Answer :
The set (x1,x2)+(y1,y2)=(x1-x2, y1+y2) is not a vector space.
A set is a vector space if it satisfy all the addition operation axiom and scalar multiplication axiom.
Addition operation:
Commutativity -
(x1, x2) + (y1, y2) equals (x1 - x2, y1 + y2) equals (y1, y2) + (x1, x2)
Associativity -
(x1, x2) plus ((y1, y2) plus (z1, z2)) = (x1, x2) + (z1 + y1, z2 + y2) = (x1+ y1 + z1, x2 + y2 + z2) =
((x1, x2) + (y1, y2)) + (z1, z2)
Zero element -
(0, 0) → (x1, x2) + (0, 0) = (x1, x2)
Inverse element -
(x1, x2) adding (-x1, -x2) = (0, 0)
Scalar multiplication:
Compatibility -
a(b (x, y)) = a(bx, 0) = (abx, 0) = b(ax, 0) = b(a(x, y))
Identity element -
1(x, y) = (x, 0) ≠ (x, y) [Identity element doesn’t exist for this operation.)
Distributivity law -
a((x1, x2) + (y1, y2)) = a(x1 + y1, x2 + y2) = (a(x1 + y1), 0) = a(x1, x2) + a(y1, y2)
Distributivity law -
(a + b)(x, y) = ((a + b)x, 0) = (ax, 0) + (bx, 0) = a(x, y) + b(x, y)
The scalar multiplication postulate is not fulfilled in this space (this operation does not have the identity element). It is therefore not a vector space.
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