Math 221: LINEAR ALGEBRA §6-1. Vector Spaces - Examples and Basic Properties Le Chen 1 Emory University, 2020 Fall (last updated on 08/27/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary.
In these lecture notes, arbitrary vectors are generally denoted with lower case What is a vector space? Definition Let V be a nonempty set of objects (elements) with two operations. ◮ Vector Addition: for any v , w ∈ V, the sum u + v ∈ V. (V is closed under vector addition.) ◮ Scalar Multiplication: for any v ∈ V and k ∈ R , the product k v ∈ V. (V is closed under scalar multiplication.) Then V is a vector space if it satisfies the Axioms of Addition and the Axioms of Scalar Multiplication that follow. In this case, the elements of V are called vectors. Notation. boldface letters. When written by hand, you can use the notation � v for v .
Axioms of Addition Axioms of Addition A1. Addition is commutative. u + v = v + u for all u , v ∈ V . A2. Addition is associative. ( u + v ) + w = u + ( v + w ) for all u , v , w ∈ V . A3. Existence of an additive identity. There exists an element 0 in V so that u + 0 = u for all u ∈ V . A4. Existence of an additive inverse. For each u ∈ V there exists an element − u ∈ V so that u + ( − u ) = 0 .
Axioms of Scalar Multiplication Axioms of Scalar Multiplication S1. Scalar multiplication distributes over vector addition. a ( u + v ) = a u + a v for all a ∈ R and u , v ∈ V . S2. Scalar multiplication distributes over scalar addition. ( a + b ) u = a u + b u for all a , b ∈ R and u ∈ V . S3. Scalar multiplication is associative. a ( b u ) = ( ab ) u for all a , b ∈ R and u ∈ V . S4. Existence of a multiplicative identity for scalar multiplication. 1 u = u for all u ∈ V .
M 0 0 M Example R n with matrix addition and scalar multiplication is a vector space.
0 0 M Example R n with matrix addition and scalar multiplication is a vector space. Example M mn , the set of all m × n matrices (of real numbers) with matrix addition and scalar multiplication is a vector space. It is left as an exercise to verify the eight vector space axioms.
Example R n with matrix addition and scalar multiplication is a vector space. Example M mn , the set of all m × n matrices (of real numbers) with matrix addition and scalar multiplication is a vector space. It is left as an exercise to verify the eight vector space axioms. Notes. ◮ Notation: the m × n matrix of all zeros is written 0 or, when the size of the matrix needs to be emphasized, 0 mn . ◮ The vector space M mn “is the same as” the vector space R mn . We will make this notion more precise later on. For now, notice that an m × n matrix has mn entries arranged in m rows and n columns, while a vector in R mn has mn entries arranged in mn rows and 1 column.
What needs to be shown is closure under addition (for all v w . showing the existence of an additive identity and additive inverses in the set ), as well as v , and closure under scalar multiplication (for all v ), and w , v as an identity element. two distributive properties, the associative property, and has that addition is commutative and associative, and that scalar multiplication satisfjes the , it is automatic Since we are using the matrix addition and scalar multiplication of M M may be described as follows: The matrices in Problem Let V be the set of all 2 × 2 matrices of real numbers whose entries sum to zero. We use the usual addition and scalar multiplication of M 22 . Show that V is a vector space.
What needs to be shown is closure under addition (for all v w . Since we are using the matrix addition and scalar multiplication of M showing the existence of an additive identity and additive inverses in the set ), as well as v , and closure under scalar multiplication (for all v ), and w , v as an identity element. two distributive properties, the associative property, and has that addition is commutative and associative, and that scalar multiplication satisfjes the , it is automatic Problem Let V be the set of all 2 × 2 matrices of real numbers whose entries sum to zero. We use the usual addition and scalar multiplication of M 22 . Show that V is a vector space. Solution The matrices in V may be described as follows: � � � � � a b � V = ∈ M 22 � a + b + c + d = 0 � c d .
