I use linear algebra quite a lot in applications, but I do not have a very strong abstract algebra background (i.e. around the level of an intro course, just knowing the basics of rings, groups, ideals, the first isomorphism theorem).

Of course, the latter is far more general, so I was interested in how it can given insight into the former. For instance, I thought it was interesting to look at the set of rotations as the group $SO(3)$.

So, essentially I'm curious as to what insights one can glean from "viewing linear algebra though an abstract algebra lens". Not necessarily practical tools, but rather more important are ones that aid in understanding or intuition (i.e. provide something new).

As a particular example, what are the relations of vector spaces to these abstract structures, and is viewing them from that point of view helpful?

  • 6
    $\begingroup$ Modules are generalizations of vector fields. Roughly speaking, when you talking vector spaces in linear algebra it is over a "field". Fields are subsets of rings. Modules are objects over rings. Pretty much everything is in correspondence. I think Dummit is a good reference for this. I am unfortunately too lazy to write out the details, but I am sure someone will provide a concise summary. $\endgroup$
    – IAmNoOne
    Dec 17, 2016 at 4:38
  • $\begingroup$ Additionally, if $L$ is a field with a subfield $K$ (it is typically said that $L$ is an extension of $K$), we can view $L$ as a vector space over $K$. This leads to some interesting results in the theory of fields. $\endgroup$
    – Aweygan
    Dec 17, 2016 at 4:49
  • $\begingroup$ I'm somewhat like you -- I use simple linear algebra quite often, but the last time I thought about rings and modules was decades ago. I never had the sense that I was missing out on any insights or elucidations. And a quick read of the answers below didn't change my mind. $\endgroup$
    – bubba
    Dec 18, 2016 at 14:16

8 Answers 8


The "Rank-Nullity" Theorem from Linear Algebra can be viewed as a corollary of the First Isomorphism Theorem, which may be more intuitive.

Suppose $T:V\to V$ is a linear transformation. Then by First Isomorphism Theorem, $V/\ker T\cong T(V)$.

So $\dim V-\rm{Null}(T)=\rm{Rank}(T)$, which is the Rank-Nullity Theorem.

This may be more intuitive than the traditional Linear Algebra proof of Rank-Nullity Theorem (see https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem).

  • 3
    $\begingroup$ The proof of the rank nullity theorem includes a proof of the first isomorphism theorem for vector spaces, but we are only interested in the dimension of things because that's the only isomorphism invariant for vector spaces. =) $\endgroup$
    – Pedro
    Dec 17, 2016 at 17:48
  • 1
    $\begingroup$ Very neat! Is there a geometric interpretation of $V/\text{ker}T \cong T(V)$? $\endgroup$ Dec 17, 2016 at 21:14
  • 4
    $\begingroup$ @user3658307 Yes. Writing $V/\ker T$ is saying "Look at $V$, but treat everything in $\ker T$ as $0$. As the kernel is a subspace, looking at $V/\ker T$ is looking at $V$ but squishing the subspace $\ker T$ down to $0$. The claim is that this should be isomorphic to the image of $T$, which makes sense (to me at least) when you use this geometric argument. (To be clear, the isomorphism is $T$ because everything in $(V/\ker T)$ gets sent to a distinct point in $\text{im }T$, as $T$ has trivial kernel, which in vector spaces means it's injective). $\endgroup$ Dec 18, 2016 at 1:02

There are tons of ways that abstract algebra informs linear algebra; here is just one example. Suppose you have a vector space $V$ over a field $k$ with a linear map $T:V\to V$. Given a polynomial $p(x)$ with coefficients in $k$, you get a linear map $p(T):V\to V$. This makes $V$ a module over the ring $k[x]$ of polynomials with coefficients in $k$: given $p(x)\in k[x]$ and $v\in V$, the scalar multiplication $p(x)\cdot v$ is just $p(T)v$. In particular, multiplication by $x$ corresponds to the linear map $T$.

Conversely, given a $k[x]$-module $V$, it is a $k$-vector space by considering multiplication by constant polynomials, and multiplication by $x$ gives a $k$-linear map $T:V\to V$. So any $k[x]$-module $V$ can be thought of as a vector space together with a linear map $V\to V$, and this is inverse to the construction described in the previous paragraph.

