Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Many introductory books on vector spaces mention that the scalars need not be reals, and might even have sections discussing complex vector spaces or vector spaces over the integers mod 2. I have never seen any such book mention that all of the theory goes through as well if one restricts the scalars to be just rational numbers. Perhaps this is because there is a dearth of interesting problems about such vector spaces accessible at this level that couldn't simply be discussed in the context of real scalars.

I wonder if there is an interesting introductory-level problem or topic about vector spaces that would be most naturally conducted by allowing rational number scalars. Does anyone know of such, perhaps one with a number-theoretic aspect?

(By introductory: I envision a first course on linear algebra, including non-math majors. They would be seeing vector spaces (and that level of abstraction) for the first time. Perhaps they would be seeing matrix multiplication for the first time. Usually, in my experience, such courses primarily use the real numbers as scalars.)

share|cite|improve this question
My first thought is that most of algebraic number theory takes place in vector spaces over $\mathbb{Q}$; but I am hard pressed right now to come up with something striking that might appeal to the level you indicate. Perhaps some of Euler's solutions to diophantine problems using quadratic extensions of $\mathbb{Q}$, but it seems a bit of a stretch. – Arturo Magidin Nov 30 '10 at 15:34
This is basically Arturo's comment, but tell your students work out as an exercise that if $r$ is a real number, then we can define a vector space over $\mathbb Q$ by $V := \mathbb Q[r]$, and that $r$ being algebraic means exactly that $V$ is finite dimensional. People interested in maths have usually heard about algebraic numbers, but it's a pleasant surprise to see them pop up naturally (?) for the first time. – Gunnar Þór Magnússon Nov 30 '10 at 20:01
up vote 4 down vote accepted

One standard example of using $\mathbb{Q}$-vector spaces is to construct discontinuous additive functions $f: \mathbb{R} \to \mathbb{R}$. The approach is to choose a $\mathbb{Q}$-basis for $\mathbb{R}$. For each $x \in \mathbb{R}$, define $f(x)$ to be the sum of the coefficients in this basis. This is linear but discontinuous (it maps into $\mathbb{Q}$).

share|cite|improve this answer
discontinuous $\mathbb{Q}$-linear functions $f$, right? Maybe it's an easier sell just to say discontinuous additive functions. (This seems to be in a similar spirit to my answer, in which we exploit $\mathbb{Q}$-vector spaces to understand the structure of $\mathbb{R}$ as an abelian group.) – Pete L. Clark Nov 30 '10 at 14:18
@Pete: Yes, of course. – Akhil Mathew Nov 30 '10 at 15:00

Continuing Akhil's answer, let's prove a theorem of Dehn: if a rectangle is tiled by squares, the the ratio of the lengths of its sides is rational.

Suppose to the contrary that the sides of the rectangle $x,y$ are not rationally dependent. Then we can find some linear homomorphism $f\colon \mathbb{R} \rightarrow \mathbb{Q}$ such that $f(x) = 1$ and $f(y) = -1$.

We define the $f$-area $A(R)$ of a rectangle $R$ with edge lengths $h,v$ to be $f(h)f(v)$. If a rectangle $R$ is tiled by rectangles $R_i$ forming a grid, then from linearity it immediately follows that $$A(R) = \sum_i A(R_i).$$

Denote the big rectangle $R$ and the squares $S_i$. Take your tiling and extend all the lines to form a grid inside the rectangle. Denote the grid rectangles by $G_j$. Then $$A(R) = \sum_j A(G_j) = \sum_i A(S_i).$$ Since a square has both sides equal, $A(S_i) \geq 0$. On the other hand, by construction $A(R) < 0$. This contradiction shows that the two sides of the big rectangle are, in fact, rationally dependent.

Instead of taking a linear mapping from $\mathbb{R}$ to $\mathbb{Q}$ we could take a linear mapping from a smaller, finite dimensional domain by only considering the lengths in the grid $G_j$ - then everything becomes beginner's linear algebra.

share|cite|improve this answer

An abelian group is torsionfree and divisible iff it has the (unique) structure of a vector space over $\mathbb{Q}$.

Using this observation the classification of vector spaces over a field in terms of their dimension gives rise to a classification of torsionfree divisible abelian groups. In particular, since if $V$ is a vector space over a field $K$ with $\operatorname{dim} V> \# K$ we have $\operatorname{dim} V = \# V$, we find that any two uncountable torsionfree divisible abelian groups of the same cardinality are isomorphic. In particular $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as abelian groups.

