In the first chapter, we are going to cover the definitions of vector spaces, linear maps, and subspaces. In addition, we are introducing several important concepts such as basis, dimension, direct sum, matrix representations of linear maps, and kernel and image for linear maps. We shall prove the dimension theorem for linear maps that relate the dimension of the domain to the dimensions of kernel and image. We give an account of Gauss elimin Ation and how it ties in with the more abstract theory. This will be used to define and compute the characteristic polynomial in Chap.2.
It is important to note that Sects. and then go back to the missed sections.
As induction is going to play a big role in many of the proofs, we have chosen to say a few things about that topic in the first section.
1.1 Induction andWell-Ordering
A fundamental property of the natural numbers , i.e., the positive integers
, that will be used throughout the book is the fact that they are well ordered . This means that any nonempty subset S has a smallest element s min S such that s min s for all s S . Using the natural ordering of the integers, rational numbers, or real numbers, we see that this property does not hold for those numbers. For example, the half-open interval
does not have a smallest element.
In order to justify that the positive integers are well ordered, let S be nonempty and select k S . Starting with 1, we can check whether it belongs to S . If it does, then s min=1. Otherwise, check whether 2 belongs to S . If 2 S and 1 S ,then we have s min=2. Otherwise, we proceed to check whether 3 belongs to S . Continuing in this manner, we must eventually find k 0 k ,such that k 0 S ,but 1,2,3,, k 01 S . This is the desired minimum: s min= k 0.
We shall use the well-ordering of the natural numbers in several places in this text. A very interesting application is to the proof of the prime factorization theorem: any integer2 is a product of prime numbers. The proof works the following way. Let S be the set of numbers which do not admit a prime factorization. If S is empty, we are finished; otherwise, S contains a smallest element n = s min S . If n has no divisors, then it is a prime number and hence has a prime factorization. Thus, n must have a divisor p >1. Now write n = p q . Since p , q < n both numbers must have a prime factorization. But then also n = p q has a prime factorization. This contradicts that S is nonempty.
The second important idea that is tied to the natural numbers is that of induction . Sometimes, it is also called mathematical induction so as not to confuse it with the inductive method from science. The types of results that one can attempt to prove with induction always have a statement that needs to be verified for each number n . Some good examples are
.
Every integer2 has a prime factorization.
Every polynomial has a root.
The first statement is pretty straightforward to understand. The second is a bit more complicated, and we also note that in fact, there is only a statement for each integer2. This could be finessed by saying that each integer n +1, n 1 has a prime factorization. This, however, seems too pedantic and also introduces extra and irrelevant baggage by using addition. The third statement is obviously quite different from the other two. For one thing, it only stands a chance of being true if we also assume that the polynomials have degree1. This gives us the idea of how this can be tied to the positive integers. The statement can be paraphrased as: Every polynomial of degree1 has a root. Even then, we need to be more precise as x 2+1 does not have any real roots.
In order to explain how induction works abstractly, suppose that we have a statement
for each n . Each of the above statements can be used as an example of what
can be. The induction process now works by first ensuring that the anchor statement is valid. In other words, we first check that
is true. We then have to establish the induction step . This means that we need to show that if
is true, then
is also true. The assumption that
is true is called the induction hypothesis . If we can establish the validity of these two facts, then
must be true for all n . This follows from the well-ordering of the natural numbers. Namely, let
. If S is empty, we are finished, otherwise, S has a smallest element k S . Since 1 S , we know that k >1. But this means that we know that
is true. The induction step then implies that
is true as well. This contradicts that S is nonempty.
Let us see if we can use this procedure on the above statements. For 1, we begin by checking that
. This is indeed true. Next, we assume that
and we wish to show that
Using the induction hypothesis, we see that