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Conditions for Existence of Finite Bases in Vector Spaces

This section explores the criteria for the existence of finite bases in vector spaces. We establish that a vector ( v ) in a vector space ( V ) must satisfy certain conditions related to linear independence and spans. Theorem 1 outlines the equivalence of linear independence and non-inclusion in spans. Theorem 2 specifies that any linearly independent set is part of a basis, and any spanning set includes a basis. Examples illustrate how to identify bases for polynomial spaces and vector spaces defined by given vectors.

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Conditions for Existence of Finite Bases in Vector Spaces

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  1. 6.4 Finite Dimensional Spaces

  2. Goal in this section • We have so far been able to find bases if they exist, but here, we will establish the conditions which must hold in order for a finite basis to exist.

  3. Theorem 1 Let {v1,v2,…,vn} be a linearly independent set of vectors in vector space V. Then the following conditions are equivalent for a vector v in V: 1. {v,v1,v2,…,vn} is linearly independent 2. v does not lie in span {v1,v2,…,vn} Proof: Given (1), and suppose that v lies in span {v1,v2,…,vn} which means v = a1v1 + a2v2+…+anvn. Then v - a1v1 - a2v2-…-anvn=0 is a non-trivial combination which contradicts (1), so (1) must imply (2).

  4. Proof of Theorem 1(cont) Given (2), and suppose that {v,v1,v2,…,vn} is linearly dependent, so av + a1v1 + a2v2+…+anvn=0 w/ a≠0, then v = (-a-1a1)v1 + (-a-1a2)v2+…+ (-a-1an)vn which would mean that v lies in span {v1,v2,…,vn} which contradicts (2). So a=0, and thus a1v1 + a2v2+…+anvn=0. And this implies that a1=a2=…=an=0 since the set {v1,v2,…,vn} is linearly independent. This shows that (2) implies (1). 

  5. A use of Theorem 1 We can use Theorem 1 to create larger linearly independent sets by finding addition vectors which are not in the span of the vectors we already have.

  6. Example 1 Find a basis of P3 containing the linearly independent set {1+x,1+x2} Solution: All linear combinations of the elements will create polynomials with degree no greater than 2 which means that x3 is not in their span. So {1+x,1+x2,x3} is linearly independent. We could also show that 1 is not in span {1+x,1+x2,x3} so {1,1+x,1+x2,x3} is linearly independent. This set must also span P3 since it has 4 elements and we know that dim P3 = 4. Therefore, this is a basis.

  7. Theorem 2 Let V≠0 be a space spanned by n vectors. 1. Each set of linearly independent vectors in V is part of a basis of V. 2. Each spanning set for V contains a basis of V. 3. V has a basis, and dim V ≤ n.

  8. Proof of Theorem 2 Proof: (1) Take S={v1,v2,…,vk} a linearly independent subset of V. If V=span {v1,v2,…,vk}then we have a basis. If the set does not span V, then there exists some vector in V (call it vk+1) not in span {v1,v2,…,vk}. Then {v1,v2,…,vk,vk+1} is linearly independent, and either this spans V, or we can repeat the process to add another vector and create a larger set of linearly independent vectors. Eventually, we will either arrive at a set that spans V, and thus have a basis, or a finite basis does not exist for V.

  9. Proof of Theorem 2 Proof: (2) Let V = span {v1,v2,…,vm}, where we may assume that each vi ≠ 0. If {v1,…,vm} is L.I., then the set is a basis. If not, one of the vectors must be a linear combination of the others, so we eliminate that vector (say v1). The V=span {v2,…,vm}Then the set is either LI and we have a basis, or we repeat the process. Eventually we reach a linearly independent set (a single vector is L.I.) and we will thus have a basis.

  10. Proof of Theorem 2 Proof: (3) V has a spanning set of n vectors is given, and by (2), we know that each spanning set for V contains a basis. So the basis must contain ≤ n vectors. 

  11. Corollary to Theorem 2 A nonzero vector space is finite dimensional iff it can be spanned by finitely many vectors.

  12. Example 2 Find a basis of U=span{(1,-1,3,2),(0,-1,2,1),(2,1,0,1)} Solution: When put into row-echelon form, we get: So R3 is a linear combination of R1 and R2, but R1 and R2 are linearly independent. So U = span{R1,R2} and it is a LI set, so we have a basis.