[Added: In model-theoretic terms, we have established that the elementary theory of torsionfree divisible abelian groups is uncountably categorical. Since it has no finite models, by Vaught's Test this theory is complete. See these notes, especially Theorem 6. (This is one of the most inexpensive examples of a complete theory that I know.) Now I feel honorbound to advise you not to mention this in a linear algebra class!]

Whether one would or could want to introduce any of this in a linear algebra class is highly dubious (a graduate-level linear algebra class, perhaps). However, I can't help but recall that Paul Halmos was very fond of the conclusion in the last paragraph and apparently liked to include as a problem on his exams: "Can $\mathbb{R}$ be given the structure of a vector space over $\mathbb{C}$?" He describes this in his autobiography.

Although I am a number theorist, no number-theoretic applications are springing to mind. A little voice in my head is saying "Hilbert 90", but that's probably not a direction you want to go in either.

You may well know this already, but there are lots of cool combinatorial problems which -- in a highly non-obvious way! -- can be solved using linear algebra over finite fields, especially $\mathbb{F}_2$. I recommend L. Babai's manuscript "Linear algebra methods in combinatorics".

share|cite|improve this answer
On the other hand, no matter what else is said, in practice almost all the examples and problems in a linear algebra book are being done over $\mathbb{Q}$. How often does one see problems involving matrices with entries like $\sqrt{2}$ or $\log \pi$? It is certainly worth pointing out that in solving linear systems with rational coefficients one is guaranteed solutions with rational coefficients. (Cough -- Hilbert 90 -- cough!) – Pete L. Clark Nov 30 '10 at 13:31

One nontrivial example of vector spaces over $\mathbb{Q}$ being used to solve a problem is the construction of the Dehn invariant in the resolution of Hilbert's third problem.

Edit: After reading your comments on Alex Bartel's answer, I guess this is not really "introductory level" as there are two nontrivial $\mathbb{Q}$-vector spaces involved and one has to tensor them...

I think if you want to impress on your students that stuff like the rank-nullity theorem goes through over an arbitrary field, you should be looking for combinatorial applications over $\mathbb{F}_2$. If your students can digest the definitions, these are very elegant examples of just how general the theory is.

On the other hand, much of the theory covered in a standard textbook - eigenvalues, eigenvectors - does not go through over an arbitrary field. Perhaps this will all just be unnecessarily confusing.

share|cite|improve this answer
What's wrong with eigenvalues and eigenvectors over arbitrary fields? – Alex B. Nov 30 '10 at 14:28
@Alex: they only always exist if the field is algebraically closed? I don't think an introductory linear algebra course for non-math majors is the best place to start talking about algebraic closures. – Qiaochu Yuan Nov 30 '10 at 14:38
@Alex: If $F$ is not algebraically closed, a linear transformation over $F$ need not have any $F$-rational eigenvectors. It is arguable whether this is in any way pathological -- is there something pathological about rotations in $\mathbb{R}^2$? -- but it certainly makes things more complicated than if the field is algebraically closed. – Pete L. Clark Nov 30 '10 at 14:41
The point I was trying to make with my provocative comment is that the fact that a matrix might not be triangularisable over an arbitrary field is a good thing, because is forces you to develop criteria for when it is. That was the point I was making, when I said that restricting to real or complex numbers obfuscates matters: what is it about the complex numbers that makes matrices triangularisable? If you work over an arbitrary field, it is clear: you want the minimal polynomial to split into linear factors and this happens to always be possible over $\mathbb{C}$. – Alex B. Dec 1 '10 at 1:59

The question is kind of general, but here is an attempt at an answer:

  1. In the general introductory books on vector spaces, one usually tries to develop as much of the theory as possible without specifying the field of scalars at all. This has a good reason: the main advantage in talking about finite dimensional vector spaces rather than tuples of numbers is that one gets rid of baggage, as it were, and leaves only the essential parts of a theory. The same reasoning suggests to not specify the field. This way, the relevant structure becomes visible and is not obstructed by unnecessary details. There is no need to specify the field, if you want to define vectors, linear maps, bases, matrices, change of basis, determinants etc. In the usual order of teaching these topics, the first place when it becomes necessary to specify the field is when one starts talking about inner products, because to define non-degeneracy, one needs a notion of positivity. To sum up point one: the reason that books rarely say "by the way, this works also for the rationals" is that most books state and prove as much as possible by starting "Let $V$ be a vector space over a field $F$..." They don't feel that they have to remind the reader that $\mathbb{Q}$ is one such field. Indeed, there are soooo many others out there.

