Chapter Two Vector Spaces Vector space Linear combinations

Chapter Two: Vector Spaces Vector space ~ Linear combinations of vectors. I. Definition of Vector Space II. Linear Independence III. Basis and Dimension • • Topic: Fields Topic: Crystals Topic: Voting Paradoxes Topic: Dimensional Analysis Ref: T. M. Apostol, “Linear Algebra”, Chap 3, Wiley (97)

I. Definition of Vector Space I. 1. Definition and Examples I. 2. Subspaces and Spanning Sets

Algebraic Structures Ref: Y. Choquet-Bruhat, et al, “Analysis, Manifolds & Physics”, Pt I. , North Holland (82) Structure Internal Operations Scalar Multiplication Group * No Ring, Field *, No Module / Vector Space + Yes Algebra +, * Yes Field = Ring with idenity & all elements except 0 have inverses. Vector space = Module over Field.

I. 1. Definition and Examples Definition 1. 1: (Real) Vector Space ( V, ; R ) A vector space (over R) consists of a set V along with 2 operations ‘ ’ and ‘ ’ s. t. (1) For the vector addition : v, w, u V a) v w V ( Closure ) b) v w = w v ( Commutativity ) c) ( v w ) u = v ( w u ) ( Associativity ) d) 0 V s. t. v 0 = v ( Zero element ) e) v V s. t. v ( v) = 0 ( Inverse ) (2) For the scalar multiplication : v, w V and a, b R, [ R is the real number field (R, +, ) f) a v V ( Closure ) g) ( a + b ) v = ( a v ) (b v ) ( Distributivity ) h) a ( v w ) = ( a v ) ( a w ) i) ( a b ) v = a ( b v ) ( Associativity ) j) is 1 always v = written v Ø as + so that one writes v + w instead of v w Ø and are often omitted so that one writes a b v instead of ( a b ) v

Definition in Conventional Notations Definition 1. 1: (Real) Vector Space ( V, + ; R ) A vector space (over R) consists of a set V along with 2 operations ‘+’ and ‘ ’ s. t. (1) For the vector addition + : v, w, u V a) v + w V ( Closure ) b) v + w = w + v ( Commutativity ) c) ( v + w ) + u = v + ( w + u ) ( Associativity ) d) 0 V s. t. v + 0 = v ( Zero element ) e) v V s. t. v v = 0 ( Inverse ) (2) For the scalar multiplication : v, w V and a, b R, [ R is the real number field (R, +, ) ] f) a v V ( Closure ) g) ( a + b ) v = a v + b v ( Distributivity ) h) a ( v + w ) = a v + a w i) ( a b ) v = a ( b v ) = a b v ( Associativity ) a) 1 v = v

Example 1. 3: R 2 is a vector space if with Proof it yourself / see Hefferon, p. 81. Example 1. 4: Plane in R 3. The plane through the origin P is a subspace of R 3. Proof it yourself / see Hefferon, p. 82. is a vector space.

Example 1. 5: Let & be the (column) matrix addition & scalar multiplication, resp. , then ( Zn, + ; Z ) is a vector space. ( Zn, + ; R ) is not a vector space since closure is violated under scalar multiplication. Example 1. 6: Let then (V, + ; R ) is a vector space. Definition 1. 7: A one-element vector space is a trivial space.

Example 1. 8: Space of Real Polynomials of Degree n or less, Pn E. g. , The kth component of a is Pn is a vector space with vectors Vector addition: i. e. , Scalar multiplication: Zero element: Inverse: Pn is isomorphic to Rn+1 with i. e. ,

Example 1. 9: Function Space The set { f | f : N → R } of all real valued functions of natural numbers is a vector space if Vector addition: Scalar multiplication: Zero element: Inverse: f ( n ) is a vector of countably infinite dimensions: f = ( f(0), f(1), f(2), f(3), … ) E. g. , ~

Example 1. 10: Space of All Real Polynomials, P P is a vector space of countably infinite dimensions. Example 1. 11: Function Space The set { f | f : R → R } of all real valued functions of real numbers is a vector space of uncountably infinite dimensions.

Example 13: Solution Space of a Linear Homogeneous Differential Equation is a vector space with Vector addition: Scalar multiplication: Zero element: Inverse: Closure: → Example 14: Solution Space of a System of Linear Homogeneous Equations

Remarks: • Definition of a mathematical structure is not unique. • The accepted version is time-tested to be most concise & elegant. Lemma 1. 16: Lose Ends In any vector space V, 1. 0 v=0. 2. ( 1 ) v + v = 0. 3. a 0=0. 4. v V and a R. Proof: 1. 2. 3.

