I refer to the textbook for the basic notion and properties of vector spaces. A field is a triple satisfying the axioms listed in pages 2, 3 of the textbook as properties of see also https://caicedoteaching.wordpress.com/2009/02/11/305-4-fields/ and surrounding lectures, in this blog. I am writing this note mostly to record the exercise, the question, and the statement of Steinitz lemma, so I am not recording the proofs we discussed in lecture.
The following example, that I want to leave as a (voluntary) exercise, is due to John Conway.
Exercise 1 Define Nim-addition and Nim-multiplication on as follows:
- is the result of adding without carrying the binary expansions of and . For example,
- is computed by applying the following rules:
- is commutative.
- distributes over
- Letting we have and for
Show that is a field.
In lecture, the basic properties of vector spaces were presented. Recall the following notions:
Definition 1 Let be a vector space over a field and let
- The span of is the collection of all finite linear combinations of elements of The empty combination is understood as the vector We say that spans iff and that spans (or that is spanning) iff spans
- The set is independent iff for all vectors we have that
- The space is finite dimensional iff there is a finite spanning set Otherwise, is infinite dimensional.
- is a basis for iff it is independent and spans.
The following is very useful; in many algebraic context, it is taken as a basic property that any decent independence relation must satisfy:
Lemma 2 (Steinitz exchange lemma) For any subset of a vector space and any vectors if
Corollary 3 If a subset of a vector space is independent, and is such that then is independent.
Theorem 4 Any vector space admits a basis.
Remark 1 The result is true for all vector spaces, whether they are finite dimensional or not, but the argument in the infinite dimensional case involves Zorn’s lemma. In lecture, we only proved the finite dimensional case. The argument shows that any spanning set contains a basis. The proof we gave in lecture does not seem to adapt to the infinite dimensional case. In particular, in principle we could need to deal with the following pathological situation:
Question 1 Can we have a vector space and a set of vectors such that the are pairwise distinct and yet
for all ?
(I encourage you to think about this question, although just as with the exercise, I am not requesting that you turn it in.)
Steinitz exchange lemma can be used to prove the following fundamental result, with the proof of which we will begin the next lecture.
Remark 2 If is infinite dimensional, and are bases for then it is still the case that in the sense that there is a bijection between and
Definition 6 In light of Lemma 5 and the subsequent remark, we can define the dimension of a vector space as the size of any basis for
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