Math 221: LINEAR ALGEBRA §6-4. Vector Spaces - Finite Dimensional Spaces Le Chen 1 Emory University, 2020 Fall (last updated on 08/18/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary.
fjnite basis. We will learn how to extend any independent set in such a vector To show that any vector space that is spanned by a fjnite set of vectors has a space to a basis of the vector space, and how to cut down any spanning set of such a vector space to a basis of the vector space. An earlier result from R n Let U be a subspace of R n , U � = { 0 } . Then 1. U has a basis, and dim ( U ) ≤ n . 2. Any independent subset of U can be extended (by adding vectors) to a basis of U . 3. Any spanning set of U can be cut down (by deleting vectors) to a basis of U .
To show that any vector space that is spanned by a fjnite set of vectors has a fjnite basis. We will learn how to extend any independent set in such a vector space to a basis of the vector space, and how to cut down any spanning set of such a vector space to a basis of the vector space. An earlier result from R n Let U be a subspace of R n , U � = { 0 } . Then 1. U has a basis, and dim ( U ) ≤ n . 2. Any independent subset of U can be extended (by adding vectors) to a basis of U . 3. Any spanning set of U can be cut down (by deleting vectors) to a basis of U . Goal of this section
is a basis of 0 . is the empty sum, which is equal is independent and spans 0 , independent. Since is the empty sum, and hence all the coeffjcients are equal to zero. Thus 0 . Furthermore, there are no coeffjcients in to zero. Therefore, span . The only linear combination of vectors of or, more commonly, The empty set is the set with no vectors in it, denoted 0 0 0 M Definition A vector space V is finite dimensional if it is spanned by a finite set of vectors. Otherwise it is called infinite dimensional.
is a basis of 0 . . The only linear combination of vectors of is independent and spans 0 , independent. Since is the empty sum, and hence all the coeffjcients are equal to zero. Thus 0 . Furthermore, there are no coeffjcients in to zero. Therefore, span is the empty sum, which is equal or, more commonly, The empty set is the set with no vectors in it, denoted 0 Definition A vector space V is finite dimensional if it is spanned by a finite set of vectors. Otherwise it is called infinite dimensional. Example R n , P n and M mn are all examples of finite dimensional vector spaces, while P is an infinite dimensional vector space. The zero vector space, { 0 } , is also finite dimensional, since it is spanned by { 0 } .
Definition A vector space V is finite dimensional if it is spanned by a finite set of vectors. Otherwise it is called infinite dimensional. Example R n , P n and M mn are all examples of finite dimensional vector spaces, while P is an infinite dimensional vector space. The zero vector space, { 0 } , is also finite dimensional, since it is spanned by { 0 } . The empty set spans { 0 } and is independent The empty set is the set with no vectors in it, denoted {} or, more commonly, ∅ . The only linear combination of vectors of ∅ is the empty sum, which is equal to zero. Therefore, span ( ∅ ) = { 0 } . Furthermore, there are no coeffjcients in the empty sum, and hence all the coeffjcients are equal to zero. Thus ∅ is independent. Since ∅ is independent and spans { 0 } , ∅ is a basis of { 0 } .
, and then the v The proof of this lemma involves repeated application of the is independent. , and it follows that is independent, 0 . Since v v v , so , a contradiction. Therefore span i.e., u v v u implying that v v v u , then 0 . If u v v v Suppose that Lemma (Independent Lemma) Let V be a vector space and S = { v 1 , v 2 , . . . , v k } an independent subset of V. If u is a vector in V, but u �∈ span ( S ) , then S ′ = { v 1 , v 2 , . . . , v k , u } is independent.
, and then the v The proof of this lemma involves repeated application of the is independent. , and it follows that is independent, 0 . Since v v v , so , a contradiction. Therefore span i.e., u v v u implying that v v v u , then If Lemma (Independent Lemma) Let V be a vector space and S = { v 1 , v 2 , . . . , v k } an independent subset of V. If u is a vector in V, but u �∈ span ( S ) , then S ′ = { v 1 , v 2 , . . . , v k , u } is independent. Proof. Suppose that a 1 v 1 + a 2 v 2 + · · · + a k v k + a u = 0 .