Exercises 2. I. 1. 1. At this point “the same” is only an intuition, but nonetheless for each vector space identify the k for which the space is “the same” as Rk. (a) (b) (c) The 2 3 matrices under the usual operations The n m matrices (under their usual operations) This set of 2 2 matrices 2. (a) (b) Prove that every point, line, or plane thru the origin in R 3 is a vector space under the inherited operations. What if it doesn’t contain the origin?

I. 2. Subspaces and Spanning Sets Definition 2. 1: Subspaces For any vector space, a subspace is a subset that is itself a vector space, under the inherited operations. Note: A subset of a vector space is a subspace iff it is closed under & . → It must contain 0. (c. f. Lemma 2. 9. ) Example 2. 2: Plane in R 3 Proof: is a subspace of R 3. Let → with → QED

Example 2. 3: The x-axis in Rn is a subspace. Proof follows directly from the fact that Example 2. 4: • { 0 } is a trivial subspace of Rn. • Rn is a subspace of Rn. Both are improper subspaces. All other subspaces are proper. Example 2. 5: Subspace is only defined wrt inherited operations. ({1}, ; R) is a vector space if we define 1 1 = 1 and a 1=1 a R. However, neither ({1}, ; R) nor ({1}, + ; R) is a subspace of the vector space (R, + ; R).

Example 2. 6: Polynomial Spaces. Pn is a proper subspace of Pm if n < m. Example 2. 7: Solution Spaces. The solution space of any real linear homogeneous ordinary differential equation, L f = 0, is a subspace of the function space of 1 variable { f : R → R }. Example 2. 8: Violation of Closure. R+ is not a subspace of R since ( 1) v R+ v R+.

Lemma 2. 9: Let S be a non-empty subset of a vector space ( V, + ; R ). W. r. t. the inherited operations, the following statements are equivalent: 1. S is a subspace of V. 2. S is closed under all linear combinations of pairs of vectors. 3. S is closed under arbitrary linear combinations. Proof: See Hefferon, p. 93. Remark: Vector space = Collection of linear combinations of vectors.

Example 2. 11: Parametrization of a Plane in R 3 is a 2 -D subspace of R 3. i. e. , S is the set of all linear combinations of 2 vectors (2, 1, 0)T, & ( 1, 0, 1)T. Example 2. 12: Parametrization of a Matrix Subspace. is a subspace of the space of 2 2 matrices.

Definition 2. 13: Span Let S = { s 1 , …, sn | sk ( V, +, R ) } be a set of n vectors in vector space V. The span of S is the set of all linear combinations of the vectors in S, i. e. , with Lemma 2. 15: The span of any subset of a vector space is a subspace. Proof: Let S = { s 1 , …, sn | sk ( V, +, R ) } and QED Converse: Any vector subspace is the span of a subset of its members. Also: span S is the smallest vector space containing all members of S.

Example 2. 16: For any v V, span{v} = { a v | a R } is a 1 -D subspace. Example 2. 17: Proof: The problem is tantamount to showing that for all x, y R, unique a, b R s. t. i. e. , Since has a unique solution for arbitrary x & y. QED

Example 2. 18: P 2 Let Question: = subspace of P 2 ? Answer is yes since and Lesson: A vector space can be spanned by different sets of vectors. Definition: Completeness A subset S of a vector space V is complete if span S = V.

Example 2. 19: All Possible Subspaces of R 3 Planes thru 0 Lines thru 0 See next section for proof.

Exercises 2. I. 2 1. Consider the set under these operations. (a) (b) (c) any Show that it is not a subspace of R 3. (Hint. See Example 2. 5). Show that it is a vector space. ( To save time, you need only prove axioms (d) & (j), and closure under all linear combinations of 2 vectors. ) Show that any subspace of R 3 must pass thru the origin, and so subspace of R 3 must involve zero in its description. Does the converse hold? Does any subset of R 3 that contains the origin become a subspace when given the inherited operations?

2. (a) (b) (c) (d) 3. Because ‘span of’ is an operation on sets we naturally consider how it interacts with the usual set operations. Let [S] Span S. If S T are subsets of a vector space, is [S] [T] ? Always? Sometimes? Never? If S, T are subsets of a vector space, is [ S T ] = [S] [T] ? Is the span of the complement equal to the complement of the span? Find a structure that is closed under linear combinations, and yet is not a vector space. (Remark. This is a bit of a trick question. )