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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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

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 For v= Now by continuity of addition on the real numbers

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

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

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

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

Properties:3 Show = .( v) .(v  ( v)) by inverse under  Therefore,
by 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 ??

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 <W,> is a sub-group of <V, > 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

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)= u by 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

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

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)

Linear Combinations Definition If then where
is a linear combination of S

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

Linear Independence Definition A subset of 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

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 If is 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,

Matrices:2 Proof (continued) If then 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 matrix are linearly dependent if and only if

Basis Definition A set 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 V S

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

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

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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