So vector spaces $V$ together with a chosen linear map $V\to V$ are essentially the same thing as $k[x]$-modules. This is really powerful because $k[x]$ is a very nice ring: it is a principal ideal domain, and there is a very nice classification of all finitely generated modules over any principal ideal domain. This gives us a classification of all linear maps from a finite-dimensional vector space to itself, up to isomorphism. When you represent linear maps by matrices, "up to isomorphism" ends up meaning "up to conjugation". So this gives a classification of all $n\times n$ matrices over a field $k$, up to conjugation by invertible $n\times n$ matrices, called the rational canonical form. In the case that $k$ is algebraically closed (for instance, $k=\mathbb{C}$), you can go further and get the very powerful Jordan normal form from this classification.

Now of course, these canonical forms for matrices can be obtained without all this language of abstract algebra: you can formulate the arguments in this particular case purely in the language of matrices if you really want to. But the general framework provided by abstract algebra provides a lot of context that can make these ideas easier to understand (for instance, you can think of this classification of matrices as being very closely analogous to the classification of finite abelian groups, since that is just the same result applied to the ring $\mathbb{Z}$ instead of $k[x]$). It also provides a framework to generalize these results to more difficult situations. For instance, if you want to consider a vector space together with two linear maps which commute with each other, that is now equivalent to a $k[x,y]$-module. There is not such a nice classification in this case, but the language of rings and modules lets you formulate and think about this question using the same tools as when you had just one linear map.


A vector space over a field $k$ is a similar construct to what's known as a module over a ring $R$. The idea is very similar - we want somewhere where we can add elements together, and multiply by elements from some other space. Only here, the other space is a ring. An example of this would be $k[t]$ (polynomials of arbitrary degree with coefficients in a field $k$).

As an example, there's a concept of a Smith Normal Form in Linear Algebra. The idea of this is if $A$ is a $m\times n$ matrix, then we can find invertible $m\times m,n\times n$ matrices $S,T$ such that $SAT$ is:

  1. Diagonal

  2. The diagonal elements of the diagonal matrix ($a_1,a_2,\dots$) satisfy $a_i\mid a_{i+1}$ for "small enough" $i$ (essentially, some of the $a_i$ might be zero, we want to ignore these).

Moreover, the diagonal elements are unique up to "multiplication by units". Over a field, this is rather boring, as all non-zero elements of a field are invertible (which is what it means to be a unit). But, the Smith Normal Form is true for many rings (any that are a PID), so we can do work over integer matrices and compute the smith normal form, and the diagonal elements are unique up to multiplication by $-1$ (the only non-identity unit in $\mathbb Z$). We could even do this for matrices with elements in $k[t]$! (although I'm not sure if it'd be useful).

This kind of idea tends to be true for plenty of things in Linear Algebra. You get taught a version for vector spaces, but often it's implicitly true over more general rings (as a field is just a special kind of ring).

  • 2
    $\begingroup$ This seems more like getting insight into abstract algebra from linear algebra than the other way around! $\endgroup$
    – user14972
    Dec 17, 2016 at 16:05
  • $\begingroup$ Note that $\mathbb{Z}[t]$ is not a PID, and (I am pretty sure) Smith Normal form does not apply to matrices over $\mathbb{Z}[t]$. $\endgroup$ Dec 18, 2016 at 4:55
  • $\begingroup$ @PeteL.Clark My mistake, thought that UFD$\implies PID$ instead of the other way around. $\endgroup$ Dec 18, 2016 at 4:57

The other responders above (below?)have given several excellent examples of how more general algebraic principles can be used to clarify linear algebra,but there's one I'm surprised no one's brought up. Actually,that's not entirely true-several HAVE mentioned it-they just couched it in different terms then the one I have in mind.

Consider a group action on a set:

Def: For any group G and X is a set then $\phi : G \times X \to X$ and such that $\phi(e,x)=x$ and $\phi(g,\phi(h,x))=\phi(gh,x) $ for every $x \in X$ and every $g,h \in G$. Then $\phi$ is called a group action on X.

Then consider the definition of an R-module over a ring R.

