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Vector Spaces A set V is called a vector space over a set K denoted V(K) if is an Abelian group, is a field, and For every element v V and K there exists.

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Presentation on theme: "Vector Spaces A set V is called a vector space over a set K denoted V(K) if is an Abelian group, is a field, and For every element v V and K there exists."— Presentation transcript:

1 Vector Spaces A set V is called a vector space over a set K denoted V(K) if is an Abelian group, is a field, and For every element v V and K there exists an element.v V called the scalar multiple of v by satisfying (i) (ii) (iii) (iv) Notation : 0 K denotes the additive identity under +, K, - denotes the inverse of under + denotes the identity under v denotes the inverse of v under

2 Examples Example 1 (Polynomials of degree n) V = set of all polynomials of order n The additive operation on vectors is defined as follows: V(K) is a vector space Proof v= For Now by continuity of addition on the real numbers

3 Examples:1 Proof continued Closure is trivial Associativity of follows from associativity of normal addition therefore Similarly for properties (ii)-(iv) Hence, is an Abelian group

4 Examples:2 Example 2 (n dimensional vectors) Example 3 (Complex Numbers) Example 4 (Matrices) is the set of n m matrices is the set of n n matrices

5 Properties of Vector Spaces Theorem Proof Identity under + by axiom (ii) of vector spaces But Identity under Therefore by the cancellation law for K 0 V v

6 Properties:2 Theorem (i) (ii) (- ).( v) =.v Proof (i) by previous theorem and inverse under + Therefore, by axiom (ii) of vector spaces Also =.( v)v) (.v) by inverse under Therefore (.v) by the cancellation law for Show

7 Properties:3 =.( v)v).(v ( v)) by inverse under Therefore,.v. vby axiom (i) vector spaces Now by previous theorem and axiom (iii) by above theorem.v. v Also.v (.v) by inverse under Therefore,.v. v=.v (.v). v= (.v) By the cancellation law for (ii) proof ??

8 Subspaces Definition Let W V such that W then W(K) is a subspace of V(K) if W is a vector space over K with the same definition of and scalar multiple as V Clearly to show that W(K) is a subspace of V(K) we need only show that is a sub-group of and that Characterisation Theorem A non-empty subset W of V is a subspace of V iff.u v W for all K, u,v W Proof ( ) trivial since.u W

9 Subspaces:2 Proof (continued) ( ) Taking =1 then u,v W by axiom (iv) Taking =-1 then u,v W and by a previous theorem (-1).u= (1.u)=uby axiom (iv) Therefore u u W by inverse under Therefore K, u W taking gives.u W Therefore, taking =-1 gives (1.u) W u W by a previous theorem Hence, is a subgroup and as required

10 Examples of Subspaces Example 1 Let Then W(K) is a subspace of V(K) W(K)W(K) V(K)V(K)

11 Examples of Subspaces:2 Example 1 (continued) proof (i) Clearly W and W V (ii) For R and u,v W such that and then Example 2 (i) Clearly W and W V and 2 R but W(K) is not a subspace of V(K)

12 Linear Combinations Definition Ifthenwhere is a linear combination of S

13 Linear Combinations:2 Theorem Forthen is a subspace of V Proof sinceand by a previous theorem and hence If then and for some Then by axiom (i) by axiom (iii) by axiom (ii)

14 Linear Independence Definition A subsetof V is linearly independent iff Otherwise if there exist one such that then are linearly dependent Example 1 is linearly dependent over R since Example 2 is linearly independent since

15 Linear Dependence in Matrices Theorem If and are the n column vectors of A then is linearly dependent over K if and only if Proof Ifis linearly dependent over K, then there exists (not all zero) such that Without loss of generality assume then Hence, performing a column operation where is added to column 1 gives a matrix with zero first column. Hence,

16 Matrices:2 Proof (continued) Ifthen the system of equations has a non-trivial solution, But this is the same as saying that there exist (not all zero) such that By considering the transpose of A we obtain Corollary The n rows of a matrixare linearly dependent if and only if

17 Basis A set Definition is a basis for V iff (i) S is linearly independent over K (ii) Condition (ii) means that S is a spanning set for V Definition (Finitely Generated) A vector space V is said to be finitely generated if it has a Basis S with a finite number of elements S V

18 Examples of Basis Let Example 1 Then is a basis for V Linear independence: Spanning: is also a basis for V Proof ??

19 Examples of Basis:2 Example 2 Let thenis a basis where Example 3 then is a basis

20 Dimension Theorem Every Basis of a finitely generated vector space has the same number of elements. Definition (Dimension) The number of elements in a basis for a finitely generated vector space V is called the dimension of V and denoted dim V. Examples then is a basis dim(V) = 3 Let then is a basis and dim(V)=4


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