This page covers section 4.2 (“Linear Independence and Bases”).
Throughout, is a vector space over a field .
Definition: Let and . Any element of the form is a linear combination over of the .
Definintion: If is a subset of , then is the set of all linear combinations of finite sets of elements of .
Lemma 4.2.1: If , then is a subspace of .
Lemma 4.2.2: If , then
Definition: is finite-dimensional if there exists a finite subset such that .
Definition: The elements of the set are linearly dependent over if there exist , not all zero, such that . Otherwise, the elements of the set are linearly independent.
Lemma 4.2.3: If are linearly independent, then any element in their span has exactly one representation for some .
Theorem 4.2.1: If , then they are either (1) linearly independent or (2) some is a linear combination of the preceding .
Corollary 1: Let and . If are linearly independent, then we can construct a linearly independent subset whose span is also . In other words, the generating set of can be pared down to a linearly independent generating set.
Corollary 2: If is finite-dimensional, then it contains a finite set of linearly independent elements with .
Definition: is a basis of if the elements of are linearly independent and . The definition of basis that I give is the classic one (a minimal, spanning set); Herstein’s definition appears to allow for sets which merely contain a conventional basis. I may just be misreading the text.
Corollary 3: If and , then contains a basis of . In particular, (which is finite-dimensional) has a basis with finitely many elements.
Lemma 4.2.4: If are a basis of and if are linearly independent, then . The proof of this result is subtle and, in my view, inelegant. Is there a nicer proof?
Corollary 1: If is finite-dimensional, then any two bases of have the same number of elements.
Corollary 2: if and only if . This relies on an exercise that an isomorphism maps one basis into another.
Corollary 3: If is finite-dimensional, then for a unique . Any basis of has elements.
Definition: The integer in Corollary 3 is the dimension of over , denoted .
Corollary 4: Any two finite-dimensional vector spaces and over of the same dimension are isomorphic.
Lemma 4.2.5: If is finite-dimensional and are linearly independent, then we can find vectors such that is a basis of . That is, any linearly independent set can be extended to a basis of the space.
Lemma 4.2.6: If is finite-dimensional and if is a subspace, then is finite-dimensional, and .
Corollary: If are finite-dimensional subspaces of (not necessarily finite-dimensional), then is finite-dimensional and .
The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book.
(a) A linear combination in is of the form where and . Because also, we have that and hence .
(b) An element is of the form where or . Let be the set of such that , and let , the leftovers. Then we have
The first term in the sum belongs to and the second term belongs to . Hence .
The opposite inclusion is clear by the same type of argument.
© By definition .
Elements of are of the form . Then an arbitrary element in is a linear combination of those, i.e. of the form
with and . Of course, this may be rewritten as
from which we see that . Hence and the result is proven.
(a) A linear combination of these vectors is of the form
with . If this is to equal the zero vector, then the first, third and fourth components fix each of to be zero. Therefore, the set is linearly independent.
(b) If the field is of characteristic , then is the zero vector, and the set of vectors is linearly dependent.
Otherwise, implies because fields have no zero divisors. Then forces all three coefficients to be zero, and the vectors are linearly independent.