Def: Suppose that R is a ring and 1$ \in$ R is its multiplicative identity. A left R-module M consists of an Abelian group (M, +) and an operation ⋅ : R × M → M such that for all r, s in R and x, y in M, we have:

1 )r$\cdot$(x+y)=r$\cdot$ x+ r$\cdot$y

2) (r+s)$\cdot$x=r$\cdot$x+s$\cdot$x

3) (rs)$\cdot$x=r$\cdot$ (s$\cdot$x)

4) 1 $\cdot$ x=x

(A right module is defined similarly.)

Looking carefully at this definition, we notice that if we rewrite the scalar action as $L_r$ so that L$_r$(x) = r ⋅ x, and L for the map that takes each r to its corresponding map $L_r$,then (1) states that every $L_r$ is a group homomorphism of M by compatibility.

Also, (2)-(4) assert that the map L : R → End(M) given by r ↦ $L_r$ is a ring homomorphism from R to the endomorphism ring End(M). But this means every left (right) module is a ring action on an Abelian group!

Therefore,every vector space can be thought of as a group action in which the Abelian structure of the field of scalars "acts" on the Abelian group of vectors via multiplication.

Another observation worth mentioning is that if R is a field and G is a group, then a group representation of G is a left module over the group ring R[G].Representation theory is a major branch of abstract algebra with enormous utility in many areas of both pure and applied mathematics where the structure of a group can be analyzed by specific group actions on the a given vector space.

How's that?

  • $\begingroup$ Thanks for expanding on the module aspect that others mentioned! $\endgroup$ Dec 18, 2016 at 15:10

Jordan normal form is about how a matrix can be almost diagonalized. The proof is technical and derives from the exact same more general theorem that yields the Fundamental Theorem of Finitely Generated Abelian Groups.


The study of spectral theory generalizes the investigation of eigenvalues, diagonalization, Jordan normal form etc. An important class of objects there are the C*-algebras.

  • $\begingroup$ Actually, I'd be very interested in the analogue of spectral theory in abstract algebra, if you (or anyone) would like to mention a bit more about it. :) $\endgroup$ Dec 18, 2016 at 15:07
  • $\begingroup$ In any unital algebra you can define the spectrum of an element. The algebra of linear maps on a finite dimensional space is just a special instance. One can for instance consider bounded operators on an arbitrary Hilbert space and one obtains a spectral theorem that is similar to that for normal matrices. $\endgroup$ Dec 18, 2016 at 18:33

I think circulant matrices provide a good example of crossroads between linear algebra and abstract algebra.

For a first approach, see paragraph "Analytic Interpretation" in the Wikipedia article about circulant matrices: (https://en.wikipedia.org/wiki/Circulant_matrix)

Let us take the case of $3 \times 3$ circulant matrices that can be decomposed in the following way:


Thus any circulant matrix is a a matrix polynomial in $C$ (as introduced in formula above).

Moreover, it is connected with harmonic analysis, a major branch of abstract algebra, because it is diagonalized by Fourier matrix $F_n$:

Let us take an example ; let $\omega=e^{i 2\pi/3}$ be a primitive cubic root of unity.

Let us consider matrices :

$$C=\begin{pmatrix}1&2&4\\4&1&2\\2&4&1\end{pmatrix} \ \ \& \ \ F_3=\tfrac{1}{\sqrt{3}}\begin{pmatrix}1&1&1\\1&\omega&\omega^2\\1&\omega^2&\omega\end{pmatrix} \ \ \& \ \ D=\begin{pmatrix}7&0&0\\0&-2+\sqrt{3}i&0\\0&0&-2-\sqrt{3}i\end{pmatrix}$$

With these matrices, one can write:


Moreover the diagonal elements of $D$ constitute the DFT (Discrete Fourier Transform) of the first column of $C$. (see for example : (https://math.stackexchange.com/q/2629712)).

As a corollary (because of the correspondence between product and cyclic convolution $\star$ by Fourier Transform), denoting by $C_V$ the cyclic matrix associated with first $V$, we have :

$$C_V \times C_{V'}=C_{V \star V'}.$$


The above answers were all excellent. BUT I recommend you find a decent linear algebra text and carefully go through the axioms of a vector space (I think there are 23 +/- n). Each of them is algebraic in nature: you are working with a field of scalars and (essentially) a group of vectors. When you do complicated things in LA, like finding eigenvalues, you are using a multitude of abstract algebra. The fun is in the "discovery"


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .