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 fails. The set of all quadratic functions whose graphs pass through the point (0, 5) with the standard operations O 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 the commutative property of addition is not satisfied. 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 a scalar identity does not exist.

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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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 fails.
The set of all quadratic functions whose graphs pass through the point (0, 5) with the standard operations
O 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 the commutative property of addition is not satisfied.
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 a scalar identity does not exist.
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 fails. The set of all quadratic functions whose graphs pass through the point (0, 5) with the standard operations O 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 the commutative property of addition is not satisfied. 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 a scalar identity does not exist.
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