Does the following set of vectors constitute a vector space? Assume \"standard\"

Question

Does the following set of vectors constitute a vector space? Assume "standard" definitions of the operations.

The set of vectors in the first quadrant of the plane.

(1 pt) Does the following set of vectors constitute a vector space? Assume "standard" definitions of the operations. The set of vectors in the first quadrant of the plane. A. Yes B. No If not, which condition(s) below does it fail? (Check all that apply) A. Vector spaces must be closed under addition B. Vector spaces must be closed under scalar multiplication C. There must be a zero vector D. Every vector must have an additive inverse E. Addition must be associative F. Addition must be commutative G. Scalar multiplication by 1 is the identity operation H. The distributive property 1. Scalar multiplication must be associative J. None of the above, it is a vector space

Explanation / Answer

The Set Of Vectors in First quadrant of the plane is not a vector Space.Here V={(a,b):a>0,b>0}

The set is not closed under multiplication l.For example, take vector v=(u,v) where u and v > 0. Take scalar c which can be any integer. Try multiplying -c*(u,v)=(-cu,-cv) which is not in the first quadrant(multiplication by a negative scalar)Therefore It is not Closed Under multiplication

The set also does not contain additive inverse.As the Additive identity is (0,0) If we take an arbitrary (a,b) SO additive inverse of (a,b)is (-a,-b) which is not in the first quadrant.Therefore It is not A vector Space

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