3
$\begingroup$

I'm a high school student taking classes at a local college, and because of this I've taken classes in an unusual order. In particular, I took abstract algebra I (focused on group theory) last semester before exposure to linear algebra, and am now starting a linear algebra course. I greatly enjoyed abstract algebra, and I'm curious which concepts will transfer over to linear algebra; any perspectives or connections that could aid in understanding would be helpful. Thanks!

$\endgroup$
  • 2
    $\begingroup$ Linear Algebra is an interesting subject. It is pretty elementary, but it can be approached by several different point of views, in several different levels of abstraction and/or interests. Your question is quite broad, since you can approach Linear Algebra from a heavy algebra background, with Category Theory, with no algebra background etc. It really depends on the taste and the interests. $\endgroup$ – Aloizio Macedo Jan 14 '16 at 2:08
4
$\begingroup$

You might want to take a look at Dummit and Foote's Abstract Algebra book. They cover linear algebra (vector spaces), but only after covering module theory, and also groups rings and other stuff. I think an approach similar to theirs might be what you are looking for.

$\endgroup$
4
$\begingroup$

I recommend Linear Algebra done right. While it demands a lot of mathematical maturity, upper level linear algebra textbooks tend to start with "vector spaces," and deal with objects in this space (kind of like groups and their elements.) Understanding vectors this way (as opposed to multidimensional vector [maybe some kind of multivariable/ode perspective) might take away some of the more difficult applications of matrices and such. It would be interesting to see how your understanding would shift if it began theoretically and then went on to prove some things about determinants and such. Good luck! You're very advanced

$\endgroup$
4
$\begingroup$

Well, quite a bit of the algebraic concepts, and structures will. First and foremost, a vector space is an abelian group, so much of what you learned about abelian groups will still apply.

However, with abelian groups, typically we only act on them using integers (in multiplicative notation we say the integer $n$ acts on an abelian group element $g$ by:

$n \cdot g = g^{n}$; however in abelian groups we often use $+$ as the operation sign, so we have:

$n \cdot g = g+g+\cdots +g$ ($n$ times), for $n > 0$.

$(-n) \cdot g = -(n \cdot g)$ (for $-n < 0$), and:

$0\cdot g = 0$ (the identity of the group on the right).

There are some other rules that go with this, we want the action to give a homomorphism: $G \to G$, and we want there to be a ring-homomorphism:

$\Bbb Z \to \text{Hom}(G,G)$ (here $\text{Hom}(G,G)$ is the ring of endomorphisms of an abelian group $G$).

In so doing, we can talk about $\Bbb Z$-linear combinations of abelian group elements, like:

$2g_1 + 3g_2 - 5g_3$ (in multiplicative notation, this would be $g_1^2g_2^3g_3^{-5}$).

Well, it's natural instead of just taking "discrete" integral amounts of abelian group elements, to "scale them" by any kind of field element, like real numbers, for example. That gets you to vector spaces (I'm over-generalizing here a bit, to give you a "wide-angle" view).

It turns out that direct sums of finite copies of a given field form the "basic" vector spaces (it's not hard to see that the additive group of a field is an abelian group, so so is the direct sum of a finite number of copies). Since we can also multiply in a field, we can "scale" an array of field elements, by scaling each "coordinate" (this roughly corresponds to the geometric notion of "stretching/shrinking").

As with abelian groups, there is a version of the fundamental homomorphism theorem (called the rank-nullity theorem), and one speaks of subspaces (akin to subgroups) and quotient spaces. Homomorphisms are called $F$-linear maps, and isomorphisms are still nice things to have. Instead of kernels/normal subgroups, you have null spaces, but the general idea is the same sort of thing.

One caveat: in abstract algebra, there's a lot of diversity in the "character" of groups from one group to the next (some are non-abelian, some have lots of subgroups, some only a few, a group might be cyclic, or a permutation group, or a $p$-group, all of which might have widely differing properties). Vector spaces are sort of "boring" structurally, one vector space of dimension $n$ is pretty much like any other. As a result, there's much more emphasis on computation in linear algebra, which can make it seem a bit less lustrous than abstract algebra.

But stick with it, just as the really cool stuff in groups happens once you know about homomorphisms, a similar thing happens in linear algebra-the $F$-linear maps are where it really gets interesting (and, as luck would have it, these have a non-abelian multiplication, so prior experience with groups will keep you from making some typical beginner's mistakes). These are often presented "concretely" as matrices (just as some groups are presented "concretely" as permutation groups), but it's the algebraic properties that give the most insight.

$\endgroup$
1
$\begingroup$

I've read that it's best to take linear algebra before abstract algebra. Just because a lot of the concepts from linear algebra (matrices, vectors, determinants) etc. apply to abstract algebra. The things that you will perhaps use the most are vectors and system of equations.

But maybe linear algebra may help you brush up on concepts that you might not have understood before you started abstract algebra.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.