In the basics we saw that vector spaces (v.s.) are defined over a field. Field is the keyword. Recall that examples of field include real numbers ‘r’, complex numbers ‘c’, rational numbers ‘q’. Let’s take the last example of q. q consists of numbers of the form a/b where a and b are integers like -2, 3, 5, 10 etc.. Examples of numbers that are not rational include square root of 2, pi, e etc.. v.s. of dimension n are sets of n – tuple of numbers where each number in the tuple comes from the underlying field. The question is, why do we need a field, that is, why can’t we just define it over the set of integers ‘z’. Recall also that z shares all properties with q except that not all elements have a multiplicative inverse i.e., there is no number for 21 in z so that the product of both of these numbers equal 1 but if we ask for an element in q, then we have 1/21 * 21 = 1. But that’s ok because nothing is stopping me from defining n – tuples from z. For example let’s look at z^2 which consists of numbers of the form (a,b) where a and b are integers. Well, the world didn’t come to an end anyway, in fact, we can see that if we take two 2- tuples of this form and add them coordinate-wise, we get a number of the same form with different integers. Secondly, if we multiply (coordinate-wise) a 2-tuple with an integer we still end with a different 2-tuple where each coordinate is again an integer, so what’s the problem?
To answer that question, we need a little bit of machinery unfortunately, here we go. We’re going to define what it means for two 2-tuples to be linearly independent or rather linearly dependent. I’m just going to say what it means right away. Two 2-tuples are linearly dependent if we can multiply one of the two by a scalar to get the other, that is, 2(1,2) = (2,4), so (1,2) and (2,4) are linearly dependent, that’s it! A counter example will be (1,0) and (0,1) since we can’t obtain one of them from the other by multiplying any number. I cheated right there because I didn’t tell you what a scalar is. A scalar is the underlying set on which the 2-tuple is defined, that is, it can be z or q. Note that if it’s q then the set of tuple is called as a v.s. whereas if it’s z then it’s not a v.s. (it’s called as a z-module). Of course, one might immediately see in the first example that (2,4) times any integer is never equal to (1,2) and conclude that both the tuples are linearly independent, this is exactly one of the problems with z-modules. Just to finish off this, we say that a set consisting of 2-tuples is linearly independent if each tuple is linearly independent from the rest, that is, any tuple cannot be written as a scaled sum of the rest.
Observe that the 2-tuple set is much bigger than the underlying scalars be it q or z. It is much bigger than twice the size of the underlying scalars which is easy to see. We can see that by the following construction. Take a piece of paper, draw to lines that are perpendicular to each other. Mark the origin and choose a specific distance to be 1 unit. Then the horizontal and the vertical lines denote the real numbers. Now we will mark 1, 2, 3,… and so on, on both of these axes. Then z^2 consists of all the points that have integer coordinates; r^2 consists of everything on the paper. If we imagine the integers on the horizontal line to be denote z as a set of integers, we can easily see that are more than 2 such sets. To be a bit fancy, z^2 is often referred to as lattice and some might say that z^2 is embedded in r^2 (or q^2). So what we have done is enlarged our space by a huge amount just by including a coordinate. But to store any point or tuple we just need to store two numbers, that is, a tuple (3,4) can be written as 3(1,0) + 4(1,0). What’s the big deal? (1,0) and (0,1) are called as the generating set, that is, any tuple can be stored in terms of these two tuples or in other words (1,0) and (0,1) generate (or span) z^2. Note also that the same two vectors span r^2 and q^2.
We have seen two concepts viz., linear independence (or dependence) and generating set or spanning set. The key property of v.s. that z-modules don’t have is that the minimal generating set is equal to the maximal independent set. Let me explain this. Let’s say that we have a set that consists of a 2-tuple, we add a 2-tuple that is linear independent and continue to do is. In r^2, let’s say we pick (2,3) to be the first tuple, and we have added (4,5), it’s easy to see that these two 2-tuples are linearly independent because multiplying one 2-tuple by any real number will never result in the other 2-tuple. But the problem is that we can’t add any more 2-tuples because -5/2(2,3) + 3/2(4,5) = (1,0); 2(2,3) -1(4,5) = (0,1), which means that any 2-tuple can be written as a scaled sum of the 2-tuples that we already have. A fancier way of saying this is that any 2-tuple can be written as a linear combination of (2,3) and (4,5) in r^2. In other words, in r^2 these two 2-tuples are linearly independent and they span the entire space r^2. Now one can easily see that (2,3) and (4,5) are linearly independent even in the z-module as well. But the problem is that they don’t span z^2, for example, they cannot generate (1,0) anymore. So we can continue adding more elements to our set to make linearly independent sets. This means that the size of maximal independent set is not 2 anymore for z-module whereas for v.s. of 2 – dimensions, once we choose two linearly independent tuples, we can’t find any more linearly independent tuples. Though in this case we can see that we can pick (1,0) and (0,1) to generate z^2. But the point is that you can have a different linearly independent set generating z^2 and I can have a different generating set and it might not be clear how to convert one to other. (1,0) and (0,1) are also called as basis or standard basis tuples.
This phenomenon is often referred to as duality, that is, maximum of something is equal to the minimum of something else. v.s. have this duality property whereas z-modules and many others don’t have this, that is, for z-modules there is nothing called as basis tuples whereas it is easy to choose standard bases for v.s.. We can easily extend this to any finite dimensional v.s.. Now it’s easy to see why v.s. are defined over a field, stare at the examples for a while if you’re still not convinced.