  2. There are however areas neighbouring linear algebra where the field of scalars becomes important and in many of them, $\mathbb{Q}$ is one of the fields that receive special attention. One example is: classical representation theory is usually first developed over $\mathbb{C}$. But when all the necessary theory is in place, one starts wondering about fields of definition. Questions like "which representations are defined over $\mathbb{Q}$" and "how many irreducible rational representations does a finite group have" are very natural and very non-trivial.

  3. Since your thoughts are already going in that direction, I will also mention, that some theories such as number theory are even more interested in free $\mathbb{Z}$-modules than in vector spaces. Those are even more special (and difficult) than rational vector spaces. Such beasts arise naturally in algebraic number theory. For example Galois groups of finite extensions of number fields naturally act on many free $\mathbb{Z}$-modules, such as rings of integers, units therein (modulo torsion) and many more. Determining the structure of these actions is a very active area of research, but you will have difficulty introducing that in an early undergraduate course. Sometimes, determining the integral structure is so difficult that one is forced to tensor with $\mathbb{Q}$ and turn these modules into rational vector spaces, which brings us back to where you started from.

So to summarise all this: the very basic theory of linear algebra tries to be oblivious to the field itself. The more advanced theory is very much interested in the field of definition. $\mathbb{C}$ tends to be the easiest to understand, then comes $\mathbb{R}$, the rationals are much more difficult than these two. So I guess, the reason that you don't see $\mathbb{Q}$ often in introductory undergraduate books is that, as soon as the field actually matters, $\mathbb{Q}$ is a difficult one to work with.

share|cite|improve this answer
Hmmm, I must take issue with "most books state and prove as much as possible by starting 'Let V be a vector space over a field F...'" You haven't been reading the same introductory books that I have. Most first courses in linear algebra, in my experience, work over the reals for the most part (exceptions often made, of course, when computing eigenvalues). They differ in whether they start with a big focus on matrices or jump quickly into abstract vector spaces, but the assumption is that most science/engineering students won't ever need to know what a field is – Barry Smith Nov 30 '10 at 14:05
I trust you on this one, I seem to have taken care to avoid those. I am teaching an (advanced) linear algebra course this semester and I have certainly not gone near books that work only over the reals. In my opinion, there is nothing gained, except for confusion over which results depend on the special nature of the reals and which ones don't. – Alex B. Nov 30 '10 at 14:08
"C tends to be the easiest to understand, then comes R, the rationals are much more difficult than these two." From a very advanced perspective, this is true. From an introductory class perspective, most of my students don't know about complex exponentials, and many of them could not describe plainly in words the difference between real numbers and rational numbers. From this perspective, Q should seem to be the MOST accessible type of scalar. – Barry Smith Nov 30 '10 at 14:09
@Barry: it would probably be a good idea to say something in the original question about the level of your students, then. – Qiaochu Yuan Nov 30 '10 at 14:13
I agree with Qiaochu. My answer was certainly written with a different premise. A simple example from a much less advanced area that most of the answers here: to prove that self-adjoint complex operators are diagonalisable is much easier than to prove the same for real self-adjoint operators, because for the latter, you first need to show that all the eigenvalues are real. That's a typical "field of definition problem". Self-adjoint operators of rational vector spaces can still be defined, but are so badly behaved compared to the first 2, that I don't think any introductory book mentions them. – Alex B. Nov 30 '10 at 14:26

I just thought of one that is at about the level I envisioned.

Let V be a two-dimensional Q-subspace of the Q-vector space R.

(a) Show that if V is closed under multiplication of real numbers, and if the nonzero vectors in V are closed under division of real numbers, then V has a basis of the form {1, \alpha} where alpha is a root of a quadratic polynomial whose coefficients are integers.

(b) Show that under the same hypotheses as in part (a), V has a basis of the form {1, \sqrt{N}} where N is a positive integer.

(c) Show that if V has a basis of the form {1,\sqrt{N}} where N is a positive integer, then V is closed under multiplication of real numbers and the nonzero vectors of V are closed under division of real numbers.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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