  13. Theorem 3 Let V be a vector space, and assume that dim V = n>0. 1. Any set of n L.I. vectors in V is a basis (it must span V) 2. Any spanning set of n nonzero vectors in V is a basis (it must be linearly independent) Proof: (1) If n independent vectors did not span V, then there would exist a vector in V not in the span of these vectors, so we could create a set of n+1 linearly independent vectors in V, which would either span V, or not. Continue until we reach a spanning set (or it is not spanned by a finite set). Regardless, the basis would include more than n vectors which contradicts the initial statement.

  14. Proof of Theorem 3 (cont) (2) If the n vectors in a spanning set are not linearly independent, then there is a smaller spanning set which is linearly independent, and thus the basis would be less than n which contradicts the initial statement. Now if we know the dimension of a set is n, we can easily show that a set of n vectors is a basis (just need to show that it either spans the space, or that it is linearly independent).

  15. Example Show that {(1,0,0,0), (1,1,0,0),(1,1,1,0),(1,1,1,1)} is a basis for 4. We know that dim 4 = 4. We know that {(1,0,0,0), (0,1,0,0),(0,0,1,0),(0,0,0,1)} spans 4, and it is easy to show that {(1,0,0,0), (0,1,0,0),(0,0,1,0),(0,0,0,1)} lies in span {(1,0,0,0), (1,1,0,0),(1,1,1,0),(1,1,1,1)} So {(1,0,0,0), (1,1,0,0),(1,1,1,0),(1,1,1,1)} spans 4 and is thus a basis.

  16. Theorem 4 Let V be a vector space of dimension n and let U and W be subspaces of V. Then: 1. U is finite dimensional and dim U ≤ n. 2. Any basis of U is part of a basis for V. 3. If U  W and dim U = dim W, then U = W.

  17. Theorem 4 Proof (1) If U=0, then dim U=0 If U≠0, we pick u1≠0 in U. Then if U = span{u1}, then dim U = 1, otherwise, pick u2 in U not in span{u1}. If U=span{u1,u2} then dim U=2, Continue until we get a set which spans U which must happen by the time we get up to n different vectors since U only contains n linearly independent vectors.

  18. Theorem 4 Proof (2) From theorem 2, we know that a set of linearly independent vectors in V is part of a basis of V. A basis of U will consist of a set of linearly independent vectors in V. Therefore, any basis of U is part of a basis of V.

  19. Theorem 4 Proof (3) Let dim U = dim W = m. Then any basis of U {u1,u2,…,um} is a set of m linearly independent vectors which are all contained in W (since we said UW). Since we have m LI vectors in W and dim W=m, the basis for U is also a basis for W. Therefore {u1,u2,…,um} spans both U and W, so W = span {u1,u2,…,um} =U. 

  20. Example If a is a number, let W denote the subspace of all polynomials in Pnwith a as a root: W={p(x) | p(x) is in Pn and p(a) = 0} Show that {(x-a),(x-a)2,…,(x-a)n} is a basis of W. The elements in the set are all in Pn and are linearly independent. Then say U=span {(x-a),(x-a)2,…,(x-a)n} . If this set also spans W, then it is a basis of W. UWPn , dim U=n, dim Pn=n+1 So n≤dimW≤n+1 So W=U or W = Pn. We know W≠Pn, so W = U, w/ basis {(x-a),(x-a)2,…,(x-a)n} 

  21. Example Show that the only subspaces of 3 are 0, 3, lines through the origin and planes through the origin. Let U be a subspace of 3 . Since dim 3 =3, dim U=0,1,2 or 3. If dim U = 0, then U=0, and if dim U = 3, U = 3 . If dim U = 1, let {d} be a basis. Then U= d = {td | t } which is just the set of all lines through the origin.

  22. Example (cont) If dim U = 2, let {u1,u2} be a basis. Choose n = u1 x u2 which gives a normal to the two vectors. We will need to show that U is the plane P through the origin with normal vector n: P = {v | v•n = 0} We could show that this set P is a subspace of 3 . Since u1 and u2 are orthogonal to n (as is any linear combination of the vectors), U  P 3 . Since dim U = 2, and dim 3 =3, dim P = 2 or 3, so P=U or P= 3. P= 3 imlies that every vector in 3 is in P which can only happen if v • n = 0 implies that n=0 but n≠0, so P=U. 

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