# Vector bases dilemma

1. May 11, 2005

### Kocur

I have been thinking about bases of vector spaces and have come to some strange results.

I mean, we cannot present vectors without giving a base first. That is okay . But which base shall we use for presenting the base used? How can we identify "the real" elements (vectors) behind our representation ?

Is it possible?

Kocur.

2. May 11, 2005

### mathwonk

a vector space is a set V of objects that can be added and multiplied by scalars.

one such gadget is R^n. a basis is nothing but an isomorphism of your space V with some R^n. that lets you "present" the vectors in your space by means of rows of numbers (a1,...,an).

saying you already have a basis is like saying you are starting from the space V = R^n, and hence your vectors already have assigned numbers or coordinates in terms of the standard basis (1,0,...,0),.(0,1,0,...,0,),....(0,....,0,1).

You can either continue to use the "standard basis", i.e. the identity map from R^n to itsaelf, or some other basis, i.e. some other identification of R^n with R^n, hence someother way of choosing the coordinates of your vectors.

i.e. if i want to use (1,1) and (0,1) as a basis for R^2, then the vector formerly known as (a,b) = a(1,0) + b (0,1), would now have coordinates

x(1,1) + y(0,1) where i need (x,x) + (0,y) = (a,b), so i need

x+y = b, and x = a, hence i in the new basis i have coordinates x = a and y = b-a.

so now i am calling that old vector (a,b) by the new name (a,b-a).

confusing isnt it?

3. May 12, 2005

### matt grime

yes we can talk about vectors without reference to a basis, though it is often eqasier to fix some universal choice of basis. Contrast the Bourbaki approach to the basis approach.

It is perfectly possible, for instance to talk about the vector space of polynomials of degree at most n without reference to a basis.

4. May 12, 2005

### snoble

This conversation reminds me of a quote by John H Conway that a friend sent me once. It was part of his discussion on the importance of foundations in mathematics. I unfortunately do not have the bibliographic reference at hand but here's the quote.

"The situation is analogous to the theory of vector spaces. Once
upon a time these were collections of n-tuples of numbers, and the
interesting theorems were those that remained invariant under linear
transformations of these numbers. Now even the initial definitions
are invariant, and vector spaces are defined by axioms rather than
as particular objects. However, it is proved that every vector
space has a base, so that the new theory is much the same as the
old. But now no particular base is distinguished, and usually
arguments which use particular bases are cumbrous and inelegant
compared to arguments directly in terms of the axioms."
-John H Conway

Just thought someone else might find that interesting too,
Steven

5. May 12, 2005

### matt grime

Usually I would post a link, but I think this (in the spirit of Steven's post) is also another good article in the same vein.

From http://www.dpmms.cam.ac.uk/~wtg10/vspaces.html

Why study finite-dimensional vector spaces in the abstract if they are all isomorphic to Rn?

Here are several (closely related) reasons.

* Thinking of a vector space as Rn encourages us to think of an individual vector as a string of numbers. It is often more illuminating, however, to think of a vector geometrically - as something like a magnitude and a direction. This is true particularly with vectors that come from physics.
* To turn such a vector into a string of numbers one must first choose a coordinate system, and very often there is no choice that is obviously best. In such circumstances, choosing coordinates is necessarily unnatural' and non-canonical', and therefore offensive to the delicate aesthetic sensibilities of the pure mathematician.
* There are many important examples throughout mathematics of infinite-dimensional vector spaces. If one has understood finite-dimensional spaces in a coordinate-free way, then the relevant part of the theory carries over easily. If one has not, then it doesn't.
* There is often a considerable notational advantage in the coordinate-free approach. For example, it is a lot easier to write (and read) v than (v1,v2,...,vn). To give another example, a simple looking equation like Av=b can turn out to stand for a system of m equations in n unknowns.

Let me give two examples of vector spaces that illustrate some of the above points. First, the set of all continuous functions defined on the closed interval [0,1] can be made into a vector space in a very natural way. This vector space is infinite-dimensional (which simply means not finite-dimensional). If you do not immediately know how to prove this then it is a good exercise.

Of more relevance to the question that heads this page is the following finite-dimensional vector space V. Start with the space R3 and let V be the subspace consisting of all vectors (x1,x2,x3) such that x1+x2+x3=0. This subspace is two-dimensional, but there is no single basis that stands out as being the most natural. This example can of course be generalized to subspaces of Rn defined by simultaneous equations.

