Tuesday, August 17, 2010

What is a basis?

Consider a vector. Just to make things concrete, consider a vector on the 2-D plane. In fact, let's consider this one (call it v⃑):
It's a vector, to be sure, but it's hardly clear how one is supposed to work with it. It doesn't make sense to pull out a ruler and pencil every time we want to add our vector to something; mathematics is supposed to be a model of the world, and thus we should be able to understand things about that model without recourse to physical measurements. To solve this problem for vectors on the plane, we can introduce two new vectors, x̂ and ŷ, then use vector addition to write v⃑ as a sum:
Now we can write v⃑ = ax̂ + bŷ̂, which doesn't at first seem to buy us much. Note, however, that we can write any vector on the 2D plane as a sum of these two new vectors in various linear combinations. Mathematically, we write this as ℝ² = span {x̂, ŷ}. Whenever a space can be written this way for some set of vectors B, we say that B is a basis for the space.

Once we have a basis picked out, we can work with the coefficients (a and b in our example) instead of the vector itself, as they completely characterize the vector. For example, adding vectors becomes a matter of adding their respective coefficients.

In spaces other than the 2-D plane, we can also apply the same idea to find bases for representing vectors. Consider, for instance, the space of column vectors such as [a; b] (pretend they're stacked in a column, OK?). Then, a perfectly fine basis would be the set:
It's easy to see that we can write any other 2-dimensional column vector as a sum of the form a[1; 0] + b[0; 1] = [a; 0] + [b; 0] = [a; b].

A point that can get lost in this kind of discussion, however, is that there's absolutely nothing special about the bases I've given here as examples. We could just as well used [1; 1] and [1; -1] as a basis for column vectors, or just as well used a different pair of vectors in the plane:
Put differently, a basis is a largely arbitrary choice that you make when working with vectors. The relevant operations work regardless of what basis you use, since each of the vectors in a basis can itself be expanded. For example, [1; 0] = ½([1;1] + [1; -1]) and [0; 1] = ½([1; 1] - [1; -1]), so that we have a way of converting from a representation in the {[1; 0], [0; 1]} basis to the {[1; 1], [1; -1]} basis.

While there is much, much more to be said on the topic of bases for vectorspaces, I'm happy to say a few words about bases. As we shall see when we get into discussing linear operations, the existence of bases for vectorspaces is a large part of what gives us so much power in linear algebra. We shall need this power in the quantum realm, as linear algebra may well be said to be the language of quantum mechanics. Hopefully I'll get a few more words in on the subject before my vacation!

No comments: