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Rather than use the standard definitions of addtion and sscalar multiplication in R, suppose these two operations are defined as follows. With these new definitions, is Ra vector space ustity your answers ex, Y) (ox, y, 0) O The set is a vector space. O The set is not a vector space because the assodiative property of addition is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the assodiative property of multiplication is not satisfied. O The set is not a vector space because the mutipicative identity property is not satisfied. (0) )- a )- (0, 0, 0) e, V. )- (o, y, a) O The set isa vector space. O The set is not a vector space because the commutative property of addition is not satisfied. O The set is not a vector space because the additive identity property is not satisfed. O The set is not a vector space because it is not closed under scalar mutiplication. O The set ia net a vector space because the mutipicative identity preperty is not satisfied. , V. ) - (or, y, a) O The set isa vector space. O The set is not a vecter space because the additive identity property is not satisfed. O The set is not a vecter space because the additive inverse property is not satisfed. O The set is not a vector space because it is not closed under scalar mutiplication. O The set is not a vector space because the distributive property is not satisfed. x, v. ) - (or e-, y+ -3, a - 3) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfed. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the distributive property is not satisfied. O The set is not a vector space because the multiplicative identity property is not satisfied.

Rather than use the standard definitions of addtion and sscalar multiplication in R, suppose these two operations are defined as follows. With these new definitions, is Ra vector space ustity your answers ex, Y) (ox, y, 0) O The set is a vector space. O The set is not a vector space because the assodiative property of addition is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the assodiative property of multiplication is not satisfied. O The set is not a vector space because the mutipicative identity property is not satisfied. (0) )- a )- (0, 0, 0) e, V. )- (o, y, a) O The set isa vector space. O The set is not a vector space because the commutative property of addition is not satisfied. O The set is not a vector space because the additive identity property is not satisfed. O The set is not a vector space because it is not closed under scalar mutiplication. O The set ia net a vector space because the mutipicative identity preperty is not satisfied. , V. ) - (or, y, a) O The set isa vector space. O The set is not a vecter space because the additive identity property is not satisfed. O The set is not a vecter space because the additive inverse property is not satisfed. O The set is not a vector space because it is not closed under scalar mutiplication. O The set is not a vector space because the distributive property is not satisfed. x, v. ) - (or e-, y+ -3, a - 3) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfed. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the distributive property is not satisfied. O The set is not a vector space because the multiplicative identity property is not satisfied.

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.1: Vector Spaces And Subspaces
Problem 2EQ
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Rather than use the standard definitions of addition and scalar multiplication in R, suppose these two operations are defined as follows. With these new definitions, is Ra vector space? ustify your answers. () ( Y. ) - 2) - * *. n* * 2) x. Y. 2) - (ox, cy, 0) O The set is a vector space. O The set is not a vector space because the associative property of addition is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the associative property of multiplication is not satisfied. O The set is not a vector space because the multiplicative identity property is not satisfied. (6) . Ya. 72) - . V 2) = (0, 0. 0) c, Y. 2) - (ox, y, cz) O The set is a vector space. O The set is not a vector space because the commutative property of addition is not satisfied. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the multiplicative identity property is not satisfied. () (. M- ) - ( Ya- a) - ( - * 7. Yn -+ 7.-+ 7) C. Y. 2) - (o cy, ca) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because the additive inverse property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the distributive property is not satisfied. () (. Ys a) + (. ) - - 3, V * V2+3. +2 - 3) ex, Y. 2) - (ox + 3e- 3, cy + - 3, 2+ 3 - 3) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the distributive property is not satisfied. O The set is not a vector space because the multiplicative identity property is not satisfied.
Transcribed Image Text:Rather than use the standard definitions of addition and scalar multiplication in R, suppose these two operations are defined as follows. With these new definitions, is Ra vector space? ustify your answers. () ( Y. ) - 2) - * *. n* * 2) x. Y. 2) - (ox, cy, 0) O The set is a vector space. O The set is not a vector space because the associative property of addition is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the associative property of multiplication is not satisfied. O The set is not a vector space because the multiplicative identity property is not satisfied. (6) . Ya. 72) - . V 2) = (0, 0. 0) c, Y. 2) - (ox, y, cz) O The set is a vector space. O The set is not a vector space because the commutative property of addition is not satisfied. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the multiplicative identity property is not satisfied. () (. M- ) - ( Ya- a) - ( - * 7. Yn -+ 7.-+ 7) C. Y. 2) - (o cy, ca) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because the additive inverse property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the distributive property is not satisfied. () (. Ys a) + (. ) - - 3, V * V2+3. +2 - 3) ex, Y. 2) - (ox + 3e- 3, cy + - 3, 2+ 3 - 3) O The set is a vector space. O The set is not a vector space because the additive identity property is not satisfied. O The set is not a vector space because it is not closed under scalar multiplication. O The set is not a vector space because the distributive property is not satisfied. O The set is not a vector space because the multiplicative identity property is not satisfied.
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