6. May 12, 2005

### mathwonk

in the spirit of matt's last example, it was interesting to me to note that the process of reduction to reduced echelon form, consist in choosing a basis for this subspace as follows:

since x1+x2+x3=0 is a 2 dimensional subspace of R^3, it must project isomorphically onto at elast one 2 dimensional coordinate plane, either (x1,x2), (x1,x3), or (x2,x3). If as in this case it projects isomorphically onto more than one of these, one chooses the one whose corrdinates are largest, lexicographically, i.e. (x2,x3). i.e. one lets the preferred basis of the space x1+x2+x3=0, be the unique pair of vectors in that subspace which project onto (0,1,0) and (0,0,1), the standard basis of the (x2,x3) plane.

i.e. when one solves the system by row reduction, one changes it to

x1 = -x2 - x3, and lest x2, x3 take any values. then the solution set of x1+x2+x3=0,

is expressed as (-s-t,s,t) = s(-1,1,0) + t(-1,0,1), where {(-1,1,0), (-1,0,1)} is the basis of the solution set that projects to the standard basis of the (x2,x3) plane.

This is indicative of the fact that it is difficult to specify a (finite dimensional) space in which there is not some way, however unnatural, to specify a preferred basis.

as to the continuous functions on the real line however, that is another matter, and i nkow of no natural way.

this is probably why the theory of hilbert space is so useful, i.e. fourier series, and other methods of expressing functions as infinite linear combinations of fucntions.

i.e. there are natural ways of designating dense independent subsets of functions in many useful spaces, but not vector bases.

7. May 12, 2005

### matt grime

Here's one I like from representation theory - it is slightly off topic, but it's quite nice as an example as to why choosing bases is a tricky prospect, and I think Steven or Roy may find it interesting if they've never seen it before, though I suspect they have.

Suppose we've got an equilateral triangle embedded in the x-y plane it's center at the origin. The symmetry group of it can be realized as 2x2 matrices. Wrt the standard basis the matrices are in O(2) and have entries like sqrt(3)/2 since they are rotations by 2pi/3 radians.

If however one uses u,v and w as the vertices of the traingle, then u and v are a basis and -u-v=w. With respect to this basis the elements of the matrices are only over 1 or -1, and in partilcular we can forget they are real numbers and pretend they're in a field of char 2 or 3 and thus we get the modular reps of D_3 in interesting characteristics - reducing sqrt(3)/2 in either char 2 or 3 isn't exactly possible or useful respectively.

I'm sure Roy and Steven have seen this idea - it's in the spirit of picking a lattice or talking about fields of definition.

8. May 12, 2005

### mathwonk

as to influential quotes on abstract versus numerical study of vectors, here is one from the great mathematician and pedagogue emil artin: (Geometric Algebra, pages 13-14):

"Thm. The ring Hom(V,V) is isomorphic to the ring of n by n matrices.
..........

Mathematical education is still suffering from the enthusiasm which the discovery of this isomorphism has aroused. the result has been that geometry was eliminated and replaced by computations. Instead of the intuitive maps of a space preserving addition and multiplication by scalars (these maps have an immediate geometric meaning), matrices have been introduced. From the innumerable absurdities - from a pedagogical point of view-let me point out one example and contrast it with the direct description.

Matrix method: A product of a matrix A and a vector X (which is then an n tuple of numbers) is defined; it is also a vector. Now the poor student has to swallow the following definition:

A vector is called an eigenvector if a number c exists such that

AX = cX.

Going through the formalism, the characteristic equation, one then ends up with theorems like: If a matrix A has n distinct eigenvalues, then a matrix D can be found such that DAD^(-1) is a diagonal matrix.

The student will of course learn all this since he will fail the course if he does not.

Instead one should argue like this: Given a linear transformation f of the space V into itself, does there exist a line which is kept fixed by f? In order to include the eigenvalue 0 one should then modify the question by asking whether a line is mapped into itself. This means of course for a vector spanning the line that

f(X) = cX.

Having thus motivated the probleml, the matrix A describing f will enter only for a moment for the actual computation of c. It should disappear again. Then one proves all the customary theorems without ever talking of matrices and asks the question: Suppose we can find a basis of eigenvectors; what does this imply for the geometric description of f? Well the space is stretched in the various directions of the basis by factors which are the eigenvalues. Only then does one ask what this means for the desription of f by a matrix in terms of a basis. We have obviously the diagonal form.

.......It is my experience that proofs involving matrices can be shortened by 50% if one throws the matrices out. Sometimes it cannot be done; a determinant may have to be computed."

Last edited: May 12, 2005
9. May 12, 2005

### mathwonk

no i did not know that example matt. thats really neat! im pretty ignorant of group reps but have encountered problems with characteristic 2 in various settings, especially taylor series and symmetric functions. this is a good lesson in not just throwing in the towel and saying "let the charac be 0."

10. May 13, 2005

### matt grime

Well, to follow up on that, it's possible to pick another basis so that the rep is still defined of a field of char 3, but when you consider it as a rep in char 3 it is not isomorphic to the other choice. I can't recall how to do it off the top of my head.

11. May 13, 2005

### Kocur

Well, I would like to thank you all for your input. Now, I have a lot of stuff to think about again it seems:).