Topics in Algebra, Chapter 4.2

2014-07-04 math algebra topics-in-algebra

This page covers section 4.2 (“Linear Independence and Bases”).

Topics covered: 4.2

Throughout, V is a vector space over a field F.

  • Definition: Let v1,,vnV and α1,,αnF. Any element of the form iαivi is a linear combination over F of the {vi}.

  • Definintion: If SV is a subset of V, then span(S) is the set of all linear combinations of finite sets of elements of S.

  • Lemma 4.2.1: If SV, then span(S) is a subspace of V.

  • Lemma 4.2.2: If S,TV, then

    1. ST implies span(S)span(T).
    2. span(ST)=span(S)+span(T).
    3. span(span(S))=span(S).
  • Definition: V is finite-dimensional if there exists a finite subset SV such that V=span(S).

  • Definition: The elements of the set {v1,,vn}V are linearly dependent over F if there exist α1,,αnF, not all zero, such that iαivi=0. Otherwise, the elements of the set are linearly independent.

  • Lemma 4.2.3: If v1,,vnV are linearly independent, then any element u in their span has exactly one representation u=iαivi for some αiF.

  • Theorem 4.2.1: If v1,,vnV, then they are either (1) linearly independent or (2) some vk is a linear combination of the preceding v1,,vk1.

  • Corollary 1: Let S={v1,,vn}V and W=span(S). If v1,,vk are linearly independent, then we can construct a linearly independent subset {v1,,vk,vi1,,vir}S whose span is also W. In other words, the generating set of W can be pared down to a linearly independent generating set.

  • Corollary 2: If V is finite-dimensional, then it contains a finite set S={v1,,vn} of linearly independent elements with V=span(S).

  • Definition: SV is a basis of V if the elements of S are linearly independent and V=span(S). 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 S={v1,,vn}V and V=span(S), then S contains a basis of V. In particular, V (which is finite-dimensional) has a basis with finitely many elements.

  • Lemma 4.2.4: If v1,,vnV are a basis of V and if w1,,wmV are linearly independent, then mn. The proof of this result is subtle and, in my view, inelegant. Is there a nicer proof?

  • Corollary 1: If V is finite-dimensional, then any two bases of V have the same number of elements.

  • Corollary 2: FnFm if and only if m=n. This relies on an exercise that an isomorphism maps one basis into another.

  • Corollary 3: If V is finite-dimensional, then VFn for a unique nN. Any basis of V has n elements.

  • Definition: The integer n in Corollary 3 is the dimension of V over F, denoted n=dimV.

  • Corollary 4: Any two finite-dimensional vector spaces V and W over F of the same dimension are isomorphic.

  • Lemma 4.2.5: If V is finite-dimensional and {v1,,vn}V are linearly independent, then we can find vectors vn+1,,vn+rV such that {v1,,vn,vn+1,,vn+r} is a basis of V. That is, any linearly independent set can be extended to a basis of the space.

  • Lemma 4.2.6: If V is finite-dimensional and if WV is a subspace, then W is finite-dimensional, dimWdimV and dimV/W=dimVdimW.

  • Corollary: If A,B are finite-dimensional subspaces of V (not necessarily finite-dimensional), then A+B is finite-dimensional and dim(A+B)=dimA+dimBdim(AB).

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.


Herstein 4.2.1

Prove Lemma 4.2.2: Let V a vector space over field F and let S,TV.

(a) Show that ST implies that span(S)span(T).

(b) Show that span(ST)=span(S)+span(T).

© Show that span(span(S))=span(S).

(a) A linear combination in span(S) is of the form s=iαisi where αiF and siS. Because siT also, we have that sspan(T) and hence span(S)span(T).

(b) An element uspan(ST) is of the form u=i=1nαiui where uiS or uiT. Let AS be the set of i such that uiS, and let AT={1,,n}AS, the leftovers. Then we have u=jASαjuj+jATαjuj.

The first term in the sum belongs to span(S) and the second term belongs to span(T). Hence span(ST)span(S)+span(T).

The opposite inclusion is clear by the same type of argument.

© By definition span(S)span(span(S)).

Elements of span(S) are of the form iαisi. Then an arbitrary element z in span(span(S)) is a linear combination of those, i.e. of the form z=i=1nβij=1miαijsij

with βi,αijF and sijS. Of course, this may be rewritten as z=i=1nj=1mi(βiαij)sij

from which we see that zspan(S). Hence span(span(S))span(S) and the result is proven.


Herstein 4.2.2

(a) Prove that (1,1,0,0), (0,1,1,0) and (0,0,0,3) in F4 are linearly independent over F where F=R4.

(b) What conditions on the characteristic of F would make the vectors linearly dependent?

(a) A linear combination of these vectors is of the form a(1,1,0,0)+b(0,1,1,0)+c(0,0,0,3)=(a,a+b,b,3c).

with a,b,cR. If this is to equal the zero vector, then the first, third and fourth components fix each of a,b,c to be zero. Therefore, the set is linearly independent.

(b) If the field is of characteristic 3, then (0,0,0,3) is the zero vector, and the set of vectors is linearly dependent.

Otherwise, 3c=0 implies c=0 because fields have no zero divisors. Then (a,a+b,b,3c)=0 forces all three coefficients to be zero, and the vectors are linearly independent.

comments powered by Disqus