What needs to be shown is closure under addition (for all v w . that addition is commutative and associative, and that scalar multiplication satisfjes the showing the existence of an additive identity and additive inverses in the set ), as well as v , and closure under scalar multiplication (for all v ), and w , v Problem Let V be the set of all 2 × 2 matrices of real numbers whose entries sum to zero. We use the usual addition and scalar multiplication of M 22 . Show that V is a vector space. Solution The matrices in V may be described as follows: � � � � � a b � V = ∈ M 22 � a + b + c + d = 0 � c d . Since we are using the matrix addition and scalar multiplication of M 22 , it is automatic two distributive properties, the associative property, and has 1 as an identity element.
that addition is commutative and associative, and that scalar multiplication satisfjes the Problem Let V be the set of all 2 × 2 matrices of real numbers whose entries sum to zero. We use the usual addition and scalar multiplication of M 22 . Show that V is a vector space. Solution The matrices in V may be described as follows: � � � � � a b � V = ∈ M 22 � a + b + c + d = 0 � c d . Since we are using the matrix addition and scalar multiplication of M 22 , it is automatic two distributive properties, the associative property, and has 1 as an identity element. What needs to be shown is closure under addition (for all v , w ∈ V , v + w ∈ V ), and closure under scalar multiplication (for all v ∈ V and k ∈ R , k v ∈ V ), as well as showing the existence of an additive identity and additive inverses in the set V .
Suppose Since and Solution (continued) ◮ Closure under addition � w 1 � w 2 x 1 � x 2 � A = B = y 1 z 1 y 2 x 2 are in V . Then w 1 + x 1 + y 1 + z 1 = 0 , w 2 + x 2 + y 2 + z 2 = 0 , and � w 1 � w 2 � w 1 + w 2 � � � x 1 x 2 x 1 + x 2 A + B = + = . y 1 z 1 y 2 z 2 y 1 + y 2 z 1 + z 2 ( w 1 + w 2 ) + ( x 1 + x 2 ) + ( y 1 + y 2 ) + ( z 1 + z 2 ) = ( w 1 + x 1 + y 1 + z 1 ) + ( w 2 + x 2 + y 2 + z 2 ) = 0 + 0 = 0 , A + B is in V , so V is closed under addition.
Since Solution (continued) ◮ Closure under scalar multiplication � w � x Suppose A = is in V and k ∈ R . y z Then w + x + y + z = 0 , and � w � kw � � x kx kA = k = . y z ky kz kw + kx + ky + kz = k ( w + x + y + z ) = k (0) = 0 , kA is in V , so V is closed under scalar multiplication.
Solution (continued) ◮ Existence of an additive identity The additive identity of M 22 is the 2 × 2 matrix of zeros, � 0 � 0 0 = ; 0 0 Since 0 + 0 + 0 + 0 = 0 , 0 is in V , and has the required property (as it does in M 22 ).
Since Solution (continued) ◮ Existence of an additive inverse � w � x Let A = be in V . y z Then w + x + y + z = 0 , and its additive inverse in M 22 is � − w − x � − A = . − y − z ( − w ) + ( − x ) + ( − y ) + ( − z ) = − ( w + x + y + x ) = − 0 = 0 , − A is in V and has the required property (as it does in M 22 ).
det det det det Example Let �� a � a � � b b � � � V = � a , b , c , d ∈ R and = 0 . . � c d c d We use the usual addition and scalar multiplication of M 22 . Then V is not vector space.
det det Example Let �� a � a � � b b � � � V = � a , b , c , d ∈ R and = 0 . . � c d c d We use the usual addition and scalar multiplication of M 22 . Then V is not vector space. For example, if � 1 � 1 � � 1 0 A = and B = , 0 0 1 0 then det ( A ) = 0 and det ( B ) = 0 , so A , B ∈ V.
det Example Let �� a � a � � b b � � � V = � a , b , c , d ∈ R and = 0 . . � c d c d We use the usual addition and scalar multiplication of M 22 . Then V is not vector space. For example, if � 1 � 1 � � 1 0 A = and B = , 0 0 1 0 then det ( A ) = 0 and det ( B ) = 0 , so A , B ∈ V. However, � 2 � 1 A + B = , 1 0 and det ( A + B ) = − 1 , so A + B �∈ V, i.e., V is not closed under addition.
Definitions Let P be the set of all polynomials in indeterminate x, with coefficients from R , and let p ∈ P . Then n � a i x i p ( x ) = i =0 for some integer n.
Definitions Let P be the set of all polynomials in indeterminate x, with coefficients from R , and let p ∈ P . Then n � a i x i p ( x ) = i =0 for some integer n. ◮ The degree of p is the highest power of x with a nonzero coefficient. Note that p ( x ) = 0 has undefined degree.
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