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6.4 Finite Dimensional Spaces

6.4 Finite Dimensional Spaces. 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. Theorem 1.

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6.4 Finite Dimensional 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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