, and then the implying that The proof of this lemma involves repeated application of the Lemma (Independent Lemma) Let V be a vector space and S = { v 1 , v 2 , . . . , v k } an independent subset of V. If u is a vector in V, but u �∈ span ( S ) , then S ′ = { v 1 , v 2 , . . . , v k , u } is independent. Proof. Suppose that a 1 v 1 + a 2 v 2 + · · · + a k v k + a u = 0 . If a � = 0 , then a u = − a 1 v 1 − a 2 v 2 − · · · − a k v k , u = − a 1 a v 1 − a 2 a v 2 − · · · − a k a v k , i.e., u ∈ span ( S ) , a contradiction. Therefore a = 0 , so a 1 v 1 + a 2 v 2 + · · · + a k v k = 0 . Since S is independent, a 1 = a 2 = · · · = a k = 0 , and it follows that S ′ is independent. �
, and then the implying that The proof of this lemma involves repeated application of the Lemma (Independent Lemma) Let V be a vector space and S = { v 1 , v 2 , . . . , v k } an independent subset of V. If u is a vector in V, but u �∈ span ( S ) , then S ′ = { v 1 , v 2 , . . . , v k , u } is independent. Proof. Suppose that a 1 v 1 + a 2 v 2 + · · · + a k v k + a u = 0 . If a � = 0 , then a u = − a 1 v 1 − a 2 v 2 − · · · − a k v k , u = − a 1 a v 1 − a 2 a v 2 − · · · − a k a v k , i.e., u ∈ span ( S ) , a contradiction. Therefore a = 0 , so a 1 v 1 + a 2 v 2 + · · · + a k v k = 0 . Since S is independent, a 1 = a 2 = · · · = a k = 0 , and it follows that S ′ is independent. � Lemma Let V be a finite dimensional vector space. If U is any subspace of V, then any independent subset of U can be extended to a finite basis of U.
implying that Lemma (Independent Lemma) Let V be a vector space and S = { v 1 , v 2 , . . . , v k } an independent subset of V. If u is a vector in V, but u �∈ span ( S ) , then S ′ = { v 1 , v 2 , . . . , v k , u } is independent. Proof. Suppose that a 1 v 1 + a 2 v 2 + · · · + a k v k + a u = 0 . If a � = 0 , then a u = − a 1 v 1 − a 2 v 2 − · · · − a k v k , u = − a 1 a v 1 − a 2 a v 2 − · · · − a k a v k , i.e., u ∈ span ( S ) , a contradiction. Therefore a = 0 , so a 1 v 1 + a 2 v 2 + · · · + a k v k = 0 . Since S is independent, a 1 = a 2 = · · · = a k = 0 , and it follows that S ′ is independent. � Lemma Let V be a finite dimensional vector space. If U is any subspace of V, then any independent subset of U can be extended to a finite basis of U. The proof of this lemma involves repeated application of the Independent Lemma , and then the Fundamental Theorem.
dim dim Constructing bases from independent sets Theorem Let V be a finite dimensional vector space spanned by a set of m vectors. 1. V has a finite basis, and dim ( V ) ≤ m.
dim dim Constructing bases from independent sets Theorem Let V be a finite dimensional vector space spanned by a set of m vectors. 1. V has a finite basis, and dim ( V ) ≤ m. 2. Every independent subset of V can be extended to a basis of V by adding vectors from any fixed basis of V.
Constructing bases from independent sets Theorem Let V be a finite dimensional vector space spanned by a set of m vectors. 1. V has a finite basis, and dim ( V ) ≤ m. 2. Every independent subset of V can be extended to a basis of V by adding vectors from any fixed basis of V. 3. If U is a subspace of V, then (i) U is finite dimensional and dim ( U ) ≤ dim ( V ) ; (ii) every basis of U is part of a basis of V.
the second part of this theorem. elements, so dim of (again by a previous ). Since is an independent subset of , has at most dim dim be any nonzero vector of . (ii) If 0 , then any basis of suffjces. Otherwise, any basis of can be extended to a basis of : because is independent, we apply and extend x to a basis . Otherwise, choose x to is not a basis , then there is a vector b (which exists by the fjrst part of this theorem), and let be an independent subset of . If dim of that is not in span Let ; b is a larger independent subset of . Repeat. (i) If 0 , then dim denote an arbitrary fjxed basis of Proof. 1. If V = { 0 } , then V has dimension zero, and the (unique) basis of V is the empty set. Otherwise, choose any nonzero vector x in V and extend { x } to a fjnite basis B of V (by a previous Lemma ). By the Fundamental Theorem , B has at most m elements, so dim ( V ) ≤ m .
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