Determine whether the set, together with the indicated operations, is a vector space. If it is not, then identify one of the vector space axioms that falls. The set of all first-degree polynomial functions ax, a 0, whose graphs pass through the origin with the standard operations The set is a vector space. O The set is not a vector space because it is not closed under addition. O The set is not a vector space because an additive identity does not exist. O The set is not a vector space because an additive inverse does not exist. O The set is not a vector space because the distributive property of scalar multiplication is not satisfied.

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.1: Vector Spaces And Subspaces
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Determine whether the set, together with the indicated operations, is a vector space. If it is not, then identify one of the vector space axioms that falls.
The set of all first-degree polynomial functions ax, a 0, whose graphs pass through the origin with the standard operations
The set is a vector space.
O The set is not a vector space because it is not closed under addition.
O The set is not a vector space because an additive identity does not exist.
O The set is not a vector space because an additive inverse does not exist.
O The set is not a vector space because the distributive property of scalar multiplication is not satisfied.
Transcribed Image Text:Determine whether the set, together with the indicated operations, is a vector space. If it is not, then identify one of the vector space axioms that falls. The set of all first-degree polynomial functions ax, a 0, whose graphs pass through the origin with the standard operations The set is a vector space. O The set is not a vector space because it is not closed under addition. O The set is not a vector space because an additive identity does not exist. O The set is not a vector space because an additive inverse does not exist. O The set is not a vector space because the distributive property of scalar multiplication is not satisfied.
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