote in the other post, a field is a ring such that not(0=1) and
every nonzero element has a multiplicative inverse.
In the econtext of vector spaces, this is where your scalars come from.
If you keep the axioms for a vector space but allow your scalars to
come from a ring without insisting that the ring must be a field,
you obtain the notion of a module.

>
> We did vectors, then vector spaces, then subspaces. Then spans, dependence,
> bases, matrices, determinants, then eigenvalues and vectors.
> It's odd because Schaum's Outlines does it in a different order.
> And Cliff's does an even different order.


I wonder how the curriculum is designed at your department.
Did all the students in the current "linear algebra" course take
"algebraic structures" in the previous term? If so, I would expect
the current course to utilize this previous knowledge to some extent.

>
>>By the time we met vector spaces, the notion of image and kernel as well
>>as the proof of the first homomorphism theorem

>
> "Homomorphism theorem"? Is this what's called "Dimension Theorem" now?


Not quite. Depending on the text it might also run under the name
"first isomorphism theorem":

| Suppose V and W are vector spaces and f: V ---> W is a linear map.
| Then there is a unique linear map g: V/ker(f) ---> W such that
| f = g o q (o means composition) where q : V ---> V/ker(f)
| is the canonical map with q(v) = v + ker(f).
| The map g is injective. If f is surjective then g is also surjective.

As a special case one has V/ker(f) isomorph to f(V).
If you already know how to calculate dimensions of quotient spaces
you can read off dim(V) - dim(ker(f)) = dim f(V) from the above.

The point is, that the theorem also holds if you replace
"vector space" by "abelian group" and "linear map" by "group homomorphism".
The same works for "groups" and "group homomorphism".
The same works for "rings" and "ring homomorphism".
There is even a version for "sets" and "maps".

In all these cases, the map g is constructed from f in the same way.

For this to make sense one needs to check how "quotient structures"
are constructed and what "kernel of a homomorphism" is supposed to mean.

The case of sets is the only one that looks a bit odd at first, but it is
in fact the basic stage of these constructions:

Given a map f: A ---> B between two sets, define
the relation ker(f) on A ( = subset of A x A ) by
(x,y) in ker(f) <==> f(x) = f(y) ( for x, y in A )

You can check that this indeed defines an equivalence relation.
For any x in A, let [x] := { y in A | f(x)=f(y) } be the
equivalence class of x with respect to ker(f).

Consider the set A/ker(f) := { [x] | x in A }
of all equivalence classes and define q: A ---> A/ker(f)
by q(x) := [x] (for x in A).

Now you can try to define g: A/ker(f) ---> B by
g([x]) := f(x).

Because of [x]=[y] ==> f(x)=f(y) the above really defines a map
from A/ker(f) to B.
Because of f(x)=f(y) ==> [x]=[y] this map is injective.

Just via the construction one has f(x) = g(q(x)) for all x in A.

Believe it or not, this was the main part.

How can you build the other theorems from this?

Suppose you have a group homomorphism f: A ---> B between to groups.
We already have q : A ---> A/ker(f) and g: A/ker(f) ---> B
from the above.
So far A/ker(f) is just given as a set, therefore it does not yet make
sense to speak of group homomorphisms q or g.
However there is exactly one way to give a group structure to A/ker(f)
such that q is a group homomorphism: try to define [x][y] := [xy] .
One can check [x]=[x'] and [y]=[y'] ==> [xy]=[x'y'], hence
the above attempt indeed defines a group multiplikation on A/ker(f).
If 1 is the neutral element of A, then [1] is the neutral element of A/ker(f).
Indeed N:=[1] is a normal subgroup of A and the equivalence class [x]
is exactly the coset Nx (= xN).
For this reason one does not need to consider the relation ker(f)
when dealing with groups; it is sufficient to deal with the normal
subgroup { x in A | f(x)=1 }.

The group case also covers the case of abelian groups.

In the case of a ring homomorphism f: A ---> B between to groups
the procedure is the same. Equip the abelian group A/ker(f) with
the only possible ring operations, such that the group homomorpohism
q : A ---> A/ker(f) is also a ring homomorphism.
Here I:=[0] is an ideal of A and the equivalence class of [x]
is exactly the coset x+I.

Likewise one can deal with the case of linear maps between two
vector spaces.

The above was a little condensed and hides a lot of fine print that
needs to be checked. Most likely your current course only showed the
construction for the case of vector spaces. But perhaps you can spot
in that proof a few of the ingredients from above.

Marc


Date Subject Author
4/24/04
Read Re: Failing Linear Algebra:
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4/24/04
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